diff --git "a/data_all_eng_slimpj/shuffled/split2/finalzzeuvi" "b/data_all_eng_slimpj/shuffled/split2/finalzzeuvi" new file mode 100644--- /dev/null +++ "b/data_all_eng_slimpj/shuffled/split2/finalzzeuvi" @@ -0,0 +1,5 @@ +{"text":"\\section{Introduction}\\label{S0}\n\nMany function spaces of practical interest are also algebras and\/or lattices. An important example is $C(X)$, the space of continuous real-valued functions on a compact Hausdorff space. \nHence, a natural problem in the course of understanding the structure of the space $C(X)$ is to characterize the algebraic and\/or lattice isomorphisms on it.\nThe classical solutions were given by Gelfand and Kolmogorov \\cite{GK} and Kaplansky \\cite{K} respectively.\nAn in depth exposition of investigations into the algebraic structure of $C(X)$ and much more can be found in the classic monograph \\cite{GJ}.\nSubsequent research has tied these two strands together in the form of the disjointness structure of the function space $C(X)$.\nSpecifically, algebraic or lattice homomorphisms are disjointness preserving; they map disjoint functions to disjoint functions. An algebraic or lattice isomorphism is biseparating; that is, it is a bijection $T$ so that both $T$ and $T^{-1}$ are disjointness preserving. Moreover, generalization to disjointness preserving or biseparating maps allows for extension to function spaces that are neither algebras nor lattices, and even to vector-valued functions.\nCopius research has been devoted to the study of disjointness preserving and biseparating maps on various function spaces; see, e.g., \\cite{A}-\\cite{AJ2}, \\cite{BBH}, \\cite{GJW}, \\cite{HBN}-\\cite{J-V W}, \\cite{L}.\nAs far as the authors are aware of, the study of biseparating maps thus far has been confined to linear or at least additive maps. Since additive bijective maps are ${\\mathbb Q}$-linear, such maps are not far removed from the linear world.\nIn this paper, we initiate the study of general nonlinear biseparating maps on spaces of vector-valued functions.\nThe following example shows that the definition of ``biseparating'' needs to be adjusted in order to obtain meaningful results.\n\n\\bigskip\n\n\\noindent{\\bf Example}. Let $A$ be the set of all functions $f\\in C[0,1]$ so that the set $\\{t\\in [0,1]: f(t) \\neq 0\\}$ is dense in $[0,1]$.\nLet $T:C[0,1]\\to C[0,1]$ be a map such that $T$ maps $A$ bijectively onto itself and that $Tf = f$ if $f\\neq A$.\nThen $T$ is a bijection so that $f\\cdot g = 0 \\iff Tf\\cdot Tg = 0$.\n\n\\bigskip\n\nThe example shows that the definition of ``biseparating'' used for linear or additive maps is too weak when applied to general nonlinear maps. In the next section, we propose a revised definition of ``biseparating'' for nonlinear maps. The definition reduces to the usual one for additive maps. Moreover, with the revised definition, a satisfactory theory of nonlinear biseparating maps arise, subject to some mild assumptions. See the paragraph preceding Lemma \\ref{l1.21}. The theory of nonlinear biseparating maps is somewhat related to the theory or order isomorphisms developed in \\cite{LT}. It also partly generalizes the notion of ``nonlinear superposition operators''. We refer to \\cite{AZ} for a comprehensive study of the latter types of operators.\n\nLet us give an overview of the content of the paper. As mentioned, the definition of ``nonlinear biseparatimg maps'' is given in \\S \\ref{s1}. Under the mild assumptions of ``basic'' and ``compatible'', the fundamental characterization theorem of nonlinear biseparating maps (Theorem \\ref{t5}) is obtained. The theorem shows that a nonlinear bijective operator is biseparating if and only if it is ``locally determined''. For an exposition of some applications of locally determined operators to operator functional equations, particularly on spaces of differentiable functions, refer to \\cite{KM}.\nThe characterization theorem applies in particular to a number of familiar (vector-valued) function spaces such as spaces of continuous, uniformly continuous, Lipschitz and differentiable functions.\nWe would like to point out a general resemblance of Theorem \\ref{t5} with the fundamental characterization theorem for ``nonlinear order isomorphisms'' \\cite[Theorem 2.11]{LT}.\nIndeed, our study of nonlinear biseparating maps is motivated and informed by the study of nonlinear order isomorphisms at various points. However, the lack of an order makes many of the arguments more difficult in the present case, especially for uniformly continuous and Lipschitz functions.\nFor further information on nonlinear order isomorphisms on function spaces, we refer to \\cite{LT} and the references therein.\n\n\\S \\ref{s2} studies nonlinear biseparating maps between spaces of vector-valued continuous or bounded continuous functions.\nOne of the main results is Theorem \\ref{t2.8}, which shows that if $X,Y$ are realcompact spaces and $E,F$ are Hausdorff topological vector spaces, and there is a biseparating map $T:C(X,E)\\to C(Y,F)$, then $X$ and $Y$ are homeomorphic. With reference to the classical Gelfand-Kolmogorov and Kaplansky theorems, one sees that one needs rather much less than the full algebraic or lattice structure of $C(X)$ to determine the topology of $X$.\n\nFrom \\S\\ref{s3} onwards, we focus on metric spaces $X$ and $Y$.\nIn the course of \\S\\ref{s3} and \\S\\ref{s4}, full representations of biseparating maps between spaces of continuous, uniformly continuous and Lipschitz functions defined on metric spaces are obtained. See Propositions \\ref{p4.2}, \\ref{p4.3} and \\ref{p4.4}. \\S\\ref{s5} revisits spaces of continuous functions, this time defined on metric spaces.\nComplete characterizations of nonlinear biseparating maps are obtained; see Theorems \\ref{t5.4} and \\ref{t5.5}.\nWe also prove an automatic continuity result Theorem \\ref{t5.6}.\n\n\\S7 is concerned with nonlinear biseparating maps between spaces of uniformly continuous functions.\nCharacterization of such maps is carried out in two stages. First it is shown that a biseparating map induces a uniform homeomorphism of the underlying metric spaces. The second part involves solving the ``section problem'': determining the maps $\\Xi$ so that $Sf(x) = \\Xi(x,f(x))$ is uniformly continuous whenever the input function $f$ is uniformly continuous. Refer to Theorems \\ref{t6.7.1} and \\ref{t6.7.2}.\nFrom these characterization theorems, one can also obtain an automatic continuity result (Theorem \\ref{t6.9}).\nA classical result of Atsuji \\cite{At} and Hejcman \\cite{H}, rediscovered in \\cite{O'F}, states that all uniformly continuous functions on a metric space $X$ are bounded if and only if $X$ is Bourbaki bounded (see definition in \\S7.2). Theorem \\ref{t6.10} generalizes this result. It shows that there is a biseparating map from $U(X,E)$ onto a space $U_*(Y,F)$ (the space of bounded uniformly continuous functions) if and only if $X$ is Bourbaki bounded.\n\n\n\\S \\ref{s8} focuses on spaces of Lipschitz functions. First it is shown that we may reduce to considering spaces $\\operatorname{Lip}(X,E)$, where $X$ is bounded metric (Proposition \\ref{p6.2}).\nMaking use of the Baire Catergory Theorem and some intricate combinatorial arguments, it is then shown that a biseparating map between vector-valued Lipschitz spaces defined on bounded complete metric spaces induces a Lipschitz homeomorphism between the underlying metric spaces (Theorem \\ref{t7.5}).\nNext, the section problem for Lipschitz functions is solved (Theorem \\ref{t7.7}), which enables the characterization of nonlinear biseparating maps between spaces of Lipschitz functions (Theorem \\ref{t7.8}).\nSuppose that $\\Xi$ is a ``Lipschitz section'', i.e., the function $\\Xi(x,f(x))$ is a Lipschitz function of $x$ whenever $f$ is Lipschitz. It is known that even if $x_0$ is an accumulation point, the function $\\Xi(x_0,\\cdot)$ need not be continuous with respect to the second variable. Nevertheless, exploiting the Baire Category Theorem, we show that $\\Xi(x_0,\\cdot)$ is continuous on an open dense set if $x_0$ is an accumulation point (Theorem \\ref{t7.11}).\n\nThe final section \\S \\ref{s9} determines the biseparating maps that act between a space of uniformly continuous functions on the one hand and a space of Lipschitz functions on the other. The main results (Theorems \\ref{t6.6} and \\ref{t6.7}) show that there is a certain rigidity, so that the existence of such maps imply very strong conditions on the underlying metric spaces.\n\nTo end this introduction, we note that linear or nonlinear biseparating maps acting between spaces of differentiable functions seem to be rather more difficult to deal with. A notable achievement in this regard is the paper by Araujo \\cite{A3}. We intend to address some of the problems raised therein in a future paper.\n\n\n\\section{Generalities}\\label{s1}\n\nLet $X, Y$ be sets and let $E, F$ be (real or complex) vector spaces. \nSuppose that $A(X,E)$ is a vector subspace of $E^X$ and $A(Y,F)$ is a vector subspace of $F^Y$.\nIf $f\\in A(X,E)$, let the {\\em carrier} of $f$ be the set \n\\[C(f) = \\{x\\in X: f(x) \\neq 0\\}.\\] \nSet ${\\mathcal C}(X) = \\{C(f):f\\in A(X,E)\\}$.\nFor functions $f,g,h\\in A(X,E)$, say that $f$ and $g$ are {\\em disjoint with respect to} $h$, $f\\perp_h g$, if $C(f-h) \\cap C(g-h) = \\emptyset$. We abbreviate $\\perp_0$ as $\\perp$.\nThe {\\em support} of a function $f\\in A(X,E)$ is the set\n\\[ \\widehat{C}(f) = X\\backslash \\bigcup\\{C(g): g\\in A(X,E), f\\perp g\\} = X\\backslash \\bigcup\\{C\\in {\\mathcal C}(X): C \\cap C(f) = \\emptyset\\}.\\]\nObviously $C(f) \\subseteq \\widehat{C}(f)$. Furthermore, if $f_1,f_2\\in A(X,E)$ and $f_1=f_2$ on $C(f)$, then $f_1 = f_2$ on $\\widehat{C}(f)$.\n Set ${\\mathcal D}(X) = \\{\\widehat{C}(f): f\\in A(X,E)\\}$. Similar definitions apply to $A(Y,F)$.\n A map $T:A(X,E) \\to A(Y,F)$ is {\\em biseparating} if it is a bijection and for any $f,g,h \\in A(X,E)$,\n\\[ f\\perp_hg \\iff Tf\\perp_{Th}Tg.\\]\nFor the remainder of the section, let $T:A(X,E)\\to A(Y,F)$ be a given biseparating map.\nThe following proposition, although simple to state and easy to prove, turns out to be key to understanding biseparating maps.\n\n\\begin{prop}\\label{p1}(Araujo's Lemma, cf \\cite[Lemma 4.2]{A})\n\nFor any $f,g,h\\in A(X,E)$, \n\\[ \\widehat{C}(f-h) \\subseteq \\widehat{C}(g-h) \\iff \\widehat{C}(Tf-Th) \\subseteq \\widehat{C}(Tg-Th).\\]\n\\end{prop}\n\n\\begin{proof}\nSuppose that $\\widehat{C}(f-h) \\subseteq \\widehat{C}(g-h)$.\nAssume that there exists $z\\in \\widehat{C}(Tf-Th) \\backslash \\widehat{C}(Tg-Th)$.\nThere exists $v\\in A(Y,F)$ so that $v\\perp Tg-Th$ and $z\\in C(v)$.\nSince $z\\in \\widehat{C}(Tf-Th)$, $v\\not\\perp Tf-Th$. \nSet $u = T^{-1}(v+Th) \\in A(X,E)$. Then $v = Tu - Th$. Hence\n\\[ Tu -Th = v \\perp Tg-Th \\implies Tu \\perp_{Th} Tg \\implies u\\perp_h g\\implies u-h \\perp g-h.\\]\nTherefore,\n\\[ C(u-h) \\subseteq (\\widehat{C}(g-h))^c \\subseteq (\\widehat{C}(f-h))^c \\implies u-h \\perp f-h \\implies u\\perp_h f.\\]\nIt follows that \n\\[ Tu \\perp_{Th}Tf \\implies v = Tu-Th \\perp Tf-Th.\\]\nThis contradicts that fact that $v\\not\\perp Tf-Th$. This completes the proof for the forward implication ``$\\implies$\". The reverse implication follows by symmetry.\n\\end{proof}\n\n\n\n\n\n\\begin{prop}\\label{p2}\nLet $f\\in A(X,E)$ be given. The map \\[\\theta_f: \\widehat{C}(h) \\mapsto \\widehat{C}(T(f+h) -Tf)\\] is a well-defined bijection from ${\\mathcal D}(X)$ onto ${\\mathcal D}(Y)$ that preserves set inclusion. For any $f,g\\in A(X,E)$ and any $U\\in{\\mathcal D}(X)$, $f= g$ on $U$ if and only if $Tf = Tg$ on $\\theta_f(U)$.\n\\end{prop}\n\n\\begin{proof}\nBy Proposition \\ref{p1}, $\\widehat{C}(h_1) = \\widehat{C}(h_2)$ if and only if \n\\[ \\widehat{C}((h_1+f)-f) = \\widehat{C}((h_2+f)-f) \\iff \\widehat{C}(T(h_1+f) - Tf) = \\widehat{C}(T(h_2+f)-Tf).\\]\nThis shows that the map $\\theta_f$ is well-defined and injective.\nSince any $g\\in A(Y,F)$ can be written in the form $T(f+h) - Tf$ with $h = T^{-1}(g+Tf) -f$, $\\theta_f$ is surjective. It follows from Proposition \\ref{p1} that $\\theta_f$ preserves set inclusion.\n\nFinally, suppose that $U = \\widehat{C}(h) \\in {\\mathcal D}(X)$.\nThen $f= g$ on $U$ if and only if $g-f \\perp h = (f+h) -f$, which in turn is equivalent to the fact that $Tg-Tf \\perp T(f+h) - Tf$. The last statement is easily seen to be equivalent to the fact that $Tg-Tf = 0$ on $\\theta_f(\\widehat{C}(h))$.\n\\end{proof}\n\n\nThe idea behind Proposition \\ref{p2} is that a biseparating map gives rise to a collection of ``set movers'' $\\theta_f$. \nIn order to make the set mover $\\theta_f$ independent of the function $f$, we impose two conditions on the function space $A(X,E)$.\nSay that $A(X,E)$ is \n\\begin{enumerate}\n\\item {\\em basic} if whenever $x\\in C_1\\cap C_2$ for some $C_1, C_2\\in {\\mathcal C}(X)$, then there exists $C\\in {\\mathcal C}(X)$ so that $x\\in C \\subseteq C_1\\cap C_2$; \n\\item {\\em compatible} if for any $f\\in A(X,E)$, any $D\\in {\\mathcal D}(X)$ and any point $x\\notin D$, there exist $g\\in A(X,E)$ and $C\\in {\\mathcal C}(X)$ so that $x\\in C$ and \n\\[ g = \\begin{cases} f &\\text{on $C$},\\\\\n 0 &\\text{on $D$}.\\end{cases}\\]\n \\end{enumerate} \n\n\n\\begin{lem}\\label{l1.21}\nSuppose that $A(Y,F)$ is basic. If $f, g\\in A(Y,F)$ and $V\\in {\\mathcal D}(Y)$ are such that $f = g$ on $V$ and $V\\subseteq \\widehat{C}(g)$, then $V \\subseteq \\widehat{C}(f)$.\n\\end{lem}\n\n\\begin{proof}\nAssume otherwise. There is a point $y_1\\in V$ so that $y_1\\notin \\widehat{C}(f)$.\nThere exists $u\\in A(Y,F)$ so that $y_1\\in C(u)$ and $u\\perp f$. Say $V = \\widehat{C}(v)$.\nSince $y_1\\in \\widehat{C}(v)$, $u\\not\\perp v$. As $A(Y,F)$ is basic, there exists $C\\in {\\mathcal C}(Y)$ so that $\\emptyset\\neq C\\subseteq C(u)\\cap C(v)$.\nIf $y\\in C$, then $y \\in C(u)$ and hence $f(y) = 0$. Moreover, $y\\in C(v) \\subseteq V$ and hence $g(y) = f(y) = 0$. This proves that $C\\cap C(g) = \\emptyset$. Since $C\\in {\\mathcal C}(Y)$, it follows that $C\\cap \\widehat{C}(g) = \\emptyset$.\nThis is impossible since $C$ is a nonempty subset of $C(v)\\subseteq \\widehat{C}(v) = V\\subseteq \\widehat{C}(g)$.\n\\end{proof}\n\n\n\\begin{prop}\\label{p3}\nAssume that $A(Y,F)$ is basic. Suppose that $f, g\\in A(X,E)$ and $U \\in {\\mathcal D}(X)$. If $f = g$ on $U$, then $\\theta_f(U) = \\theta_g(U)$.\n\\end{prop}\n\n\\begin{proof}\nLet $U = \\widehat{C}(h)$. Since $f= g$ on $U\\supseteq C(h)$, $C(h)\\subseteq C(f+h-g)$.\nHence $U \\subseteq \\widehat{C}(f+h- g)$.\nThus \n\\[ \\theta_g(U) \\subseteq \\theta_g(\\widehat{C}(f+h- g)) = \\widehat{C}(T(f+h) - Tg)).\\]\nBy Proposition \\ref{p2}, $Tf = Tg$ on $\\theta_g(U)$. In other words,\n\\[ T(f+h) - Tf = T(f+h)-Tg \\text{ on } \\theta_g(U) \\subseteq \\widehat{C}(T(f+h) - Tg)).\\]\nTherefore, by Lemma \\ref{l1.21}, \\[\\theta_g(U) \\subseteq \\widehat{C}(T(f+h) - Tf) = \\theta_f(U).\\]\nThe reverse inclusion follows by symmetry.\n\\end{proof}\n\n\n\\begin{prop}\\label{p4}\nAssume that $A(Y,F)$ is both basic and compatible. Let $f,g\\in A(X,E)$ and let $U$ be a set in ${\\mathcal D}(X)$. Then $\\theta_g(U) = \\theta_f(U)$.\n\\end{prop}\n\n\\begin{proof} \nSuppose that $U = \\widehat{C}(h)$ and that there exists $y \\in \\theta_f(U) \\backslash \\theta_g(U)$.\nThen there exists $C\\in {\\mathcal C}(Y)$ so that $y \\in C$ and that $C \\cap \\theta_g(U) = \\emptyset$.\nSince $y\\in \\theta_f(U)$, $C \\cap C(T(f+h)-Tf) \\neq \\emptyset$.\nUsing the fact that $A(Y,F)$ is basic, there exist $C'\\in {\\mathcal C}(Y)$ and $z\\in Y$ so that \n\\[z\\in C' \\subseteq C \\cap C(T(f+h)-Tf).\\]\nIn particular, $z\\notin \\theta_g(U)$ and $\\theta_g(U) \\in {\\mathcal D}(Y)$.\nUse the compatibility of $A(Y,F)$ to choose $v\\in A(Y,F)$ and $C''\\in {\\mathcal C}(Y)$ so that $z\\in C''$ and that \n\\[ v = \\begin{cases} Tf-Tg &\\text{on $C''$,}\\\\\n0 &\\text{on $\\theta_g(U)$}.\\end{cases}\\]\nSince $A(Y,F)$ is basic, we may also assume that $C'' \\subseteq C'$.\nSet $k = v+Tg$. We have $k= Tg$ on $\\theta_g(U)$. By Proposition \\ref{p2}, $T^{-1}k = g$ on $U$.\nSay $C'' = C(w)$. Then $k = Tf$ on $\\widehat{C}(w)\\subseteq \\theta_f(U)$. Thus Proposition \\ref{p2} implies that \n\\[ T^{-1}k = f \\text{ on } (\\theta_f)^{-1}(\\widehat{C}(w)) \\subseteq U.\n\\]\nIt follows that $f = g$ on the set $(\\theta_f)^{-1}(\\widehat{C}(w))\\in {\\mathcal D}(X)$.\nBy Proposition \\ref{p3}, $\\widehat{C}(w) = \\theta_g((\\theta_f)^{-1}(\\widehat{C}(w)) \\subseteq \\theta_g(U)$.\nHence $z\\in C(w)\\subseteq \\theta_g(U) =\\emptyset$, contrary to the choice of $z$.\nThis proves that $\\theta_f(U) \\subseteq \\theta_g(U)$. The reverse inclusion follows by symmetry.\n\\end{proof}\n\n\n\n\n\n\n \nWe now obtain the fundamental description of biseparating maps from the foregoing propositions.\n\n\n\\begin{Def}\\label{d2.1}\nRetain the notation above. A bijection $T: A(X,E)\\to A(Y,F)$ is {\\em locally determined} if there is a bijection $\\theta:{\\mathcal D}(X)\\to{\\mathcal D}(Y)$, preserving set inclusions, so that \nfor any $f,g\\in A(X,E)$ and any $U \\in {\\mathcal D}(X)$, $f = g$ on $U$ if and only if $Tf = Tg$ on $\\theta(U)$\n\\end{Def}\n\n\\begin{lem}\\label{l2.0}\nAssume that $A(X,E)$ is basic. Let $T:A(X,E)\\to A(Y,F)$ be locally detemined, with a map $\\theta$ as given in Definition \\ref{d2.1}. If $g,h \\in A(X,E)$, then $C(Tg-Th) \\subseteq \\theta(\\widehat{C}(g-h))$.\n\\end{lem}\n\n\\begin{proof}\nSuppose that $y \\notin \\theta(\\widehat{C}(g-h))$. Choose $v_1\\in A(Y,F)$ so that \n$\\widehat{C}(v_1) = \\theta(\\widehat{C}(g-h))$.\nThere exists $v_2\\in A(Y,F)$ such that $y\\in C(v_2)$ and $v_1\\perp v_2$.\nLet $u_1 = g-h$ and $u_2\\in A(X,E)$ be such that $\\widehat{C}(u_2) = \\theta^{-1}(\\widehat{C}(v_2))$.\n\nWe claim that $u_1\\perp u_2$.\nOtherwise, there exists nonempty $C\\in {\\mathcal C}(X)$ so that $C\\subseteq C(u_1) \\cap C(u_2)$.\nHence \n\\[ \\theta(\\widehat{C}) \\subseteq \\theta(\\widehat{C}(u_1))\\cap \\theta(\\widehat{C}(u_2)) = \\widehat{C}(v_1)\\cap \\widehat{C}(v_2).\\]\nLet $\\theta(\\widehat{C}) = \\widehat{C}(v)$ for some nonzero $v\\in A(Y,F)$.\nSince $v_1\\perp v_2$, $C(v_1) \\cap \\widehat{C}(v_2) = \\emptyset$.\nTherefore, \n\\[C(v_1) \\cap C(v) \\subseteq C(v_1) \\cap \\widehat{C}(v_2) =\\emptyset.\n\\]\nHence $\\widehat{C}(v_1)\\cap {C}(v) = \\emptyset$. But $\\emptyset \\neq C(v) \\subseteq \\widehat{C}(v) = \\theta(\\widehat{C}) \\subseteq \\widehat{C}(v_1)\n$. Thus we have a contradiction. This proves the claim.\n\nFrom the claim, $g=h$ on $\\widehat{C}(u_2)$. Hence $Tg = Th$ on $\\theta(\\widehat{C}(u_2)) = \\widehat{C}(v_2)$. In particular, $y\\notin C(Tg-Th)$. This completes the proof of the lemma.\n\\end{proof}\n\n\\begin{thm}\n\\label{t5}\nSuppose that $A(X,E)$ and $A(Y,F)$ are both basic and compatible. A bijection $T:A(X,E)\\to A(Y,F)$ is a biseparating map if and only if it is locally determined.\n\\end{thm}\n\n\\begin{proof}\nAssume that $T$ is biseparating. Take any $f\\in A(X,E)$ and let $\\theta = \\theta_f$. By Proposition \\ref{p4}, $\\theta$ is independent of the choice of $f$.\nThe properties enunciated for $\\theta$ now follow from the same ones for $\\theta_f$ by Proposition \\ref{p2}. Therefore, $T$ is locally determined.\n\nConversely, suppose that $T$ is locally determined. Let $f,g,h\\in A(X,E)$ be such that $f\\perp_h g$.\nThen $f =h$ on $\\widehat{C}(g-h)$.\nTherefore, $Tf = Th$ on $\\theta(\\widehat{C}(g-h))$.\nBy Lemma \\ref{l2.0}, $Tf = Th$ on $C(Tg-Th)$. Thus $Tf \\perp_{Th} Tg$.\nSince the same argument applies to $T^{-1}$, we see that $T$ is biseparating.\n\\end{proof}\n\n\n\nLet us give some examples of function spaces that are both basic and compatible. The verifications are simple and will be omitted. \nIf $G$ is a Banach space, a {\\em bump function} on $G$ is a nonzero real-valued function on $G$ with bounded support.\n\n\n\\begin{ex}\\label{e1.7}\nLet $A(X,E)$ be any of the spaces described below. Then $A(X,E)$ is both basic and compatible.\nFurthermore, $\\widehat{C}(f) = \\overline{C(f)}$ for any $f \\in A(X,E)$. \n\\begin{enumerate}\n\\item Let $X$ be a Hausdorff completely regular topological space and let $E$ be a nonzero Hausdorff topological vector space.\n$A(X,E) = C(X,E)$ or $C_*(X,E)$, the subspace consisting of all bounded functions in $C(X,E)$. (By a bounded function, we mean a function whose image is bounded in $E$, i.e, can be absorbed by any neighborhood of $0$.)\n\\item Let $X$ be a metric space and let $E$ be a normed space. Take $A(X,E)$ to be one of the following spaces. $U(X,E)$, the space of all $E$-valued uniformly continuous functions on $X$; or $\\operatorname{Lip}(X,E)$, the space of all $E$-valued Lipschitz functions on $X$; or $U_*(X,E)$, respectively, $\\operatorname{Lip}_*(X,E)$, the bounded functions in $U(X,E)$ and $\\operatorname{Lip}(X,E)$ respectively.\n\\item Let $X$ be an open set in a Banach space $G$, $p\\in {\\mathbb N}\\cup\\{\\infty\\}$, and let $E$ be a normed space. Assume that $G$ supports a $C^p$ bump function and take $A(X,E) = C^p(X,E)$, the space of all $p$-times continuous differentiable $E$-valued functions on $X$. Alternatively, let $A(X,E) = C^p_*(X,E)$, the subspace of $C^p(X,E)$ so that \n$D^kf$ is bounded on $X$ for $k\\in \\{0\\}\\cup{\\mathbb N}$, $k\\leq p$. In the latter case, assume that $G$ supports a $C^p_*$ bump function.\n\\end{enumerate}\n\\end{ex}\n\n\n\n\n\\section{Spaces of continuous functions}\\label{s2}\n\n\nIn this section, let $X$ and $Y$ be Hausdorff completely regular topological spaces and $E$ and $F$ be nontrivial Hausdorff topological vector spaces.\nTake $A(X,E) = C(X,E)$ or $C_*(X,E)$ and $A(Y,F) = C(Y,F)$ or $C_*(Y,F)$.\nLet $T:A(X,E)\\to A(Y,F)$ be a biseparating map.\nWithout loss of generality, we may assume that $T0 =0$.\nWe retain the notation in \\S \\ref{s1}.\nThe main aim is to derive topological relationship between $X$ and $Y$ based on the map $T$.\nRecall that a Hausdorff completely regular topological space $X$ has a ``largest\" compactification, namely the Stone-\\v{C}ech compactification $\\beta X$. \nIf $V$ is a set in $Y$, denote its closures in $Y$ and $\\beta Y$ by $\\overline{V}$ and $\\overline{V}^{\\beta Y}$ respectively.\nBy Example \\ref{e1.7}, $\\widehat{C}(f) = \\overline{C(f)}$ for any $f\\in A(X,E)$.\n\n\n\n\n\n\n\\begin{lem}\\label{l2.0.1}\nLet $U_i, i=1,2$, be open sets in $\\beta X$ so that $U_i\\cap X\\in {\\mathcal C}(X)$ and that $\\overline{U_1}^{\\beta X}\\cap \\overline{U_2}^{\\beta X}= \\emptyset$. Then $\\overline{\\theta(\\overline{U_1\\cap X})}^{\\beta X} \\cap \\overline{\\theta(\\overline{U_2\\cap X})}^{\\beta X} = \\emptyset$.\n\\end{lem}\n\n\\begin{proof}\nLet $v$ be a nonzero vector in $F$. There exists $h\\in C_*(X)$ so that $h(x) =1$ for all $x\\in U_1\\cap X$ and $h(x) = 0$ for all $x\\in U_2\\cap X$. The function $f = h\\cdot T^{-1}(1\\otimes v)$ belongs to $A(X,E)$. By Lemma \\ref{t5}, $Tf = 1\\otimes v$ on $\\theta(\\overline{U_1\\cap X})$ and $Tf = 0$ on $\\theta(\\overline{U_2\\cap X})$.\nSince $F$ is a Hausdorff topological vector space, it is completely regular. \nSo there exists a continuous function $g: F\\to {\\mathbb R}$ so that $g(v) \\neq g(0)$.\nSet $k =g\\circ (Tf):Y\\to {\\mathbb R}$. Then $k$ is continuous on $Y$ has hence has a continuous extension $\\widetilde{k}:\\beta Y\\to {\\mathbb R}\\cup \\{\\infty\\}$.\nNow $k(y) = g(v)$ for all $y\\in \\theta(\\overline{U_1\\cap X})$ and $k(y) = g(0)$ for all $y\\in \\theta(\\overline{U_2\\cap X})$.\nBy continuity of $\\widetilde{k}$, the sets $\\overline{\\theta(\\overline{U_i\\cap X})}^{\\beta X}$, $ i=1,2$, must be disjoint.\n\\end{proof}\n\nFor any $x\\in \\beta X$, let ${\\mathcal F}_x$ be the family of all open neighborhoods $U$ of $x$ in $\\beta X$ so that $U\\cap X \\in {\\mathcal C}(X)$.\nDefine ${\\mathcal F}_y$ similarly for $y\\in \\beta Y$.\nWe will use the following fact which is easily deduced from the Urysohn Lemma.\nLet $U$ be an open neighborhood of a point $x$ in $\\beta X$. There exists an open neighborhood $V\\in {\\mathcal F}_x$ so that $V\\subseteq U$.\n\n\n\\begin{lem}\\label{l2.0.2}\nLet $x\\in \\beta X$, $y\\in \\beta Y$. Then $y\\in\\overline{\\theta(\\overline{U\\cap X})}^{\\beta Y}$ for all $U \\in {\\mathcal F}_x$ if and only if \n$x\\in\\overline{\\theta^{-1}(\\overline{V\\cap Y})}^{\\beta X}$ for all $V \\in {\\mathcal F}_y$.\n\\end{lem}\n\n\n\\begin{proof}\nAssume that $y\\in\\overline{\\theta(\\overline{U\\cap X})}^{\\beta Y}$ for all $U \\in {\\mathcal F}_x$.\nSuppose that there exists $V\\in {\\mathcal F}_y$ so that $x\\notin\\overline{\\theta^{-1}(\\overline{V\\cap Y})}^{\\beta X}$.\nChoose $U \\in {\\mathcal F}_x$ so that $\\overline{U}^{\\beta X} \\cap \\overline{\\theta^{-1}(\\overline{V\\cap Y})}^{\\beta X} = \\emptyset$.\nNote that by definition of $\\theta^{-1}$, $\\theta^{-1}(\\overline{V\\cap Y}) = \\overline{W_0}$ for some $W_0\\in {\\mathcal C}(X)$.\nExpress $W_0 = W\\cap X$ for some open set $W\\in \\beta X$.\nThen $\\overline{U}^{\\beta X} \\cap \\overline{W}^{\\beta X} = \\emptyset$.\nBy Lemma \\ref{l2.0.1}, $\\overline{\\theta(\\overline{U\\cap X})}^{\\beta X} \\cap \\overline{\\theta(\\overline{W\\cap X})}^{\\beta X} = \\emptyset$. By choice, $y \\in \\overline{\\theta(\\overline{U\\cap X})}^{\\beta Y}$.\nAlso, since $V\\in {\\mathcal F}_y$, \n\\[ y \\in \\overline{{V\\cap Y}}^{\\beta Y}= \\overline{\\theta(\\overline{W\\cap X})}^{\\beta Y}. \\]\nThis contradicts the disjointness of $\\overline{\\theta(\\overline{U\\cap X})}^{\\beta X}$ and $\\overline{\\theta(\\overline{W\\cap X})}^{\\beta X}$ and completes the proof of the ``only if'' part. The reverse implication follows by symmetry.\n\\end{proof}\n\n\\begin{lem}\\label{l2.0.3}\nFor any $x\\in \\beta X$, the set $\\bigcap\\{\\overline{\\theta(\\overline{U\\cap X})}^{\\beta Y}: U \\in {\\mathcal F}_x\\}$ has exactly one point in $\\beta Y$.\n\\end{lem}\n\n\\begin{proof}\nLet $U_i$, $i=1,\\dots, n$, be sets in ${\\mathcal F}_x$.\nThen $\\bigcap^n_{i=1}U_i$ is an open neighborhood of $x$ in $\\beta X$. By the remark after Lemma \\ref{l2.0.1}, \nthere exists $U\\in {\\mathcal F}_x$ so that $U \\subseteq \\bigcap^n_{i=1}U_i$.\nThen $\\overline{\\theta(\\overline{U\\cap X})}^{\\beta Y} \\subseteq \\bigcap^n_{i=1}\\overline{\\theta(\\overline{U_i\\cap X})}^{\\beta Y}$.\nSince the set on the left is nonempty, this shows that the family $\\{\\overline{\\theta(\\overline{U\\cap X})}^{\\beta Y}: U \\in {\\mathcal F}_x\\}$, which consists of closed sets in $\\beta Y$, has the finite intersection property. By compactness of $\\beta Y$, we conclude that the intersection in the statement of the lemma is nonempty.\n\nSuppose that $y_1,y_2$ are distinct points in \n $\\bigcap\\{\\overline{\\theta(\\overline{U\\cap X})}^{\\beta Y}: U \\in {\\mathcal F}_x\\}$.\nChoose sets $V_i \\in {\\mathcal F}_{y_i}$, $i=1,2$, so that $\\overline{V_1}^{\\beta _Y} \\cap \\overline{V_2}^{\\beta Y} = \\emptyset$.\n By Lemma \\ref{l2.0.2}, $x\\in \\overline{\\theta^{-1}(\\overline{V_i\\cap Y})}^{\\beta X}$, $i=1,2$.\nThis contradicts Lemma \\ref{l2.0.1} applied to the map $\\theta^{-1}$.\n\\end{proof}\n\nDefine the map $\\varphi: \\beta X\\to \\beta Y$ by taking $\\varphi(x)$ to be the unique point in $\\bigcap\\{\\overline{\\theta(\\overline{U\\cap X})}^{\\beta Y}: U \\in {\\mathcal F}_x\\}$.\nBy symmetry, we also have an analogous map $\\psi : \\beta Y\\to \\beta X$.\nNow we arrive at the first structural result on biseparating maps on vector-valued $C\/C_*$ spaces.\n\n\n\n\\begin{thm}\\label{t2.0.4}\nLet $X, Y$ be Hausdorff completely regular topological spaces and let $E,F$ be nonzero Hausdorff topological vector spaces.\nSuppose that $T:A(X,E)\\to A(Y,F)$ is a biseparating map, where $A(X,E) = C(X,E)$ or $C_*(X,E)$ and $A(Y,F) = C(Y,F)$ or $C_*(Y,F)$.\nThen there is a homeomorphism $\\varphi:\\beta X\\to \\beta Y$ so that for any $f,g \\in A(X,E)$ and any open set $U$ in $\\beta {X}$, $f = g$ on $U\\cap X$ if and only if $Tf = Tg$ on ${\\varphi}(U) \\cap Y$.\n\\end{thm}\n\n\\begin{proof}\nConsider the maps $\\varphi:\\beta X \\to \\beta Y$ and $\\psi: \\beta Y \\to \\beta X$ given above.\nBy Lemma \\ref{l2.0.2}, $\\varphi$ and $\\psi$ are mutual inverses.\nLet us show that $\\varphi$ is continuous. \nIf $\\varphi$ is not continuous at some $x_0 \\in \\beta X$, then there is a net $(x_\\alpha)$ converging to $x_0$ so that $(\\varphi(x_\\alpha))$ converges to $y_1 \\neq \\varphi(x_0)= y_0$.\nChoose $V_i \\in {\\mathcal F}_{y_i}$, $i=0,1$, so that $\\overline{V_0}^{\\beta Y} \\cap \\overline{V_1}^{\\beta Y} = \\emptyset$.\nFor a cofinal set of $\\alpha$, $V_1 \\in {\\mathcal F}_{\\varphi(x_\\alpha)}$, hence $x_\\alpha = \\psi(\\varphi(x_\\alpha)) \\in \\overline{\\theta^{-1}(\\overline{V_1\\cap Y})}^{\\beta X}$. Therefore, $x_0 \\in \\overline{\\theta^{-1}(\\overline{V_1\\cap Y})}^{\\beta X}$. Also, $V_0\\in {\\mathcal F}_{y_0}$ implies that $x_0\\in \\overline{\\theta^{-1}(\\overline{V_0\\cap Y})}^{\\beta X}$.\nThis is impossible since $\\overline{\\theta^{-1}(\\overline{V_i\\cap Y})}^{\\beta X}$, $i =0,1$, are disjoint by Lemma \\ref{l2.0.1}.\nThis completes the proof of continuity of $\\varphi$. It follows that $\\varphi$ is a homeomorphism by symmetry.\n\nLet $U$ be an open set in $\\beta X$ and suppose that $f= g$ on $U\\cap X$ for some $f,g\\in A(X,E)$. \nLet $y_0\\in \\varphi(U)\\cap Y$ and set $x_0 = \\psi(y_0)\\in U$. We wish to show that $Tf(y_0) =Tg(y_0)$.\nBy the remark preceding Lemma \\ref{l2.0.2}, we may assume that $U\\cap X \\in {\\mathcal C}(X)$. \nThen $U\\in {\\mathcal F}_{x_0}$. By definition of $\\varphi$, $y_0 = \\varphi(x_0) \\in \\overline{\\theta(\\overline{U\\cap X})}^{\\beta Y}$.\nSince $y_0\\in Y$ and $\\theta(\\overline{U\\cap X})$ is closed in $Y$, $y_0 \\in \\theta(\\overline{U\\cap X})$.\nBy Theorem \\ref{t5}, $Tf = Tg$ on $\\theta(\\overline{U\\cap X})$.\nHence $Tf(y_0) = Tg(y_0)$.\nThis proves that $Tf = Tg$ on $\\varphi(U) \\cap Y$.\nThe reverse implication follows by symmetry.\n\\end{proof}\n\n\n \n\n\n\n\n\n\n\nRecall that a Hausdorff completely regular topological space\n$X$ is {\\em realcompact} if for any $x_0\\in \\beta X\\backslash X$, there is a continuous function $f:\\beta X\\to [0,1]$ so that $f(x_0) =1 > f(x)$ for all $x\\in X$.\nFor more on realcompact spaces, refer to the classic \\cite{GJ}.\nIn particular, let $\\upsilon X$ be the Hewitt realcompactification of $X$ \\cite{GJ}. Then every $f\\in C(X)$ has a unique continuous extension to a (real valued) function $\\stackrel{\\smile}{f} \\in C(\\upsilon X)$. \nThe map $f\\mapsto\\stackrel{\\smile}{f}$ is an algebraic isomorphism and hence biseparating. Hence it is rather natural to consider realcompact spaces in the present context.\nWhen one or more of the spaces $X$ or $Y$ is realcompact, Theorem \\ref{t2.0.4} can be improved.\n\n\\begin{lem}\\label{l2.4}\\cite[Lemma 3.2]{LT}\nLet $Y$ be a realcompact space and let $y_0 \\in \\beta Y \\backslash Y$. There exist open sets $U_n$ and $V_n$ in $\\beta Y$, $n \\in {\\mathbb N}$, such that\n\\begin{enumerate}\n\\item $\\overline{U_n}^{\\beta Y} \\subseteq V_n$ for all $n$;\n\\item $y_0 \\in \\overline{\\bigcup^\\infty_{n=m}U_n}^{\\beta Y}$ for all $m$;\n\\item $Y \\cap \\bigcap^\\infty_{m=1}\\overline{\\bigcup^\\infty_{n=m}V_n}^{\\beta Y} = \\emptyset$;\n\\item $\\overline{V_n}^{\\beta Y} \\cap \\overline{V_m}^{\\beta Y} = \\emptyset$ if $n\\neq m$.\n\\end{enumerate}\n\\end{lem}\n\n\\begin{lem}\\label{l2.5.1}\nLet $E$ be a Hausdorff topological vector space and let $0 \\neq u\\in E$.\nThere is a continuous function $h:E\\to {\\mathbb R}$ so that $h(nu)= u$ for all $n\\in {\\mathbb N}$.\n\\end{lem}\n\n\\begin{proof}\nLet $U$ be a circled open neighborhood of $0$ in $E$ so that $u \\notin U + U +U$.\nThen let $V$ be an open neighborhood of $0$ so that $\\overline{V}\\subseteq U$.\nSet $V_n = nu + V$, $n\\in{\\mathbb N}$. Suppose that $m\\neq n$ and $\\overline{V_m} \\cap \\overline{V_n} \\neq \\emptyset$.\nThen there are $v_1,v_2\\in \\overline{V}$ so that $mu+v_1 = nu + v_2$.\nThus\n\\[ u = \\frac{v_2}{m-n} + \\frac{v_1}{n-m} \\in U+U \\subseteq U+U+U,\n\\]\ncontrary to the choice of $U$. This shows that $\\overline{V_m}\\cap \\overline{V_n} = \\emptyset$ if $m\\neq n$.\n\nNext, we claim that \n$\\overline{\\bigcup V_n} = \\bigcup \\overline{V_n}$.\nSuppose that $x\\in \\overline{\\bigcup V_n}$. \nChoose $n_0\\in {\\mathbb N}$ so that $\\frac{x}{n_0} \\in U$.\nFor any $n\\geq n_0$, if $V_n \\cap (x+U) \\neq \\emptyset$, then there are $v\\in V$ and $w\\in U$ so that \n$nu+ v = x+w$. Thus \n\\[ u = \\frac{x}{n} + \\frac{w}{n} -\\frac{v}{n} \\in U + U + U,\\]\na contradiction. Hence $x\\notin \\overline{\\bigcup_{n\\geq n_0}V_n}$.\nTherefore, $x\\in \\overline{\\bigcup^{n_0-1}_{n=1}V_n} = \\bigcup^{n_0-1}_{n=1}\\overline{V_n}$.\nThis proves the claim.\n\n\nSince $E$ is completely regular, for each $n\\in {\\mathbb N}$, there exists a continuous function $h_n:E\\to {\\mathbb R}$ so that $h_n(nu) = n$ and that $h_n(x) = 0$ if $x \\notin V_n$.\nDefine $h:E\\to {\\mathbb R}$ by $h(x) = h_n(x)$ if $x\\in V_n$ for some $n$ and $h(x) = 0$ otherwise.\nFrom the properties of the sets $V_n$ shown above, for each $x\\in E$, there are an open neighborhood $O$ \nof $x$ and some $n\\in {\\mathbb N}$ so that $h = h_n$ on $O$ or $h = 0$ on $O$.\nIt follows easily that $h$ is continuous.\nObviously, $h(nu)= n$ for all $n\\in {\\mathbb N}$.\n\\end{proof}\n\n\\begin{lem}\\label{l2.6}\nIf $A(Y,F) = C(Y,F)$ and $Y$ is realcompact, then $\\varphi(X)\\subseteq Y$.\n\\end{lem}\n\n\\begin{proof}\nRetain the notation of Theorem \\ref{t2.0.4}. Suppose that there exists $x_0\\in X$ so that $y_0 = \\varphi(x_0) \\in \\beta Y \\backslash Y$.\nChoose open sets $U_n, V_n$ in $\\beta Y$ using Lemma \\ref{l2.4}.\nFrom property (1) of said lemma, there exists a continuous function $f_n:\\beta Y \\to [0,1]$ so that\n$f_n =1$ on $U_n$ and $f_n = 0$ outside $V_n$.\nFix a nonzero vector $u \\in E$ and let $g_n$ be defined on $Y$ by $g_n(y) = f_n(y)T(n\\otimes u)(y)$.\nThen set $g(y) = g_n(y)$ if $y\\in V_n\\cap Y$ for some $n$ and $g(y) = 0$ if $y \\in Y \\backslash (\\bigcup^\\infty_{n=1}V_n)$.\nFix $y \\in Y$. By property (3) of Lemma \\ref{l2.4}, there exists $m$ so that $y \\notin \\overline{\\bigcup^\\infty_{n=m}V_n}^{\\beta Y}$.\nBy property (4) of Lemma \\ref{l2.4}, there exists at most one $n_0$, $1\\leq n_0< m$, so that $y\\in \\overline{V_{n_0}}^{\\beta Y}$.\nTherefore, there exists an open neighborhood $U$ of $y$ in $Y$ so that $g = g_{n_0}$ or $g =0$ on the set $U$.\nThus $g$ is continuous at $y$. Since $y\\in Y$ is arbitrary, $g\\in C(Y,F)$.\nAs $g = T(n\\otimes u)$ on $U_n\\cap Y$, by Theorem \\ref{t2.0.4}, $T^{-1}g = nu$ on ${\\varphi}^{-1}(U_n)\\cap X$.\nBy Lemma \\ref{l2.5.1}, there is a continuous function $h:E\\to {\\mathbb R}$ so that $h(nu) = n$ for all $n$.\nSet $k = h\\circ T^{-1}g \\in C(X)$. Let $m\\in {\\mathbb N}$. From the above, $k \\geq m$ on ${\\varphi}^{-1}(\\bigcup^\\infty_{n=m}U_n)\\cap X$.\nBy (2) of Lemma \\ref{l2.4}, $y_0 \\in \\overline{\\bigcup^\\infty_{n=m}U_n}^{\\beta Y}$.\nSince each $U_n$ is open in $\\beta Y$ and $\\varphi(X)$ is dense in $\\beta Y$,\n\\[y_0 \\in \\overline{(\\bigcup^\\infty_{n=m}U_n) \\cap \\varphi(X)}^{\\beta Y}.\n\\] As $\\varphi:\\beta X\\to \\beta Y$ is a homeomorphism,\n\\[x_0 =\\varphi^{-1}(y_0) \\in \\overline{\\varphi^{-1}(\\bigcup^\\infty_{n=m}U_n)\\cap X}^{\\beta X}.\n\\]\nRecall that $x_0\\in X$. By continuity of $k$, $k(x_0) \\geq m$. This is a contradiction since $k$ is real-valued and $m$ is arbitrary.\n\\end{proof}\n\n\\begin{thm}\\label{t2.8}\nLet $X$, $Y$ be realcompact spaces and let $E$ and $F$ be Hausdorff topological vector spaces.\nIf $T:C(X,E)\\to C(Y,F)$ is a (nonlinear) biseparating map, then $X$ and $Y$ are homeomorphic.\n\\end{thm}\n\n\\begin{proof}\nBy Lemma \\ref{l2.6}, $\\varphi$ maps $X$ into $Y$. By symmetry, $\\varphi$ maps $X$ onto $Y$.\nHence it is a homeomorphism from $X$ onto $Y$.\n\\end{proof}\n\nTheorem \\ref{t2.8} generalizes the same result obtained in \\cite{A, BBH} for {\\em linear} biseparating maps.\n\n\\begin{thm}\\label{t2.9}\nSuppose that $Y$ is realcompact. Let $T:C_*(X,E)\\to C(Y,F)$ be a biseparating map. Then $Y$ is compact.\n\\end{thm}\n\n\\begin{proof}\nThe proof is the same as the proof of Lemma \\ref{l2.6}.\nIf $y_0 \\in \\beta Y \\backslash Y$, then following the proof of Lemma \\ref{l2.6}, one can choose $0\\neq u\\in E$ and construct a function $g\\in C(Y,F)$ \nand a sequence of nonempty sets $(W_n)$ in $X$ so that $T^{-1}g = nu$ on $W_n$ for each $n\\in {\\mathbb N}$.\n($W_n$ is the set $\\varphi^{-1}(U_n)\\cap X$ in the proof of Lemma \\ref{l2.6}.)\nSince the set $(nu)^\\infty_{n=1}$ cannot be a bounded set in $E$, $T^{-1}g\\notin C_*(X,E)$, contrary to the assumption. Therefore, $Y = \\beta Y$ is compact.\n\\end{proof}\n\n\\noindent \n{\\bf Remark}. It is well known that $C_*(X)$ is algebraically isomorphic to $C(\\beta X)$. Hence one cannot expect $X$ to be compact in Theorem \\ref{t2.9} in general.\n\n\n\\section{Metric cases -- general results}\\label{s3}\nThroughout this section, let $X$ and $Y$ be metric spaces, $E$ and $F$ be nontrivial normed spaces and $A(X,E)$, $A(Y,F)$ be vector subspaces of $C(X,E)$ and $C(Y,F)$ respectively. Recall that a {\\em bump function} on a Banach space $G$ is a nonzero real-valued function on $G$ with bounded support. \nWhen we speak of the spaces $C^p(X,E)$ or $C^p_*(X,E)$, it will be assumed additionally that $X$ is an open set in a Banach space that supports a $C^p$, respectively, $C^p_*$, bump function. \nA sequence $(x_n)$ in $X$ is {\\em separated} if $\\inf_{n\\neq m}d(x_n,x_m) > 0$.\nThe main aim of the section is an analog of the structural result Theorem \\ref{t2.0.4} when $X$ and $Y$ are metric spaces. In this instance, we make use of completion instead of compactification.\n\n\n\\begin{prop}\\label{p3.0.1}\nLet $A(X,E)$ be one of the spaces $C(X,E)$, $C_*(X,E)$, $U(X,E)$, $U_*(X,E)$, $\\operatorname{Lip}(X,E)$, $\\operatorname{Lip}_*(X,E)$, $C^p(X,E)$ or $C^p_*(X,E)$.\nThen $A(X,E)$ has\n the following properties.\n\\begin{enumerate}\n\\item[(S1)] $A(X,E)$ is compatible.\n\\item[(S2)] For any $x\\in X$ and any $\n\\varepsilon >0$, there exists $C\\in {\\mathcal C}(X)$ so that $x\\in C$ and $\\operatorname{diam} C <\\varepsilon$. In particular, $A(X,E)$ is basic and $\\widehat{C}(f) = \\overline{C(f)}$ for all $f\\in A(X,E)$. \n\\item[(S3)] If $f\\in A(X,E)$ and $(x_n)$ is a separated sequence in $X$, then there are a sequence $(C_n)$ in ${\\mathcal C}(X)$ and a function $g\\in A(X,E)$ so that $x_n \\in C_n$ for all $n$, $g = f$ on $C_n$ for infinitely many $n$ and $g = 0$ on $C_n$ for infinitely many $n$. \n\\item[(S4)] Let $(x_n)$ and $(x'_n)$ be Cauchy sequences so that $\\inf_{m,n} d(x_m,x'_n) > 0$.\nFor any $f\\in A(X,E)$, there are sets $U,V\\in {\\mathcal C}(X)$ and a function $g\\in A(X,E)$ so that $x_n\\in U$ and $x'_n\\in V$ for infinitely many $n$, $g = f$ on $U$ and $g = 0$ on $V$.\n\\end{enumerate}\n\\end{prop}\n\n\\begin{proof}\nExcept for property (S3) for the spaces $U(X,E)$ and $\\operatorname{Lip}(X,E)$, all other verifications are straightforward and are left to the reader.\nTo verify (S3) for $A(X,E) = U(X,E)$ or $\\operatorname{Lip}(X,E)$, \nlet $(x_n)$ be a sequence in $X$ so that $\\inf_{n\\neq m}d(x_n,x_m) = 3r > 0$ and let $f\\in A(X,E)$.\n\nIn the first instance, assume that $(f(x_n))$ is a bounded sequence in $E$. \nSince $f\\in U(X,E)$, there exists $0< r' < r$ so that \\[\\sup\\{\\|f(x)\\|: x\\in \\bigcup_n B(x_n,r')\\} = M < \\infty.\\]\nFor each $n\\in {\\mathbb N}$, let $h_n, k_n:E\\to [0,1]$ be defined by \n\\[ h_n(x) = (2 - \\frac{2}{r'}d(x,x_n))^+\\wedge 1 \\text{ and } k_n(x) = (1 - \\frac{d(x,x_n)}{r'})^+.\\]\nThen $(h_n)$ is a sequence of disjoint functions. Let $h$ be the pointwise sum $\\sum h_{2n-1}$. It is easily verified that $h: E\\to [0,1]$ is a Lipchitz function with Lipschitz constant $\\frac{2}{r'}$.\nTake $g = h\\cdot f$. For each $n$, let $C_n = B(x_n,\\frac{r'}{2})$. Fix a nonzero vector $a\\in E$. Then $k_n\\otimes a \\in \\operatorname{Lip}(X,E) \\subseteq A(X,E)$. Hence\n$C_n = C(k_n\\otimes a) \\in {\\mathcal C}(X)$. Clearly $g= f$ on $C_n$ if $n$ is odd and $g = 0$ on $C_n$ if $n$ is even.\nLet us verify that $g\\in A(X,E)$.\nIndeed, suppose that $s, t\\in X$. If $s,t\\notin \\bigcup_n B(x_n,r')$, then $h(s) = h(t) =0$. Hence $\\|g(s) - g(t)\\| = 0$. Otherwise, assume without loss of generality that $t\\in \\bigcup_n B(x_n,r')$. We have\n\\begin{align*}\n\\|g(s) -g(t)\\| & \\leq |h(s)|\\, \\|f(s) - f(t)\\| + |h(s)-h(t)|\\,\\|f(t)\\| \\\\\n& \\leq \\|f(s)-f(t)\\| + \\frac{2}{r'}\\,d(s,t)\\cdot M.\n\\end{align*}\nSince $f\\in A(X,E)$, it follows that $g \\in A(X,E)$.\n\nIn the second case, assume that $(f(x_n))$ is unbounded in $E$.\nLet $t_n = \\|f(x_n)\\|$ for all $n$. By replacing $(x_n)$ by a subsequence if necessary, we may assume that $t_1> 0$ and $6t_n < t_{n+1}$ for all $n$. Define $\\gamma: [0,\\infty) \\to [0,1]$ by\n\\[ \\gamma(t) = \\begin{cases} \n (2 - \\frac{3}{t_n}|t-t_n|)\\wedge 1 &\\text{if $|t-t_n| < \\frac{2t_n}{3}$ for some odd $n$}\\\\\n 0 & \\text{otherwise}. \\end{cases}\n \\]\nDirect verification shows that if $0\\leq a\\leq b\\neq 0$, then $|\\gamma(a)-\\gamma(b)| \\leq \\frac{6|a-b|}{b}$.\n\nLet $g:X\\to E$ be given by $g(x) = \\gamma(\\|f(x)\\|)f(x)$.\nSuppose that $y,z\\in X$ with $\\|f(y)\\| \\leq \\|f(z)\\|$ and $f(z) \\neq 0$.\nThen\n\\begin{align*}\n\\|g(y)- g(z)\\| & \\leq |\\gamma(\\|f(y)\\|) - \\gamma(\\|f(z)\\|)|\\,\\|f(y)\\| + |\\gamma(\\|f(z)\\|)|\\,\\|f(y) - f(z)\\|\\\\\n& \\leq \\frac{6(\\|f(z)\\| - \\|f(y)\\|)}{\\|f(z)\\|}\\,\\|f(y)\\| + \\|f(y) - f(z)\\|\\\\\n& \\leq 7\\|f(y)-f(z)\\|.\n\\end{align*}\nThe same inequality obviously holds if $f(y) = f(z) = 0$.\nSince $f$ belongs to $A(X,E)$, so does $g$. \nAs $\\frac{t_n}{3} < \\|f(x_n)\\| < \\frac{5t_n}{3}$, there exists $C_n\\in {\\mathcal C}(X)$ so that \n\\[ x_n \\in C_n \\subseteq \\{x: \\frac{t_n}{3} < \\|f(x)\\| < \\frac{5t_n}{3}\\}.\\]\n\nFinally, $\\gamma(\\|f(x)\\|) =1$ if $x \\in C_{2n-1}$ and $\\gamma(\\|f(x)\\|) =0$ if $x \\in C_{2n}$.\nHence $g = f$ on $C_{2n-1}$ and $g=0$ on $C_{2n}$.\nThis completes the verification of property (S3) for $A(X,E) = U(X,E)$ or $\\operatorname{Lip}(X,E)$.\n\\end{proof}\n\n\n\n\n\n\n\n\n\n\n\nFor the sake of brevity, let us say that $A(X,E)$ is {\\em standard} if it satisfies properties (S1) -- (S4).\nFor the rest of the section, assume that $A(X,E)$ and $A(Y,F)$ are standard spaces and that $T:A(X,E)\\to A(Y,F)$ is a biseparating map. Without loss of generality, normalize $T$ by taking $T0=0$.\nLet $\\theta:{\\mathcal D}(X) \\to {\\mathcal D}(Y)$ be the map obtained from Theorem \\ref{t5}.\nAs in Section \\ref{s2}, we will show that $\\theta$ induces a point mapping $\\varphi$.\n\n\n\nDenote by $\\widetilde{X}$ and $\\widetilde{Y}$ the respective completions of the spaces $X$ and $Y$.\nFor any subset $U$ of $\\widetilde{X}$, denote the closure of $U$ in $\\widetilde{X}$ by $\\widetilde{U}$. Similarly for sets in $\\widetilde {Y}$.\nIf $x_0\\in \\widetilde{X}$ and $(U_n)$ is a sequence of nonempty sets in ${\\mathcal C}(X)$ so that $\\operatorname{diam} U_n \\to 0$ and $d(x_0,U_n)\\to 0$, we write $(U_n) \\sim x_0$. By condition (S2), for any $x_0\\in \\widetilde{X}$, there is always a sequence $(U_n)$ so that $(U_n)\\sim x_0$.\n\n\n\nSuppose that $y \\in \\theta(\\overline{U})\\cap V$, where $U = C(u) \\in {\\mathcal C}(X)$ and $V= C(v)\\in {\\mathcal C}(Y)$.\nSince $\\theta(\\overline{U}) = \\overline{C(Tu)}$, $C(Tu) \\cap C(v)\\neq \\emptyset$.\nThus $C(u)\\cap C(T^{-1}v)\\neq \\emptyset$. \nHence $U \\cap \\theta^{-1}(\\overline{V})\\neq \\emptyset$.\n\n\n\n\\begin{lem}\\label{l3.1}\nLet $x_0\\in \\widetilde{X}$ and assume that $(U_n)\\sim x_0$.\nTake $y_n \\in \\theta(\\overline{U_n})$ for each $n$.\n\\begin{enumerate}\n\\item If $x_0\\in X$, then $(y_n)$ is a Cauchy sequence in $Y$.\n\\item If, in additon, $A(X,E)\\subseteq U(X,E)$ and contains a nonzero constant function, then $(y_n)$ is a Cauchy sequence in $Y$ for any $x_0\\in \\widetilde{X}$.\n\\end{enumerate}\n\\end{lem}\n\n\\begin{proof}\nFirst we show that every subsequence of $(y_n)$ has a further Cauchy subsequence. Otherwise, there is a subsequence $(y'_n)$ of $(y_n)$ that is separated. \nUnder assumption (1), $x_0 \\in X$. It follows from condition (S2) that there is a function $f\\in A(X,E)$ so that $f(x_0) \\neq 0$. Under assumption (2), take $f = 1\\otimes a\\in A(X,E)$, where $a\\neq 0$. \nSince $A(Y,F)$ has property (S3), there are a subsequence of $(y_n')$, still denoted as $(y_n')$, a sequence $(V_n)$ in ${\\mathcal C}(Y)$ and a function $g\\in A(Y,E)$ so that $y'_n\\in V_n$ for all $n$, $g= Tf$ on $V_n$ for infinitely many $n$ and $g = 0$ on $V_n$ for infinitely many $n$.\nThen $T^{-1}g = f$ on $\\theta^{-1}(\\overline{V_n})$ for infinitely many $n$ and $T^{-1}g = 0$ on $\\theta^{-1}(\\overline{V_n})$ for infinitely many $n$.\nSince $y'_n \\in \\theta(\\overline{U_n}) \\cap V_n$, $U_n \\cap \\theta^{-1}(\\overline{V_n}) \\neq \\emptyset$ by the discussion just before the lemma. Choose a point $x'_n$ from the intersection. Then $(T^{-1}g)(x'_n)=f(x_n')$ for infintely many $n$ and $0$ infinitely often.\nMoreover, $(x'_n)$ converges to $x_0$ in $\\widetilde{X}$.\nUnder assumption (1),\n$x_0 \\in X$, and we have a contradiction to the continuity of $T^{-1}g$ at $x_0$. \nUnder assumption (2), $T^{-1}g \\in U(X,E)$ and $(x'_n)$ is Cauchy in $X$. Hence $((T^{-1}g)(x'_n))$ is Cauchy in $E$. This is impossible since $(T^{-1}g)(x'_n) = f(x'_n) = a$ and $(T^{-1}g)(x'_n) = 0$ both occur infinitely many times.\n\nIf the whole sequence $(y_n)$ is not Cauchy, then in view of the previous paragraph, there are subsequences $(y_{i_n})$ and $(y_{j_n})$ and $\\varepsilon >0$ so that both subsequences are Cauchy and that $d(y_{i_m},y_{j_n}) > \\varepsilon $ for all $m,n$.\nChoose the function $f$ as in the last paragraph. By property (S4), there are $U,V \\in {\\mathcal C}(Y)$ and $g\\in A(Y,F)$ so that $y_{i_n}\\in U$, $y_{j_n}\\in V$ for infinitely many $n$, $g= Tf$ on $U$ and $g = 0$ on $V$.\nThus $T^{-1}g = f$ on $\\theta^{-1}(\\overline{U})$ and $T^{-1}g = 0$ on $\\theta^{-1}(\\overline{V})$.\nThen $y_{i_n} \\in \\theta(\\overline{U_{i_n}}) \\cap U$ for infinitely many $n$ and hence $U_{i_n} \\cap \\theta^{-1}(\\overline{U}) \\neq \\emptyset$ for infinitely many $n$.\nLet $x_{i_n} \\in U_{i_n} \\cap \\theta^{-1}(\\overline{U})$. Then $(x_{i_n})$ converges to $x_0$ in $\\widetilde{X}$ and $T^{-1}g(x_{i_n}) = f(x_{i_n})$ for all $n$.\nSimilar consideration using the sequence $(y_{j_n})$ shows that there is a sequence $(x_{j_n})$ converging to $x_0$ in $\\widetilde{X}$ so that $T^{-1}g(x_{j_n}) = 0$ for all $n$.\nUnder assumption (1), \n\\[ \\lim T^{-1}g(x_{i_n}) = \\lim f(x_{i_n}) = f(x_0) \\neq 0 = \\lim T^{-1}g(x_{j_n}),\\]\ncontradicting the continuity of $T^{-1}g$ at $x_0$.\nUnder assumption (2), $f(x_{i_n}) =a\\neq 0$ by choice of $f$. Thus $T^{-1}g(x_{i_n}) = a$ and $T^{-1}g(x_{j_n}) = 0$ for all $n$, contradicting the uniform continuity of $T^{-1}g$.\n\\end{proof}\n\n\n\nSuppose that $x_0\\in \\widetilde{X}$. Let $(U_n)\\sim x_0, (V_n)\\sim x_0$ and $y_n\\in U_n, z_n\\in V_n$ for all $n$.\nThen $(U_1,V_1,U_2,V_2,\\dots)\\sim x_0$.\nBy Lemma \\ref{l3.1}, if $x_0\\in X$, then the sequence $(y_1,z_1,y_2,z_2,\\dots)$ is Cauchy.\nDefine $\\varphi: X\\to \\widetilde{Y}$ by setting $\\varphi(x) = \\lim y_n$, where $(U_n)\\sim x$ and $y_n \\in \\theta(\\overline{U_n})$ for all $n$. From the above, $\\varphi(x)$ is independent of the choices of $(U_n)$ and $(y_n)$.\nSimilarly, if $A(X,E)\\subseteq U(X,E)$ and contains a constant function $1\\otimes a$ for some $a\\in E\\backslash \\{0\\}$, then Lemma \\ref{l3.1}(2) shows that there is a well defined map $\\widetilde{\\varphi}:\\widetilde{X}\\to \\widetilde{Y}$ given by \n$\\widetilde{\\varphi}(x) = \\lim y_n$, where $(U_n)\\sim x$ and $y_n \\in \\theta(\\overline{U_n})$ for all $n$.\nClearly, in this case, $\\widetilde{\\varphi}$ extends $\\varphi$.\n By symmetry, there is also a similar map $\\psi:Y\\to \\widetilde{X}$ and a map $\\widetilde{\\psi}:\\widetilde{Y}\\to \\widetilde{X}$ under corresponding assumptions on $A(Y,F)$.\n\n\n\n\n\n\\begin{lem}\\label{l3.2}\nThe map $\\varphi$ is continuous from $X$ into $\\widetilde{Y}$. If, in addition, $A(X,E)\\subseteq U(X,E)$ and contains a nonzero constant function, then $\\widetilde{\\varphi}:\\widetilde{X}\\to \\widetilde{Y}$ is continuous on $\\widetilde{X}$.\n\\end{lem}\n\n\\begin{proof}\nWe will prove the second assertion. The first statement can be shown in the same way.\nUnder the second assumption, $\\widetilde{\\varphi}$ is well defined. Let $(x_n)$ be a sequence in $\\widetilde{X}$ that converges to a point $x_0\\in \\widetilde{X}$.\nBy definition of $\\widetilde{\\varphi}$, for each $n$, $\\widetilde{\\varphi}(x_n) = \\lim_k y_{nk}$, where $y_{nk} \\in \\theta(\\overline{U_{nk}})$ and $(U_{nk})_k \\sim x_n$.\nFor each $n$, choose $k_n$ so that $d(y_{nk_n}, \\widetilde{\\varphi}(x_n)), \\operatorname{diam} U_{nk_n}, d(U_{nk_n}, x_{n}) < \\frac{1}{n}$. Then $y_{n_{k_n}} \\in U_{nk_n}$ and $(U_{nk_n})\\sim x_0$.\nThus $\\widetilde{\\varphi}(x_0) = \\lim y_{n{k_n}} = \\lim \\widetilde{\\varphi}(x_n)$.\n\\end{proof}\n\n\n\nSuppose that $x\\in U \\in {\\mathcal C}(X)$. We can choose $(U_n)$ so that $(U_n)\\sim x$ and $U_n \\subseteq U$ for all $n$. By definition of $\\varphi$, $\\varphi(x) = \\lim y_n$ where $y_n\\in \\theta(\\overline{U_n}) \\subseteq \\theta(\\overline{U})$. Hence $\\varphi(x) \\in \\widetilde{\\theta(\\overline{U})}$.\n\n\n\\begin{lem}\\label{l3.2.1}\nAssume that $A(X,E)\\subseteq U(X,E)$ and contains a nonzero constant function.\nLet $f,g\\in A(X,E)$ and $U$ be an open set in $\\widetilde{X}$. If $f = g$ on $U\\cap X$, then $Tf = Tg$ on the set $\\widetilde{\\varphi}(U) \\cap Y$.\n\\end{lem}\n\n\\begin{proof}\nAssume that $x_0\\in U$ and $y_0 = \\widetilde{\\varphi}(x_0) \\in Y$.\nChoose $(U_n)\\sim x_0$ and let $x_n \\in U_n$.\nBy the foregoing remark, $\\varphi(x_n) \\in \\widetilde{\\theta(\\overline{U_n})}$. Pick $y_n \\in \\theta(\\overline{U_n})$.\nBy definition of $\\widetilde{\\varphi}$, $y_0 = \\widetilde{\\varphi}(x_0) = \\lim y_n$.\nFor all sufficiently large $n$, $\\overline{U_n} \\subseteq U\\cap X$.\nHence $f=g$ on $\\overline{U_n}$. By Theorem \\ref{t5}, $Tf = Tg$ on $\\theta(\\overline{U_n})$.\nIn particular, $Tf(y_n) = Tg(y_n)$.\nBy continuity of $Tf$ and $Tg$ at $y_0$, $Tf(y_0) = Tg(y_0)$.\n\\end{proof}\n\nThe following structure theorem applies to spaces of uniformly continuous functions and spaces of Lipschitz functions.\n\n\\begin{thm}\\label{t3.5}\nSuppose that both $A(X,E)$ and $A(Y,F)$ are standard subspaces of \n$U(X,E)$ and $U(Y,F)$ respectively so that both contain nonzero constant functions.\nThere is a homeomorphism $\\widetilde{\\varphi}:\\widetilde{X}\\to \\widetilde{Y}$ so that if\n $f,g\\in A(X,E)$ and $U$ is an open set in $\\widetilde{X}$, then $f=g$ on $U\\cap X$ if and only if $Tf = Tg$ on $\\widetilde{\\varphi}(U)\\cap Y$.\n \\end{thm}\n \n\\begin{proof}\nUnder the given assumptions, we have well defined continuous maps $\\widetilde{\\varphi}:\\widetilde{X}\\to \\widetilde{Y}$ and $\\widetilde{\\psi}:\\widetilde{Y}\\to \\widetilde{X}$ by Lemma \\ref{l3.2}.\nIn the next paragraph, we will show that $\\widetilde{\\psi}\\circ \\widetilde{\\varphi}$ is the identity map on $\\widetilde{X}$. With symmetry, this allows us to conclude that $\\widetilde{\\varphi}$ is a homeomorphism.\nThe final property in the statement of the theorem follows from Lemma \\ref{l3.2.1} and symmetry.\n\nLet $x_0\\in \\widetilde{X}$ and let $(U_n)\\sim x_0$.\nIt follows from (2) of Lemma \\ref{l3.1} and the definition of $\\widetilde{\\varphi}$ that $\\operatorname{diam} \\theta(\\overline{U_n})\\to 0$ and $d(\\theta(\\overline{U_n}),\\widetilde{\\varphi}(x_0)) \\to 0$. By definition of $\\theta$, there exists $V_n \\in {\\mathcal C}(Y)$ so that $\\theta(\\overline{U_n}) = \\overline{V_n}$.\nThen $(V_n)\\sim \\widetilde{\\varphi}(x_0)$.\nHence $\\widetilde{\\psi}( \\widetilde{\\varphi}(x_0))= \\lim x_n$, where $x_n \\in \\theta^{-1}(\\overline{V_n}) = \\overline{U_n}$ for all $n$.\nTherefore, $\\widetilde{\\psi}( \\widetilde{\\varphi}(x_0))= x_0$, as claimed.\n\\end{proof} \n \n \n\n\n\n\n\n\n\n\n\n\n\n\n\n\nNext, we consider the cases where one or both of $A(X,E)$ and $A(Y,F)$ is either $C, C_*$ or $C^p$.\n\n\n\\begin{lem}\\label{l3.6}\nSuppose that $A(X,E)$ is standard and $A(Y,F) = C(Y,F)$, $C_*(Y,F)$ or $C^p(Y,F)$. If $T:A(X,E)\\to A(Y,F)$ is a biseparating map, then $\\varphi(X)\\subseteq Y$.\n\\end{lem}\n\n\\begin{proof}\nSuppose on the contrary that there exists $x_0\\in X$ so that $y_0 = \\varphi(x_0) \\in \\widetilde{Y}\\backslash Y$.\nLet $(U_n)\\sim x_0$. Then $(\\theta(\\overline{U_n}))$ is a sequence of sets in $Y$, each with nonempty interior, so that $\\operatorname{diam}\\theta(\\overline{U_n}) \\to 0$ and $d(\\theta(\\overline{U_n}),y_0) \\to 0$.\nHence one can find a sequence $(g_n)$ in $C_*(Y)$, respectively $C^p(Y)$, and a sequence $(V_n)$ of nonempty sets in ${\\mathcal C}(Y)$ so that $g_n =1$ on $V_n$, $\\overline{C(g_n)}\\subseteq \\theta(\\overline{U_n})$ for all $n$, and $\\overline{C(g_m)}\\cap \\overline{C(g_n)} = \\emptyset$ if $m\\neq n$.\nAs observed in the proof of Theorem \\ref{t3.5}, $\\operatorname{diam} \\theta(\\overline{U_n})\\to 0$.\nSo $(\\overline{C(g_n}))$ is a pairwise disjoint sequence so that $\\operatorname{diam} \\overline{C(g_n)}\\to 0$ and $d(\\overline{C(g_n)},y_0) \\to 0$, where $y_0 \\notin Y$. Therefore the pointwise sum $g = \\sum g_{2n}$ belongs to $C_*(Y)$, respectively, $C^p(Y)$.\nBy condition (S2), there exists $f\\in A(X,E)$ so that $f(x_0) \\neq 0$.\nThen $h = g\\cdot Tf$ lies in $A(Y,F)$.\nSince $h = Tf$ on $\\overline{V_n}$ if $n$ is even and $h = 0$ on $\\overline{V_n}$ if $n$ is odd, and $\\overline{V_n} \\in {\\mathcal D}(Y)$,\nby Theorem \\ref{t5}, $T^{-1}h = f$ on $\\theta^{-1}(\\overline{V_n})$ if $n$ is even and \n$T^{-1}h = 0$ on $\\theta^{-1}(\\overline{V_n})$ if $n$ is odd.\nSince $\\overline{V_n} \\subseteq \\theta(\\overline{U_n})$, $\\theta^{-1}(\\overline{V_n}) \\subseteq \\overline{U_n}$.\nChoose $x_n \\in \\theta^{-1}(\\overline{V_n})$ for each $n$. Then $(x_n)$ converges to $x_0$.\nHowever, $T^{-1}h(x_n) = f(x_n)$ if $n$ is odd and $T^{-1}h(x_n) = 0$ if $n$ is even.\nAs $(f(x_n))$ converges to $f(x_0) \\neq 0$, \nthis contradicts the continuity of $T^{-1}h$ at $x_0$. This proves that $\\varphi(X) \\subseteq Y$.\n\\end{proof}\n\nThe next two results can be obtained utilizing the proof of Theorem \\ref{t3.5} and taking into account Lemma \\ref{l3.6}. \n\n\\begin{thm}\\label{t3.7}\nLet $A(X,E) = C(X,E),$ $C_*(X,E)$ or $C^p(X,E)$ and let $A(Y,F) = C(Y,F),$ $C_*(Y,F)$ or $C^q(Y,F)$.\nThere exists a homeomorphism $\\varphi: X\\to Y$ so that \n for any $f,g\\in A(X,E)$, and any open set $U$ in $X$, $f =g$ on $U$ $\\iff$ $Tf = Tg$ on $\\varphi(U)$. \n\\end{thm}\n\n\n\n\n\\begin{thm}\\label{t3.8}\nLet $A(X,E)$ be a standard vector subspace of $U(X,E)$ that contains a nonzero constant fucntion. Suppose that $A(Y,F) = C(Y,F)$, $C_*(Y,F)$ or $C^p(Y,F)$.\nThere exists a homeomorphism $\\varphi: X\\to \\varphi(X)$, where $\\varphi(X)$ is a dense subset of $Y$, and\n for any $f,g\\in A(X,E)$ and any open set $U$ in $X$, $f =g$ on $U$ $\\iff$ $Tf = Tg$ on $\\varphi(U)$. \n\\end{thm}\n\n\\begin{proof}\nWe will only prove the density of $\\varphi(X)$ in $Y$. The other parts follow from the proof of Theorem \\ref{t3.5}, using Lemma \\ref{l3.6}.\nBy Lemma \\ref{l3.2} and Lemma \\ref{l3.6}, $\\varphi$ is a continuous map from $X$ into $Y$ with a continuous extension $\\widetilde{\\varphi}:\\widetilde{X}\\to \\widetilde{Y}$.\nAlso, we have an analogous continuous map $\\psi:Y\\to \\widetilde{X}$. \nFrom the second paragraph of the proof of Theorem \\ref{t3.5}, we see that $\\widetilde{\\varphi}\\circ\\psi$ is the identity map on $Y$. Given $y\\in Y$, $\\psi(y)\\in \\widetilde{X}$. Hence there is a sequence $(x_n)$ in $X$ that converges to $\\psi(y)$.\nBy continuity of $\\widetilde{\\varphi}$ at $\\psi(y)$, $(\\varphi(x_n))=(\\widetilde{\\varphi}(x_n))$ converges to $\\widetilde{\\varphi}(\\psi(y)) =y$.\nThis proves that $y\\in \\overline{\\varphi(X)}$.\n\\end{proof}\n\n\nWe conclude this section with a remark concerning the space $C^p_*(X,E)$, where $X$ is an open set in a Banach space $G$ that supports a $C^p_*$ bump function. In general, it may not be true that all functions in $C^p_*(X,E)$ are uniformly continuous (with respect to the norm on $G$).\nOn the other hand, an easy application of the mean value inequality shows that if $X$ is open and {\\em convex} in $G$, then $C^p_*(X,E) \\subseteq U(X,E)$. \nIn particular, Theorems \\ref{t3.5} and \\ref{t3.8} apply to $C^p_*$ spaces whose domains are convex open sets.\n\n\n\n\n\\section{Pointwise representation}\\label{s4}\n\nRetain the notation of Section \\ref{s3}. That is, let $X$ and $Y$ be metric spaces, $E$ and $F$ be nontrivial normed spaces, and assume that $A(X,E)$ and $A(Y,F)$ are standard vector subspaces of $C(X,E)$ and $C(Y,F)$ respectively. Say that $A(X,E)$ has property (P) if\n\\begin{enumerate}\n\\item[(P)] For any accumulation point $x$ of $\\widetilde{X}$ and any function $f\\in A(X,E)$ so that $\\lim_{\\stackrel{z\\to x}{z\\in X}}f(z) =0$, there are open sets $U$ and $V$ in $X$ and a function $g\\in A(X,E)$ so that $x\\in \\widetilde{U} \\cap \\widetilde{V}$ and that $g =f$ on $U$ and $g = 0$ on $V$.\n\\end{enumerate}\n\n\\medskip\n\n\\noindent {\\bf Remark}. \nIf $x$ is an isolated point of $\\widetilde{X}$, then $x\\in X$. In this case, given $f\\in A(X,E)$ so that $f(x) = 0$, take $U = V= \\{x\\}$ and $g =f$. It is clear that the conditions above are fulfilled.\n\n\n\\begin{prop}\\label{p3.10}\nLet $A(X,E)$ be one of the spaces $C(X,E)$, $C_*(X,E)$, $U(X,E)$, $U_*(X,E)$, $\\operatorname{Lip}(X,E)$ or $\\operatorname{Lip}_*(X,E)$. Then $A(X,E)$ has property (P).\n\\end{prop}\n\n\\begin{proof}\nLet $x_0$ be an accumulation point of $\\widetilde{X}$ and let $f$ be a function in $A(X,E)$ so that $\\lim_{\\stackrel{x\\to x_0}{x\\in X}}f(x) =0$.\nThere is a sequence $(x_n)$ in $X$ converging to $x_0$ so that $0 < d(x_{n+1},x_0) < \\frac{d(x_n,x_0)}{3}$ for all $n$. Set $r_n = d(x_n,x_0)$ and let $\\gamma_n:[0,\\infty)\\to {\\mathbb R}$ be the function\n\\[\\gamma_n(r) = (2 - \\frac{4|r-r_n|}{r_n})^+\\wedge 1.\\]\n$(\\gamma_n)$ is a disjoint sequence of functions. Furthermore,\n\\[ |\\gamma_n(a) - \\gamma_n(b)| \\leq \\frac{4}{r_n}|a-b| \\wedge 1 \\text{ for all $a, b\\geq 0$}.\\]\nWe may assume that $\\|f(x)\\| \\leq1$ if $d(x,x_0) < \\frac{3r_1}{2}$.\nLet \n\\[ g_n(x) = \\gamma_n(d(x,x_0))f(x)\\quad \\text{and} \\quad g = \\sum g_{2n} \\ \\text{(pointwise sum).}\\]\nSince $\\gamma_n(d(\\cdot,x_0))$ is bounded Lipschitz and $f$ is bounded on the support of $g_n$, it is easy to check that $g_n\\in A(X,E)$.\nNote that\n\\[ \\|g_n\\|_\\infty \\leq \\sup\\{\\|f(x)\\| :\\frac{r_n}{2} < d(x,x_0) < \\frac{3r_n}{2}\\} \\to 0.\\]\nTherefore, if $A(X,E)$ is any of the spaces except $\\operatorname{Lip}(X,E)$ or $\\operatorname{Lip}_*(X,E)$, $g$ is the uniform limit of its partial sums and hence belongs to $A(X,E)$.\n\nNow consider the cases $A(X,E) = \\operatorname{Lip}(X,E)$ or $\\operatorname{Lip}_*(X,E)$. First of all, the function $g$ is bounded.\nLet's check that it is Lipschitz. Since $f$ is Lipschitz and $\\lim_{\\stackrel{x\\to x_0}{x\\in X}}f(x) =0$, $\\|f(x)\\| \\leq L(f)d(x,x_0)$, where $L(f)$ is the Lipschitz constant of $f$.\nFor any $n\\in {\\mathbb N}$, we claim that $g_n$ is Lipschitz with $L(g_n) \\leq 7L(f)$. Let $x,z\\in X$, $a = d(x,x_0)$, $b = d(z,x_0)$. \nIf $\\gamma_n(a) = \\gamma_n(b) =0$, then $g_n(x) - g_n(z) =0$. Otherwise, we may assume that $\\gamma_n(a) \\neq 0$, so that $\\frac{r_n}{2} < a < \\frac{3r_n}{2}$.\nThen\n\\begin{align*}\n\\|g_n(x) - g_n(z)\\| & \\leq |\\gamma_{n}(a)-\\gamma_{n}(b)|\\,\\|f(x)\\| + |\\gamma_{n}(b)|\\,\\|f(x) - f(z)\\|\\\\\n&\\leq \\frac{4}{r_{n}}|a-b|\\,L(f)a + \\|f(x) -f(z)\\|\\\\\n&\\leq 6L(f)|a-b| + L(f)d(x,z) \\leq 7L(f)d(x,z).\n\\end{align*}\nThus $L(g_n)\\leq 7L(f)$, as claimed.\nFor any $x,z\\in X$, either there exists $n$ so that $g(x) = g_{2n}(x)$, $g(z) = g_{2n}(z)$, or there are distinct $m,n$ so that $g(x) = g_{2n}(x) + g_{2m}(x)$, $g(z) = g_{2n}(z) + g_{2m}(z)$. In either case, it follows that $\\|g(x)-g(z)\\| \\leq 14L(f)d(x,z)$. This completes the proof that $g\\in \\operatorname{Lip}_*(X,E)\\subseteq A(X,E)$.\nClearly, $g= f$ on the open set \n\\[ U = \\bigcup_n \\{x\\in X: \\frac{3r_{2n}}{4}< d(x,x_0) < \\frac{5r_{2n}}{4}\\}\\]\n and $g = 0$ on the open set \n\\[V = \\bigcup_n \\{x\\in X: \\frac{3r_{2n-1}}{4}< d(x,x_0) < \\frac{5r_{2n-1}}{4}\\}.\\]\nSince $x_n\\in U$ for all even $n$, and $x_n\\in V$ for all odd $n$, $x_0 \\in \\widetilde{U} \\cap \\widetilde{V}$.\n\\end{proof}\n\nWith the help of property (P), we can improve Theorems \\ref{t3.5}, \\ref{t3.7} and \\ref{t3.8}.\nFirst we consider the case where $A(X,E)$ and $A(Y,F)$ are standard subspaces of \n$U(X,E)$ and $U(Y,F)$ respectively so that both contain nonzero constant functions.\nDenote by $\\widetilde{E}$ the completion of $E$. Since $A(X,E) \\subseteq U(X,E)$, every function $f\\in A(X,E)$ has a unique continuous extension $\\widetilde{f}:\\widetilde{X}\\to \\widetilde{E}$. \nFor each $x\\in \\widetilde{X}$, let \n\\[ \\widetilde{E}_x = \\{\\widetilde{f}(x): f\\in A(X,E)\\}.\\]\nSimilarly for $\\widetilde{F}_y$ if $y\\in \\widetilde{Y}$.\nFix a biseparating map $T:A(X,E)\\to A(Y,F)$, which we may normalize by taking $T0 = 0$.\nLet $\\widetilde{\\varphi}: \\widetilde{X}\\to \\widetilde{Y}$ be the homeomorphism given by Theorem \\ref{t3.5}, with inverse $\\widetilde{\\psi}$.\n\n\\begin{prop}\\label{p4.2}\nSuppose that $A(X,E)$ and $A(Y,F)$ are standard subspaces of \n$U(X,E)$ and $U(Y,F)$ respectively so that both contain nonzero constant functions. Assume that $A(X,E)$ has property (P).\nGiven any $y\\in \\widetilde{Y}$, there is a bijective function $\\Phi(y,\\cdot): \\widetilde{E}_{\\widetilde{\\psi}(y)}\\to \\widetilde{F}_y$ so that \n\\[ \\widetilde{Tf}(y) = \\Phi(y,\\widetilde{f}(\\widetilde{\\psi}(y))) \\text{ for all $f\\in A(X,E)$}.\\]\n\\end{prop}\n\n\\begin{proof}\nLet $y_0 \\in \\widetilde{Y}$ and $\\widetilde{\\psi}(y_0) =x_0\\in \\widetilde{X}$. For any $a\\in \\widetilde{E}_{x_0}$, fix a function $g_a\\in A(X,E)$ so that $\\widetilde{g_a}(x_0) =a$ and define $\\Phi(y_0,\\cdot): \\widetilde{E}_{x_0} \\to F$ by $\\Phi(y_0,a) = Tg_a(y_0)$.\nIf $f\\in A(X,E)$, let $a = \\widetilde{f}(x_0)$. Clearly $\\widetilde{f- g_a}(x_0) =0$. By property (P) and the remark following its definition, there are open sets $U,V$ in $X$ and a function $h\\in A(X,E)$ so that $x_0 \\in \\widetilde{U} \\cap \\widetilde{V}$, $h = f-g_a$ on $U$ and $h =0$ on $V$.\nLet $W$ be an open set in $\\widetilde{X}$ so that $W\\cap X = U$.\nSince $\\widetilde{\\varphi}$ is a homeomorphism and $y_0 = \\widetilde{\\varphi}(x_0)$, \n$y_0 \\in \\widetilde{\\varphi}(\\widetilde{U}) \\subseteq \\widetilde{\\widetilde{\\varphi}(W)}$.\nBut $\\widetilde{\\varphi}(W)$ is open in $\\widetilde{Y}$. So $y_0 \\in ({\\varphi(W)}\\cap Y)^{\\widetilde{\\ }}$.\nAs $f= h+g_a$ on $W\\cap X$, $Tf = T(h+g_a)$ on $\\widetilde{\\varphi}(W)\\cap Y$ by Lemma \\ref{l3.2.1}.\nBy continuity, $\\widetilde{Tf}(y_0) = [T(h+g_a)]^{\\widetilde{ \\ }}(y_0)$. \nSimilarly, looking at the set $V$ instead of $U$, one can show that \n\\[ [T(h+g_a)]^{\\widetilde{ \\ }}(y_0) = {Tg_a}(y_0) = \\Phi(y_0,a). \\]\n Thus $\\widetilde{Tf}(y_0) = \\Phi(y_0,a)$, as required.\n \n By symmetry, there is a function $\\Psi(x, \\cdot): \\widetilde{F}_{\\widetilde{\\varphi}(x)}\\to \\widetilde{E}_x$ so that $(T^{-1}g)^{\\widetilde{\\ }}(x) = \\Phi(x, \\widetilde{g}(\\widetilde{\\varphi}(x)))$ for all $g\\in A(Y,F)$.\n The fact that $\\Phi(y,\\cdot)$ is a bijection follows from expressing the equations $T(T^{-1}g) = g$ and $T^{-1}(Tf) = f$ in terms of the mappings $\\Phi(y,\\cdot)$ and $\\Psi(x,\\cdot)$.\n\\end{proof}\n\n\nThe next two propositions can be obtained in a similar vein. The details are omitted.\n\n\n\\begin{prop}\\label{p4.3}\nSuppose that $A(X,E) = C(X,E)$ or $C_*(X,E)$ and $A(Y,F)$ $= C(Y,F)$ or $C_*(Y,F)$. \nThere is a function $\\Phi:Y\\times E\\to F$ so that \n\\[ Tf(y) = \\Phi(y,f(\\psi(y))) \\text{ for all $f\\in A(X,E)$ and all $y\\in Y$}.\\]\n\\end{prop}\n\n\\begin{prop}\\label{p4.4}\nSuppose that $A(X,E)$ is standard subspace of \n$U(X,E)$ that contains a nonzero constant function. Assume that $A(X,E)$ has property (P).\nLet $A(Y,F) = C(Y,F)$ or $C_*(Y,F)$.\n\\begin{enumerate}\n\\item For any $y\\in Y$, there is a function $\\Phi(y,\\cdot): \\widetilde{E}_{\\psi(y)}\\to F$ so that \n\\[Tf(y) = \\Phi(y,\\widetilde{f}(\\psi(y))) \\text{ for all $f\\in A(X,E)$}.\\]\n\\item There is a function $\\Psi: X\\times F\\to E$ so that \n\\[T^{-1}g(x) = \\Psi(x,g(\\varphi(x))) \\text{ for all $g\\in A(Y,F)$ and all $x \\in X$}.\\]\n\\end{enumerate}\n\\end{prop}\n\n\n\n\n\n\n\n\n\n\\section{Spaces of continuous functions -- metric case}\\label{s5}\n\nIn this section, let $X, Y$ be metric spaces and $E,F$ be normed spaces.\nLet $A(X,E) = C(X,E)$ or $C_*(X,E)$ and $A(Y,F) = C(Y,F)$ or $C_*(Y,F)$.\nFix a biseparating map $T:A(X,E)\\to A(Y,F)$. \nBy Theorem \\ref{t3.7}, Proposition \\ref{p4.3} and symmetry, there are a homeomorphism $\\varphi:X\\to Y$ and functions $\\Phi:Y\\times E\\to F$, $\\Psi:X\\times F\\to E$ so that \n\\[ (Tf)(y)= \\Phi(y,f(\\varphi^{-1}(y))),\\quad (T^{-1}g)(x) = \\Psi(x,g(\\varphi(x)))\\]\nfor any $f\\in A(X,E)$, $g\\in A(Y,F)$, $x\\in X$ and $y\\in Y$.\nFrom the equations \n\\[ 1\\otimes a = T^{-1}(T(1\\otimes a)),\\quad 1\\otimes b = T(T^{-1}(1\\otimes b))\\]\nfor all $a\\in E$, $b\\in F$, we find that $\\Phi(y,\\cdot)$ and $\\Psi(x,\\cdot)$ are mutual inverses provided $y = \\varphi(x)$. \nThe aim of the present section is to characterize the functions $\\Phi$ that lead to biseparating maps and prove a result on automatic continuity.\nObserve that if we define $S:C(Y,F)\\to C(X,F)$ by $Sg(x) = g(\\varphi^{-1}(x))$, then $S$ is a biseparating map that also acts as a biseparating map from $C_*(Y,F)$ onto $C_*(X,F)$.\nThus characterization of biseparating maps reduces to the ``section problem'' addressed in Proposition \\ref{p5.1}.\nThe result is well known, at least in the case for for $C(X)$. See, e.g. \\cite[Chapter 9]{AZ}.\nWe omit the easy proof of the next lemma.\n\n\\begin{lem}\\label{l5.0}\nLet $(x_n)$ be a sequence of distinct points in $X$ and let $(a_n)$ be a sequence in $E$.\n\\begin{enumerate}\n\\item If $(x_n)$ has no convergent subsequence, then there is a function $f\\in C(X,E)$ so that $f(x_n) = a_n$ for all $n$. Moreover, $f$ can be chosen to be bounded if $(a_n)$ is bounded.\n\\item If $(x_n)$ converges to a point $x_0\\in X$, $x_0\\neq x_n$ for all $n$, and $(a_n)$ converges to a point $a_0\\in E$, then there exists $f\\in C_*(X,E)$ so that $f(x_n) = a_n$ for all $n$.\n\\end{enumerate}\n\\end{lem}\n\nDenote the set of accumulation points of $X$ by $X'$ and the unit balls of $E$ and $F$ by $B_E$ and $B_F$ respectively.\n\n\n\n\n\\begin{prop}\\label{p5.1}\nLet $X$ be a metric space and let $E$ and $F$ be normed spaces. Consider a function $\\Phi:X\\times E\\to F$.\n\\begin{enumerate}\n\\item The function $x\\mapsto \\Phi(x,f(x))$ belongs to $C(X,F)$ for every $f\\in C(X,E)$ if and only if $\\Phi$ is continuous at every point in $X'\\times E$.\n\\item The function $x\\mapsto \\Phi(x,f(x))$ belongs to $C_*(X,F)$ for every $f\\in C_*(X,E)$ if and only if both of the following conditions hold.\n\\begin{enumerate}\n\\item $\\Phi$ is continuous at every point in $X'\\times E$. \n\\item For any bounded set $B$ in $E$, every $(x_n) \\in \\prod_n X_n(B)$ has a subsequence that converges in $X$, where \n\\[X_n(B) = \\{x\\in X: \\Phi(x,B) \\not\\subseteq nB_F\\}.\\]\n\\end{enumerate}\n\\end{enumerate}\n\\end{prop}\n\n\n\\begin{proof}\nSuppose that $x\\mapsto \\Phi(x,f(x))$ belongs to $C(X,F)$ for every $f\\in C_*(X,E)$.\nLet $(x_0,a_0)$ be a point in $X'\\times E$ and let $((x_n,a_n))$ be a sequence in $X\\times E$ that converges to $(x_0,a_0)$.\nSince $\\Phi(x,a_n)$ is a continuous function of $x$, by making small perturbations if necessary, we may assume that $x_n \\neq x_0$ for all $n$.\nChanging to a subsequence, we may further assume that $(x_n)$ is a sequence of distinct points. By Lemma \\ref{l5.0}(2), there exists $f\\in C_*(X,E)$ so that $f(x_n) = a_n$ for all $n$. By continuity of $f$, $f(x_0) = a_0$.\nThen \n\\[ \\Phi(x_n,a_n) = \\Phi(x_n,f(x_n)) \\to \\Phi(x_0,f(x_0)) = \\Phi(x_0,a_0).\\]\nThis proves the continuity of $\\Phi$ at $(x_0,a_0)$.\nHence the ``only if'' parts in statement (1) and statement (2)(a) are verified.\nOn the other hand,\nif $\\Phi$ is continuous at any point in $X'\\times E$ and $f\\in C(X,E)$, then it is clear that $\\Phi(x,f(x))$ is a continuous function of $x\\in E$. This completes the proof of statement (1).\n\nLet us proceed to prove the necessity of condition (b) in statement (2).\nAssume that condition (b) in (2) fails. \nLet $B$ be a bounded set in $E$ and let $(x_n)$ be an element in $\\prod X_n(B)$ so that $(x_n)$ has no convergent subsequence in $X$.\nSince $X_n(B) \\subseteq X_m(B)$ if $m \\leq n$, we may replace $(x_n)$ by a subsequence to assume that all $x_n$'s are distinct.\nChoose $a_n\\in B$ so that $\\|\\Phi(x_n,a_n)\\| > n$ for all $n$.\nBy Lemma \\ref{l5.0}(1), there exists $f\\in C_*(X,E)$ so that $f(x_n) = a_n$ for all $n$.\nThen $\\|\\Phi(x_n,f(x_n))\\| = \\|\\Phi(x_n,a_n)\\| \\to \\infty$.\nThis contradicts the assumption that the function $x\\mapsto \\Phi(x,f(x))$ is bounded.\n\n\nFinally, we prove the sufficiency in statement (2). Let $f\\in C_*(X,E)$. As observed above, by (2) condition (a), $x\\mapsto \\Phi(x,f(x))$ is continuous on $X$ since $f\\in C(X,E)$.\nLet $B = f(X)$. Then $B$ is a bounded set in $E$. If $(\\Phi(x,f(x)))_{x\\in X}$ is unbounded, there is a sequence of distinct points $(x_n)$ so that $\\|\\Phi(x_n,f(x_n))\\| > n$ for all $n$.\nIn particular, $(x_n) \\in \\prod X_n(B)$. By condition 2(b), we may replace it by a subsequence to assume that $(x_n)$ converges to a point $x_0$ in $X$. In particular, $x_0\\in X'$. By assumption 2(a),\n$\\Phi(x_n, f(x_n))\\to \\Phi(x_0,f(x_0))$, contradicting the unboundedness of the sequence.\n\\end{proof}\n\nThe next two results follow immediately from the preceding discussion.\n\n\n\\begin{thm}\\label{t5.4}\nLet $X$ and $Y$ be metric spaces and let $E$ and $F$ be normed spaces.\nA map $T:C(X,E)\\to C(Y,F)$ is a biseparating map if and only if there are a homeomorphism $\\varphi:X\\to Y$ and functions $\\Phi:Y\\times E\\to F$, $\\Psi:X\\times F\\to E$ so that\n\\begin{enumerate}\n\\item $\\Phi(y,\\cdot)$ and $\\Psi(x,\\cdot)$ are mutual inverses if $\\varphi(x) = y$. \n\\item $\\Phi$ is continuous at any point in $Y'\\times E$; $\\Psi$ is continuous at any point in $X'\\times F$.\n\\end{enumerate}\n\\end{thm}\n\n\n\n\\begin{thm}\\label{t5.5}\nLet $X$ and $Y$ be metric spaces and let $E$ and $F$ be normed spaces.\nA map $T:C_*(X,E)\\to C_*(Y,F)$ is a biseparating map if and only if there are a homeomorphism $\\varphi:X\\to Y$ and functions $\\Phi:Y\\times E\\to F$, $\\Psi:X\\times F\\to E$ so that\n\\begin{enumerate}\n\\item $\\Phi(y,\\cdot)$ and $\\Psi(x,\\cdot)$ are mutual inverses if $\\varphi(x) = y$. \n\\item $\\Phi$ is continuous at any point in $Y'\\times E$; $\\Psi$ is continuous at any point in $X'\\times F$.\n\\item If $B_1$ and $B_2$ are bounded sets in $E$ and $F$ respectively, and \n\\[Z_n(B_1,B_2) = \\{x\\in X: \\Phi(\\varphi(x),B_1) \\not\\subseteq nB_F \\text{ or } B_2 \\not\\subseteq \\Phi(\\varphi(x),nB_E)\\},\\]\nthen every $(x_n) \\in \\prod_n Z_n(B_1,B_2)$ has a subsequence that converges in $X$.\n\\end{enumerate}\n\\end{thm}\n\nObserve that by condition (1) of Theorem \\ref{t5.5}, $B_2 \\not\\subseteq \\Phi(\\varphi(x),nB_E)$ if and only if $\\Psi(x,B_2)\\not\\subseteq nB_E$. Hence condition (3) in Theorem \\ref{t5.5} is a combination of condition 2(b) in Proposition \\ref{p5.1} for the maps $\\Phi$ and $\\Psi$.\n\n\\begin{prop}\\label{p5.2}\nIf there is a biseparating map $T:C(X,E)\\to C_*(Y,F)$, then $X$ are $Y$ are compact.\n\\end{prop}\n\n\\begin{proof}\nAssume otherwise. Since $X$ and $Y$ are homeomorphic,\n there is a sequence of distinct points $(x_n)$ in $X$ that has no convergent subsequence. Fix a nonzero element $b\\in F$ and let $a_n = (T^{-1}(1\\otimes nb))(x_n)$ for each $n$.\nBy Lemma \\ref{l5.0}(1), there is a function $f\\in C(X,E)$ so that $f(x_n) =a_n$ for all $n$.\nNote that \n\\[nb = T((T^{-1}(1\\otimes nb))(\\varphi(x_n)) = \\Phi(\\varphi(x_n),a_n) \\text{ for each $n$.}\\]\nThen \n\\[(Tf)(\\varphi(x_n)) = \\Phi(\\varphi(x_n),a_n) = nb \\text{ for all $n$},\\]\ncontradicting the boundedness of $Tf$.\n\\end{proof}\n\n\n\nNote that if $T:C(X,E)\\to C_*(Y,F)$ is biseparating, then $X$ is compact by Proposition \\ref{p5.2}. Hence $C(X,E) = C_*(X,E)$. Therefore, the characterization Theorem \\ref{t5.5} applies.\nWe conclude this section with an automatic continuity result. If $K$ is a compact subset of $X$, let \n\\[\\|f\\|_K = \\sup\\{\\|f(x)\\|:x\\in K\\} \\text{ for any $f\\in C(X,E)$}.\\]\n\n\n\\begin{thm}\\label{t5.6}\nLet $X$ and $Y$ be metric spaces and let $E$ and $F$ be normed spaces.\nSuppose that $A(X,E) = C(X,E)$ or $C_*(X,E)$, $A(Y,F) = C(Y,F)$ or $C_*(Y,F)$. \nLet $T:A(X,E)\\to A(Y,F)$ be a biseparating map. For any compact subset $K$ of $Y'$, any $f\\in A(X,E)$, and any $\\varepsilon > 0$, there exists $\\delta >0$ so that \n\\[ g\\in A(X,E),\\ \\|g-f\\|_{\\varphi^{-1}(K)} <\\delta \\implies \\|Tg-Tf\\|_K < \\varepsilon.\\]\n\\end{thm}\n\n\\begin{proof}\nSuppose that $(g_n) \\subseteq A(X,E)$ and that $\\|g_n-f\\|_{\\varphi^{-1}(K)} \\to 0$. It suffices to show that a subsequence of $(\\|Tg_n-Tf\\|)$ converges to $0$.\nPick $(y_n) \\subseteq K$ so that $\\|Tg_n-Tf\\|_K = \\|(Tg_n)(y_n) - (Tf)(y_n)\\|$ for all $n$.\nBy using a subsequence if necessary, we may assume that $(y_n)$ converges to some $y_0\\in K$.\nLet $x_n = \\varphi^{-1}(y_n)$, \n$g_n(x_n) = a_n$ , and $f(x_n) = a'_n$. Then $(x_n)$ converges to $x_0 = \\varphi^{-1}(y_0)$ and $f(x_0) = a_0$.\nSince $\\|g_n-f\\|_K \\to 0$, $(a_n)$ converges to $a_0$. \nAlso $(a_n')$ converges to $a_0$ by continuity of $f$.\nSince $y'\\in K\\subseteq Y'$, it follows fromm Proposition \\ref{p5.1}(1) that $\\Phi$ is continuous at $(y_0,a_0)$.\nTherefore,\n\\[ \n\\|(Tg_n)(y_n) - (Tf)(y_n)\\| = \\|\\Phi(y_n,a_n)- \\Phi(y_n,a'_n)\\|\\to 0.\\]\nThus $\\|Tg_n-Tf\\|_K\\to 0$.\n\\end{proof}\n\n\\section{Spaces of uniformly continuous functions}\\label{s6}\n\nIn this section, let $X$, $Y$ be complete metric spaces and let $E$, $F$ be Banach spaces.\nThe aim of this section is to characterize biseparating maps from $U(X,E)$ or $U_*(X,E)$ onto $U(Y,F)$ or $U_*(Y,F)$. By Propositions \\ref{p3.10} and \\ref{p4.2}, a biseparating map $T: U(X,E)\/U_*(X,E) \\to U(Y,F)\/U_*(Y,F)$\ncan be represented in the form \n\\[ Tf(y) = \\Phi(y,f(\\psi(y))) \\text{ for all $f\\in U(X,E)\/U_*(X,E)$ and all $y\\in Y$},\\]\nwhere $\\psi:Y\\to X$ is a homeomorphism with inverse $\\varphi$ and $\\Phi:Y\\times E\\to F$ is a function so that $\\Phi(y,\\cdot)$ is a bijection from $E$ onto $F$ for all $y\\in Y$.\nIn fact, characterizations can be obtained without completeness assumptions of $X,Y,E,F$. However, the case of complete spaces contains all pertinent ideas without the distraction of niggling details.\nCharacterizations of lattice isomorphisms and of {\\em linear} biseparating maps on spaces of uniformly continuous functions were obtained in \\cite{GaJ} and \\cite{A2} respectively.\n\n\n\n\n\n\n\n\n\n\\begin{prop}\\label{p6.3.0}\nLet $A(X,E)= U(X,E)$ or $U_*(X,E)$, and $A(Y,F)= U(Y,F), U_*(Y,F)$ or $\\operatorname{Lip}_*(Y,F)$.\nLet ${\\varphi}: {X}\\to {Y}$ be the homeomorphism associated with $T$ according to Theorem \\ref{t3.5}.\nThen ${\\varphi}$ is uniformly continuous.\n\\end{prop}\n\n\\begin{proof}\nSuppose that $\\varphi$ is not uniformly continuous. There are sequences $(x_n)$, $(x_n')$ in $X$ and $\\varepsilon >0$ so that\n$d(x_n,x'_n)\\to 0$ and that $d(\\varphi(x_n),\\varphi(x_n')) > \\varepsilon$ for all $n$.\nSet $y_n = \\varphi(x_n)$ and $y_n' = \\varphi(x_n')$.\nIn view of the continuity of ${\\varphi}$, niether $(x_n)$ nor $(x'_n)$ can have a convergent subsequence in ${X}$.\nHence we may assume that $(x_n)$ is a separated sequence.\nSince ${\\varphi}^{-1}$ is continuous, neither $(y_n)$ nor $(y'_n)$ can have a convergent subsequence in ${Y}$.\nAs we also have $d(y_n,y_n') > \\varepsilon$ for all $n$, by using subsequences if necessary, we may assume that $(y_n) \\cup (y_n')$ is a separated set.\nWithout loss of generality, take $T0 = 0$. We will use repeatedly the following formulation of Proposition \\ref{p4.2}. If $x\\in {X}$ and $f, g\\in A(X,E)$, then ${f}(x) = {g}(x)$ if and only if ${Tf}(\\varphi(x)) = {Tg}(\\varphi(x)).$\n\n\\medskip\n\n\\noindent\\underline{Case 1}. $A(Y,F) = U_*(Y,F)$ or $\\operatorname{Lip}_*(Y,F)$.\n\nFix a nonzero vector $a\\in E$ and let $b_n = {(T(1\\otimes a))}(y_n)$. Then $(b_n)$ is a bounded sequence.\nSince $(y_n)\\cup (y_n')$ is separated, one can easily construct a function $g\\in A(Y,F)$ so that ${g}(y_n) = b_n$ and ${g}(y'_n) = 0$ for all $n$. By Proposition \\ref{p4.2}, we see that $(T^{-1}g)(x_n) = a$ and $(T^{-1}g)(x'_n) = 0$. Since $T^{-1}g$ is uniformly continuous and $d(x_n,x'_n)\\to 0$, we have a contradiction.\n\n\\medskip\n\nIn the remaining cases, take $A(Y,F) = U(Y,F)$.\n\n\\medskip\n\n\\noindent\\underline{Case 2}. There exist $r>0$ and an infinite subset $N$ of ${\\mathbb N}$ so that $B(y_n,r) = \\{y_n\\}$ for all $n\\in N$.\n\nFix a nonzero vector $a \\in E$. Let $b_n = (T(1\\otimes a))(y_n)$ for each $n\\in N$. Define $g:Y \\to F$ by $g(y) = b_n$ if $y = y_n, n\\in N$, and $g(y) = 0$ otherwise.\nClearly $g\\in U(Y,F)$. \nBut by Proposition \\ref{p4.2}, for all $n\\in N$,\n\\[ (T^{-1}g)(x_n) = a \\text{ and } (T^{-1}g)(x_n') = 0.\\]\nHence $T^{-1}g$ is not uniformly continuous, contrary to the fact that $T^{-1}g \\in A(X,E) \\subseteq U(X,E)$.\n\n\\medskip\n\n\\noindent\\underline{Case 3}. For all $r>0$, $B(y_n,r) = \\{y_n\\}$ occurs for only finitely many $n$.\n\nIn this case, by using a subsequence if necessary, we may assume that there is a sequence $(y_n'')$ in $Y$ so that $0 < d(y_n,y_n'') \\to 0$.\nSet $x_n'' = {\\varphi}^{-1}(y_n'')$ for all $n$.\nTake a nonzero element $b\\in F$ and let \n$a_n =(T^{-1}(1\\otimes b))(x_n)$.\nSince $(y_n)\\cup(y_n')$ is separate\n, we can find $g\\in U(Y,F)$ so that \n ${g}(y_n) = b$ and ${g}(y_n') = 0$ for all $n$.\nBy Proposition \\ref{p4.2},\n \\[ (T^{-1}g)(x_n) = a_n \\text{ and } (T^{-1}g)(x'_n) = 0.\\]\nSince $T^{-1}g$ is uniformly continuous and $d(x_n,x_n') \\to 0$,\n\\[ a_n = (T^{-1}g)(x_n) - (T^{-1}g)(x_n') \\to 0.\n\\]\nAs $(x_n)$ is separated, $x_n'' \\neq x_n$ and $(a_n)$ is a null sequence, we may, after replacing $(x_n)$ and $(x_n'')$ with subsequences, construct a function $f\\in U_*(X,E)\\subseteq A(X,E)$ so that $f(x_n) = a_n$ and $f(x_n'') = 0$ for all $n$.\nThen ${Tf}(y_n) = {g}(y_n) = b$ and $Tf(y_n'') = 0$.\nThis is impossible since ${Tf}$ is uniformly continuous and $d(y_n,y_n'')\\to 0$.\n\\end{proof}\n\nBy Proposition \\ref{p6.3.0}, if $T: U(X,E)\/U_*(X,E) \\to U(Y,F)\/U_*(Y,F)$ is a biseparating map, then $\\varphi:X\\to Y$ is a uniform homeomorphism. \nIn this case, the map $\\widehat{T}$ given by \n\\[\\widehat{T}f(x) = Tf(\\varphi(x)) = \\Phi(\\varphi(x),f(x)) = \\widehat{\\Phi}(x,f(x))\\]\nmaps $U(X,E)\/U_*(X,E)$ onto $U(X,F)\/U_*(X,F)$, with $\\widehat{\\Phi}:X\\times E\\to F$ being a function such that $\\widehat{\\Phi}(x,\\cdot):E\\to F$ is a bijection for each $x\\in X$.\nTo complete the characterization of $T$, it suffices to determine the functions $\\widehat{\\Phi}: X\\times E\\to F$ so that $x\\mapsto \\widehat{\\Phi}(x,f(x))$ belongs to $U(X,F)\/U_*(X,F)$ for each $f\\in U(X,E)\/U_*(X,E)$.\nWe will refer to this as the ``section problem'' for uniformly continuous functions.\n\nFor any $\\varepsilon > 0$, define $d_\\varepsilon:\\widetilde{X}\\times \\widetilde{X}\\to [0,\\infty]$ by\n\\begin{equation}\\label{eq6.0} d_\\varepsilon(a,b) = \\inf\\{\\sum^n_{i=1}d(x_{i-1},x_i): n\\in {\\mathbb N}, x_0 = a, x_n = b, d(x_{i-1},x_i) \\leq \\varepsilon \\text{ for all $i$}\\},\\end{equation}\nwhere we take $\\inf \\emptyset = \\infty$.\nThe connection of the $d_\\varepsilon$ ``metrics'' with uniformly continuous functions is well known; see, e.g. \\cite{A,H, O'F}.\nIn particular, the first part of the next proposition formalizes the well known principle that uniformly continuous functions are ``Lipschitz for large distances''.\n\n\\begin{prop}\\label{p6.3}\nLet $X$ be a complete metric space and let $E$ be a Banach space.\n\\begin{enumerate}\n\\item If $f\\in U(X,E)$, then there exist $\\varepsilon > 0$ and $C<\\infty$ such that \n\\[ \\|{f}(x_1)-{f}(x_2)\\| \\leq Cd_\\varepsilon(x_1,x_2) \\text{ whenever $x_1,x_2\\in {X}$, $d(x_1,x_2) > \\varepsilon$}.\\]\n\\item If $f:X\\to E$ and there exist $\\varepsilon >0$ and $C<\\infty$ so that \n\\[ \\|f(x_1) - f(x_2)\\| \\leq Cd_\\varepsilon(x_1,x_2) \\text{ for all $x_1,x_2\\in X$},\\]\nthen $f\\in U(X,E)$.\n\\end{enumerate}\n\\end{prop}\n\n\\begin{proof}\nStatement (2) is trivial since $d_\\varepsilon(x_1,x_2) = d(x_1,x_2)$ if $d(x_1,x_2) \\leq \\varepsilon$. Let us prove statement (1).\nAssume that $f\\in U(X,E)$. There exists $\\varepsilon >0$ so that $\\|f(a)-f(b)\\| \\leq 1$ if $d(a,b) \\leq \\varepsilon$.\nLet $x_1,x_2\\in X$ be points so that $\\varepsilon < d(x_1,x_2)$ and $d_\\varepsilon(x_1,x_2) <\\infty$.\nThere are $n\\in {\\mathbb N}$, $(a_i)^n_{i=0} \\subseteq X$ so that $a_0 = x_1$, $a_n = x_2$, $d(a_{i-1}, a_i) \\leq\\varepsilon$, and \n\\[ \\sum^n_{i=1}d(a_{i-1}, a_i) \\leq 2d_\\varepsilon(x_1,x_2).\\]\nNote that since $d(x_1,x_2) > \\varepsilon$, $n \\geq 2$.\nIt is clear that we may assume that $d(a_{i-1},a_i) + d(a_i,a_{i+1}) >\\varepsilon$ for $0\\leq i< n$.\nThus \n\\[ 2d_\\varepsilon(x_1,x_2) \\geq \\sum^n_{i=1}d(a_{i-1}, a_i) \\geq \\frac{n-1}{2}\\,\\varepsilon \\geq \\frac{n\\varepsilon}{4}.\\]\nBy choice of $\\varepsilon$, $\\|f(a_{i-1}) - f(a_i)\\| \\leq 1$ for all $i$.\nHence \n\\[ \\|f(x_1)-f(x_2)\\| \\leq n \\leq \\frac{8}{\\varepsilon}\\,d_\\varepsilon(x_1,x_2).\\]\n\\end{proof}\n\n\n\nFor the rest of the section, let $\\Xi:X\\times E\\to F$ be a given function and associate with it a mapping $Sf(x) = \\Xi(x,f(x))$ for any function $f:X\\to E$.\nDenote the set of accumulation points in $X$ by $X'$.\n\n\\begin{prop}\\label{p6.4}\nIf $Sf \\in U(X,F)$ for any $f\\in U_*(X,E)$, then $\\Xi$ is continuous at any $(x_0,e_0)\\in X'\\times E$ .\n\\end{prop}\n\n\\begin{proof}\nAssume to the contrary that $\\Xi$ is discontinuous at some $(x_0,e_0)\\in X'\\times E$ .\nThere are a sequence $((x_n,e_n))^\\infty_{n=1} \\in X\\times E$ converging to $(x_0,e_0)$ and $\\varepsilon>0$\nso that $\\|\\Xi(x_n,e_n) - \\Xi(x_0,e_0)\\| > \\varepsilon$ for all $n$.\nReplacing $(x_n)$ by a subsequence if necessary, we may assume that either $(x_n)$ is a sequence of distinct points in $X\\backslash \\{x_0\\}$ or $x_n = x_0$ for all $n$. \nIn the former case, there is a function $f\\in U_*(X,E)$ so that $f(x_n) = e_n$ for all $n$ and $f(x_0) = e_0$.\nSince $Sf$ is continuous at $x_0$,\n\\[ \\Xi(x_n,e_n) = Sf(x_n) \\to Sf(x_0) = \\Xi(x_0,e_0),\\]\ncontrary to the choice of $((x_n,e_n))^\\infty_{n=1}$.\nFinally, suppose that $x_n = x_0$ for all $n$.\nFor each $n$, let $f_n$ be the constant function with value $e_n$. Then $Sf_n$ is continuous at $x_0$. Since $x_0$ is an accumulation point, there exists $x'_n$ with $0< d(x'_n,x_0) < \\frac{1}{n}$ so that \n\\[ \\|\\Xi(x'_n,e_n) - \\Xi(x_0,e_n)\\| = \\|Sf_n(x'_n) - Sf_n(x_0)\\| < \\frac{1}{n}.\\]\nBut by the previous case, $\\Xi(x'_n,e_n) \\to \\Xi(x_0,e_0)$. Thus \n$\\Xi(x_0,e_n) \\to \\Xi(x_0,e_0)$.\n\\end{proof}\n\n\nCall a sequence $((x_n,e_n))^\\infty_{n=1}\\in X\\times E$ a $u$-sequence if $(x_n)$ is a separated sequence and there are $\\varepsilon > 0$, $C<\\infty$ so that \n\\[ \\|e_n-e_m\\| \\leq Cd_\\varepsilon(x_n,x_m) \\text{ for all $m,n\\in{\\mathbb N}$}.\\]\nThe importance of $u$-sequences is captured in the next lemma.\n\n\\begin{lem}\\label{l6.4}\nLet $((x_n,e_n))^\\infty_{n=1}$ be a $u$-sequence in $X\\times E$. Then there is an infinite subset $N$ of ${\\mathbb N}$ and a uniformly continuous function $f:X\\to E$ so that $f(x_n) =e_n$ for all $n \\in N$.\n\\end{lem}\n\n\\begin{proof}\nLet $\\varepsilon$ and $C$ be as in the definition above.\nIf there is an infinite set $N$ in ${\\mathbb N}$ so that $d_\\varepsilon(x_n,x_m) = \\infty$ for all distinct $m,n\\in N$, then clearly the function defined by $f(x) = e_n$ if $d_{\\varepsilon}(x,x_n) <\\infty$ for some $n\\in N$ and $f(x) = 0$ otherwise is uniformly continuous. Obviously $f(x_n) =e_n$ for all $n\\in N$.\nThus, without loss of generality, we may assume that $d_\\varepsilon(x_m,x_n)<\\infty$ for all $m,n$.\nIf $(d_\\varepsilon(x_n,x_m))_{m,n}$ is bounded, then $(e_n)$ is a bounded sequence. Since $(x_n)$ is separated, there exists $f\\in U(X,E)$ so that $f(x_n) = e_n$ for all $n$.\n\nFinally, assume that $(d_\\varepsilon(x_n,x_m))_{m,n}$ is unbounded set in ${\\mathbb R}_+$.\nBy taking a subsequence, we may assume that $4r_n < r_{n+1}$ for all $n$, where $r_n = d_\\varepsilon(x_n,x_1)$.\nDefine $f:X\\to E$ by \n\\[ f(x) = \\begin{cases}\n e_1 + (1- \\frac{2}{r_n}d_\\varepsilon(x,x_n))^+(e_n-e_1) &\\text{if $d_\\varepsilon(x,x_n)< \\frac{r_n}{2}$, $n > 1$}\\\\\n e_1 &\\text{otherwise}.\n \\end{cases}\\]\nUsing the fact that $\\|e_n-e_1\\|\\leq Cr_n$ for all $n$, one can check that \n\\[ \\|f(a) - f(b)\\| \\leq 16Cd_\\varepsilon(a,b)\n\\]\nfor all $a,b\\in X$. By Proposition \\ref{p6.3}(2), $f\\in U(X,E)$. Clearly, $f(x_n) = e_n$ for all $n$, as required.\n\\end{proof}\n\n\\begin{prop}\\label{p6.5}\nSuppose that $Sf\\in U(X,F)$ for all $f\\in U(X,E)$.\nLet $((x_n,e_n))^\\infty_{n=1}$ be a $u$-sequence. \nAssume that $((x_n',e_n'))^\\infty_{n=1} \\in X\\times E$, $x_n\\neq x_n'$ for all $n$, and $\\lim (d(x_n,x_n') + \\|e_n-e_n'\\|) =0$, then\n\\[ \\lim\\|\\Xi(x_n,e_n) - \\Xi(x_n',e'_n)\\| = 0.\\]\n\\end{prop}\n\n\\begin{proof}\n It suffices to show that $\\liminf\\|\\Xi(x_n,e_n) - \\Xi(x_n',e'_n)\\| = 0$. By Lemma \\ref{l6.4}, there exist $f\\in U(X,E)$ and an infinite set $N$ in ${\\mathbb N}$ so that $f(x_n) = e_n$ for all $n\\in N$.\nSince $d(x_n,x_n') \\to 0$, $\\lim_{n\\in N}\\|e_n -f(x_n')\\| = 0$ and hence $\\lim_{n\\in N}\\|e_n'-f(x_n')\\| = 0$.\nAs $(x_n)$ is a separated sequence and $0 < d(x_n,x_n') \\to 0$, we can construct a uniformly continuous function $g:X\\to E$ \nsuch that $g(x_n) = 0$ and $g(x_n') = e_n'-f(x_n')$ for all sufficiently large $n\\in N$.\nThen $f+g\\in U(X,E)$, \n\\[ \\Xi(x_n,e_n) = S(f+g)(x_n) \\text{ and } \\Xi(x_n',e_n') = S(f+g)(x_n')\\] for all sufficiently large $n\\in N$.\nAs $S(f+g)\\in U(X,F)$ and $d(x_n,x_n') \\to 0$, we see that $\\lim_{n\\in N}\\|\\Xi(x_n,e_n) - \\Xi(x_n',e_n')\\| =0$.\n\\end{proof}\n\n\nWe will say that $\\Xi$ is {\\em $u$-continuous} if it satisfies the conclusion of Proposition \\ref{p6.5}.\nWe can now solve the section problem for uniformly continuous functions.\n\n\\begin{thm}\\label{t6.5}\nLet $X$ be a complete metric space, $E$ and $F$ be Banach spaces. Given a function $\\Xi:X\\times E\\to F$, associate with it a mapping $S$ by $Sf(x) = \\Xi(x,f(x))$.\nThen $S$ maps $U(X,E)$ into $U(X,F)$ if and only if \n $\\Xi$ is continuous at all $(x_0,e_0) \\in X'\\times E$ and $\\Xi$ is \n $u$-continuous.\n\\end{thm}\n\n\\begin{proof}\nThe necessity of the two conditions on $\\Xi$ follow from Propositions \\ref{p6.4} and \\ref{p6.5}.\nConversely, suppose that $\\Xi$ is continuous at any $(x_0,e_0)\\in X'\\times E$ and also $u$-continuous.\nLet $f\\in U(X,E)$. \nIf $Sf\\notin U(X,F)$, there are sequences $(x_n)$, $(x_n')$ in $X$, $d(x_n,x'_n)\\to 0$, and $\\eta >0$ so that\n\\begin{equation}\\label{eq6.1} \\|\\Xi(x_n,f(x_n)) - \\Xi(x_n',f(x_n'))\\| = \\|Sf(x_n) - Sf(x_n')\\| > \\eta \\text{ for all $n$}.\\end{equation}\nSuppose that $(x_n)$ has a subsequence that converges to some $x_0\\in X$.\nWe may assume that the whole sequence converges to $x_0$. In particular, $(f(x_n))$ and $(f(x_n'))$ converge to $f(x_0)$.\nClearly, $x_n\\neq x'_n$ for all $n$. Hence $x_0\\in X'$.\nIn this case, (\\ref{eq6.1}) violates the continuity of $\\Xi$ at $(x_0,f(x_0))$.\nFinally, assume that $(x_n)$ is a separated sequence. Choose $\\varepsilon' > 0$ so that $d(x_m,x_n) > \\varepsilon'$ if $m\\neq n$. Then let $\\varepsilon$ and $C$ be as given in condition (1) of Proposition \\ref{p6.3} for the function $f$.\nObviously, $d_\\varepsilon \\leq d_{\\varepsilon \\wedge \\varepsilon'}$. So we may assume without loss of generality that $\\varepsilon \\leq\\varepsilon'$.\nHence $(x_n,f(x_n))^\\infty_{n=1}$ is a $u$-sequence by Proposition \\ref{p6.3}(1).\nAgain, (\\ref{eq6.1}) implies that $x_n\\neq x_n'$ for all $n$.\nFurthermore, $d(x_n,x'_n)\\to 0$ and $\\|f(x_n)-f(x_n)\\| \\to 0$, the latter as a result of the uniform continuity of $f$. Therefore, \n\\[ \\lim\\|\\Xi(x_n,f(x_n)) - \\Xi(x_n',f(x'_n))\\| =0\\]\nby $u$-continuity of $\\Xi$, contradicting (\\ref{eq6.1}).\n\\end{proof}\n\nCharacterization of biseparating maps from $U(X,E)$ onto $U(Y,F)$ can be obtained by using Theorem \\ref{t6.5} together with the ``switch'' from $Y$ to $X$ described prior to Proposition \\ref{p6.3}.\n\n\\begin{thm}\\label{t6.7.1}\nLet $X, Y$ be complete metric spaces and let $E,F$ be Banach spaces.\nSuppose that $T:U(X,E)\\to U(Y,F)$ is a biseparating map. \nThen there are a uniform homeomorphism ${\\varphi}:{X}\\to {Y}$ and a function $\\Phi:Y\\times E\\to F$ so that \n\\begin{enumerate}\n\\item For each $y\\in Y$, $\\Phi(y,\\cdot):E\\to F$ is a bijection with inverse $\\Psi(x,\\cdot):F\\to E$, where $\\varphi(x) = y$.\n\\item $Tf(y) = \\Phi(y,f(\\varphi^{-1}(y)))$ and $T^{-1}g(x) = \\Psi(x,g(\\varphi(x)))$ for all $f\\in U(X,E), g\\in U(Y,F)$ and $x\\in X$, $y\\in Y$. \n\\item $\\Phi$ is continuous on $Y'\\times E$ and $\\Psi$ is continuous on $X'\\times F$.\n\\item $(x,e)\\mapsto \\Phi(\\varphi(x),e))$ and $(y,e')\\mapsto \\Psi(\\varphi^{-1}(y),e'))$ are both $u$-continuous.\n\\end{enumerate}\nConversely, assume that $\\varphi,\\Phi$ satisfy conditions (1), (3) and (4). Define $Tf(y)$ as in (2) for any $f\\in U(X,E)$ and $y\\in Y$. Then $T$ is a biseparating map from $U(X,E)$ onto $U(Y,F)$.\n\\end{thm}\n\n\n\n\\begin{lem}\\label{l6.8}\nLet $\\Xi:X\\times E\\to F$ be a given function and associate with it a mapping $Sf(x) = \\Xi(x,f(x))$ for any function $f:X\\to E$. If $Sf\\in U_*(X,F)$ for any $f \\in U_*(X,E)$, then for any separated sequence $(x_n)$ in $X$ and any bounded set $B$ in $E$, there is exists $k\\in{\\mathbb N}$ so that \n$\\bigcup_{n=k}^\\infty\\Xi(x_n,B)$ is bounded in $F$.\n\\end{lem}\n\n\\begin{proof}\nSuppose that $Sf \\in U_*(X,F)$ for any $f\\in U_*(X,E)$.\nLet $(x_n)$ be a separated sequence in $X$ and let $B$ be a bounded set in $E$.\nAssume that for any $k\\in {\\mathbb N}$, $\\bigcup_{n=k}^\\infty\\Xi(x_n,B)$\nis unbounded.\nThen there exists $(e_n)$ in $B$ so that $(\\Xi(x_n,e_n))^\\infty_{n=1}$ is unbounded.\nSince $(x_n)$ is separated and $(e_n)$ is bounded, there exists $f\\in U_*(X,E)$ so that $f(x_n) = e_n$.\nBy assumption $Sf$ is bounded. Hence $(\\Xi(x_n,e_n))^\\infty_{n=1} = (Sf(x_n))^\\infty_{n=1}$ is bounded, a contradiction.\n\\end{proof}\n\n\nWe now obtain the analog of Theorem \\ref{t6.7.1} for biseparating maps between spaces of bounded uniformly continuous functions. The details are similar to Theorem \\ref{t6.7.1}, with the extra ingredient Lemma \\ref{l6.8} for ``boundedness''.\n\n\n\\begin{thm}\\label{t6.7.2}\nLet $X, Y$ be complete metric spaces and let $E,F$ be Banach spaces.\nSuppose that $T:U_*(X,E)\\to U_*(Y,F)$ is a biseparating map. \nThen there are a uniform homeomorphism ${\\varphi}:{X}\\to {Y}$ and a function $\\Phi:Y\\times E\\to F$ so that \n\\begin{enumerate}\n\\item For each $y\\in Y$, $\\Phi(y,\\cdot):E\\to F$ is a bijection with inverse $\\Psi(x,\\cdot):F\\to E$, where $\\varphi(x) = y$.\n\\item $Tf(y) = \\Phi(y,f(\\varphi^{-1}(y)))$ and $T^{-1}g(x) = \\Psi(x,g(\\varphi(x)))$ for all $f\\in U(X,E), g\\in U(Y,F)$ and $x\\in X$, $y\\in Y$. \n\\item $\\Phi$ is continuous on $Y'\\times E$ and $\\Psi$ is continuous on $X'\\times F$.\n\\item $(x,e)\\mapsto \\Phi(\\varphi(x),e))$ and $(y,e')\\mapsto \\Psi(\\varphi^{-1}(y),e'))$ are both $u$-continuous.\n\\item Let $(x_n)$ be a separated sequence in $X$ and $y_n = \\varphi(x_n)$ for all $n$. If $B$ and $B'$ are bounded sets in $E$ and $F$ respectively, then there exists $k\\in {\\mathbb N}$ so that \n\\[ \\bigcup_{n=k}^\\infty\\Phi(y_n,B) \\text{ and } \\bigcup_{n=k}^\\infty\\Psi(x_n,B')\\]\narer bounded sets in $F$ and $E$ respectively.\n\\end{enumerate}\nConversely, assume that $\\varphi,\\Phi$ satisfy conditions (1), (3), (4) and (5). Define $Tf(y)$ as in (2) for any $f\\in U_*(X,E)$ and $y\\in Y$. Then $T$ is a biseparating map from $U_*(X,E)$ onto $U_*(Y,F)$.\n\\end{thm}\n\n\\subsection{Automatic continuity}\nAutomatic continuity results for biseparating maps acting between spaces of uniformly continuous functions can be deduced easily from the characterization theorems \\ref{t6.7.1} and \\ref{t6.7.2}.\nIf $S$ is a subset of $X$, respectively, $Y$, and $f:X\\to E$, respectively, $f:Y\\to F$, let\n\\[ \\|f\\|_S = \\sup_{s\\in S}\\|f(s)\\|.\\]\n\n\\begin{thm}\\label{t6.9}\nLet $X,Y$ be complete metric spaces and $E,F$ be Banach spaces.\nSuppose that $T$ is a biseparating map from $U(X,E)$ onto $U(Y,F)$, respectively, from $U_*(X,E)$ onto $U_*(Y,F)$. Let $T$ be represented as in theorems \\ref{t6.7.1} or \\ref{t6.7.2}.\nAssume that $f\\in U(X,E)\/U_*(X,E)$ and $S\\subseteq X'$, the set of accumulation points of $X$.\nFor any $\\varepsilon > 0$, there exists $\\delta >0$ so that if $g\\in U(X,E)\/U_*(X,E)$ and $\\|g-f\\|_S < \\delta$, then \n$\\|Tg- Tf\\|_{\\varphi(S)} < \\varepsilon$.\n\\end{thm}\n\n\\begin{proof}\nSuppose that the theorem fails. There exist $S\\subseteq X'$, $\\varepsilon >0$ and functions $(g_n)$ in $U(X,E)\/U_*(X,E)$ so that \n\\[ \\|g_n-f\\|_S\\to 0 \\text{ and } \\|Tg_n-Tf\\|_{\\varphi(S)} > \\varepsilon \\text{ for all $n$.}\\]\nChoose $(x_n) \\subseteq S$ so that $\\|Tg_n(\\varphi(x_n)) - Tf(\\varphi(x_n))\\| >\\varepsilon$ for all $n$.\nThus\n\\begin{equation}\\label{e6.5} \\|\\Phi(\\varphi(x_n), v_n) - \\Phi(\\varphi(x_n),u_n)\\| >\\varepsilon \\text{ for all $n$},\\end{equation}\nwhere $v_n = g_n(x_n)$ and $u_n = f(x_n)$.\nIf $(x_n)$ has a subsequence $(x_{n_k})$ that converges to some $x_0$, then $x_0\\in X'$.\nNote that $(u_{n_k}) = Tf(\\varphi(x_{n_k}))$ converges to $u_0 = Tf(\\varphi(x_0))$ and $\\|v_n - u_n\\| \\leq \\|g_n-f\\|_S\\to 0$ as well. Thus $(v_{n_k})$ converges to $u_0$.\nThis shows that both sequences $((\\varphi(x_{n_k}),v_{n_k}))$ and $((\\varphi(x_{n_k}),u_{n_k}))$ converge to $(\\varphi(x_0),u_0)$.\nBy condition (3) of Theorem \\ref{t6.7.1} or \\ref{t6.7.2}, $\\Phi$ is continuous at $(\\varphi(x_0),u_0)$, contradicting (\\ref{e6.5}).\n\nIf $(x_n)$ does not have a convergent subsequence, then it has a separated subsequence $(x_{n_k})$. Again, let $u_{n_k} = f(x_{n_k})$\n and $v_{n_k} = g_{n_k}(x_{n_k})$.\nSince $g_{n_k}$ and $Tg_{n_k}$ are both continuous, one can choose $x_{n_k}'\\neq x_{n_k}$\n so that\n \\[ d(x_{n_k}',x_{n_k}), \\|g_{n_k}(x'_{n_k}) - v_{n_k}\\|, \\|Tg_{n_k}(x'_{n_k}) - Tg_{n_k}(x_{n_k})\\| \\to 0.\\]\n Note that the last limit can be stated as \n \\begin{equation}\\label{e6.6} \\Phi(\\varphi(x'_{n_k}),g_{n_k}(x'_{n_k})) - \\Phi(\\varphi(x_{n_k}), v_{n_k}) \\to 0.\\end{equation}\nBy Proposition \\ref{p6.3}(1), $(x_{n_k}, u_{n_k}))$ is a $u$-sequence.\nBy (4) of Theorem \\ref{t6.7.1} or \\ref{t6.7.2}, $(x,e) \\mapsto \\Phi(\\varphi(x),e)$ is $u$-continuous.\nSince $x_{n_k}' \\neq x_{n_k}$ and \n\\[ d(x'_{n_k},x_{n_k}) + \\|g_{n_k}(x'_{n_k}) - u_{n_k}\\| \\leq \n d(x'_{n_k},x_{n_k}) + \\|g_{n_k}(x'_{n_k}) - v_{n_k}\\| + \\|g_{n_k} - f\\|_S\n \\to 0,\\]\n $u$-continuity gives \n \\begin{equation}\\label{e6.7} \\Phi(\\varphi(x'_{n_k}),g_{n_k}(x'_{n_k})) - \\Phi(\\varphi(x_{n_k}), u_{n_k}) \\to 0.\\end{equation}\nThe limits (\\ref{e6.6}) and (\\ref{e6.7}) yield \n\\[ \\Phi(\\varphi(x_{n_k}), v_{n_k}) - \\Phi(\\varphi(x_{n_k}), u_{n_k}) \\to 0,\\]\ncontrary to (\\ref{e6.5}).\n\\end{proof}\n\n\n\n\\subsection{Bourbaki boundedness} \nLet $X$ be a metric space. For any $\\varepsilon>0$, recall the ``metric'' $d_\\varepsilon$ defined by (\\ref{eq6.0}). $d_\\varepsilon$ induces an equivalence relation $\\sim_\\varepsilon$ on $X$ by $a\\sim_\\varepsilon b$ if and only if $d_\\varepsilon(a,b) < \\infty$.\nThe equivalence classes will be called {\\em $\\varepsilon$-sets}. $d_\\varepsilon$ is a proper metric (i.e., finite valued) on each $\\varepsilon$-set.\n$X$ is said to be {\\em Bourbaki bounded} if for any $\\varepsilon>0$, there are only finitely many $\\varepsilon$-sets, each of which is bounded in the $d_\\varepsilon$ metric. See \\cite{BG, B, GM}.\nA classical result of Atsuji \\cite{At} and Hejcman \\cite{H}, rediscovered in \\cite{O'F}, states that $U(X) = U_*(X)$ if and only if $X$ is Bourbaki bounded.\nThe final theorem in this section generalizes this result.\n\n\\begin{thm}\\label{t6.10}\nLet $X, Y$ be complete metric spaces and let $E, F$ be Banach spaces.\nIf there is a biseparating map from $U(X,E)$ onto $U_*(Y,F)$, then $X$ is Bourbaki bounded.\n\\end{thm}\n\nBefore proceeding to the proof of the theorem, observe that if $X$ is Bourbaki bounded, then $U(X,E) = U_*(X,E)$. This follows easily from Proposition \\ref{p6.3}(2).\n \nLet $T:U(X,E)\\to U_*(Y,F)$ be a biseparating map. By Propositions \\ref{p4.2} and \\ref{p6.3.0}, $T$ has a representation\n\\[ Tf(y) = \\Phi(y,f(\\varphi^{-1}(y))) \\text{ for all $y\\in Y$ and all $f\\in U(X,E)$},\\]\nwhere $\\varphi$ is a uniform homeomorphism and $\\Phi(y,\\cdot):E\\to F$ is a bijection for all $y\\in Y$.\nWe may and do assume that $T0 = 0$, so that $\\Phi(y,0) = 0$ for all $y$.\n\n\\begin{lem}\\label{l6.11}\nLet $X, Y$ be complete metric spaces and let $E, F$ be Banach spaces.\nIf there is a biseparating map from $U(X,E)$ onto $U_*(Y,F)$, then for any $\\varepsilon>0$, $X$ has finitely many $\\varepsilon$-sets.\n\\end{lem}\n\n\\begin{proof}\nSuppose that there exists some $\\varepsilon >0$ so that $X$ contains an infinite sequence $(X_n)^\\infty_{n=1}$ of $\\varepsilon$-sets.\nChoose $x_n \\in X_n$ for each $n$ and let $y_n =\\varphi(x_n)$.\nSince $\\Phi(y_n,\\cdot):E\\to F$ is a bijection, there exists $e_n\\in E$ so that $\\|\\Phi(y_n,e_n)\\| > n$.\nDefine $f:X\\to E$ by $f(x) = e_n$ if $x\\in X_n$, $n\\in {\\mathbb N}$ and $0$ otherwise.\nThen $f$ is uniformly continuous but $\\|Tf(y_n)\\|= \\|\\Phi(y_n,e_n)\\| > n$ for all $n$.\nThis contradicts the assumption that $Tf \\in U_*(Y,F)$.\n\\end{proof}\n\n\n\\begin{lem}\\label{l6.12}\nLet $X, Y$ be complete metric spaces and let $E, F$ be Banach spaces.\nSuppose that there is a biseparating map from $U(X,E)$ onto $U_*(Y,F)$. For any $\\varepsilon>0$, any $\\varepsilon$-set of $X$ is $d_\\varepsilon$-bounded.\n\\end{lem}\n\n\n\\begin{proof}\nDefine $\\Xi:X\\times E\\to F$ by $\\Xi(x,e) = \\Phi(\\varphi(x),e)$. The formula $Sf(x) = \\Xi(x,f(x))$ defines a biseparating map from $U(X,E)$ onto $U_*(X,F)$ so that $S0 = 0$. For each $x$, $\\Xi(x,\\cdot):E\\to F$ is a bijection. Denote its inverse by $\\Theta(x,\\cdot)$. \nSuppose that there exist $\\varepsilon>0$ and an $\\varepsilon$-set $X_0$ that is not $d_\\varepsilon$ bounded.\nFix $x_0\\in X_0$ and a sequence $(x_n)$ in $X_0$ so that $d_\\varepsilon(x_{n+1},x_0) > 3d_\\varepsilon(x_n,x_0)$ for all $n$.\nLet $a$ be a nonzero vector in $E$. \nBy Proposition \\ref{p6.3}(2), the function $f:X\\to E$ given by $f(x) = d_\\varepsilon(x,x_0)a$ belongs to $U(X,E)$.\nHence $Sf\\in U_*(X,F)$.\nIn particular, the sequence $(b_n) = (Sf(x_n))$ is bounded in $F$.\n\n\n\\medskip\n\\noindent{Claim}. There exists $m\\in {\\mathbb N}$ so that \n\\[ \\limsup_n\\|\\Theta(x_n,sb_n) - \\Theta(x_n,tb_n)\\| \\leq 1 \\text{ if $s,t\\in [0,1]$, $|s-t| \\leq \\frac{1}{m}$}.\\]\n\nFirst suppose that the claim holds.\nThen\n\\begin{align*}\n \\limsup_n\\|\\Theta(x_n,b_n)\\| & = \\limsup_n\\|\\Theta(x_n,0)-\\Theta(x_n,b_n)\\|\n \\\\&\\leq \\sum^m_{k=1}\\limsup\\|\\Theta(x_n,\\frac{k-1}{m}\\,b_n) - \\Theta(x_n,\\frac{k}{m}\\,b_n)\\| \\leq m.\n\\end{align*}\nHowever, $\\Xi(x_n,d_\\varepsilon(x_n,x_0)a) = Sf(x_n) = b_n$ for all $n$.\nHence $\\Theta(x_n,b_n) = d_\\varepsilon(x_n,x_0)a$ for all $n$.\nIn particular, $(\\Theta(x_n,b_n))$ cannot be bounded, contradicting the preceding inequality.\n\nTo complete the proof of the lemma, let us verify the claim.\nIf the claim fails, for each $m\\in {\\mathbb N}$, one can find $s_m,t_m\\in [0,1]$, $|s_m-t_m|\\leq \\frac{1}{m}$, so that\n\\[ \\limsup_n\\|\\Theta(x_n,s_mb_n) - \\Theta(x_n,t_mb_n)\\| > 1.\\]\nWe may assume that $(s_m), (t_m)$ both converge to some $t_0\\in [0,1]$.\nWithout loss of generality, \n\\[ \\limsup_n\\|\\Theta(x_n,s_mb_n) - \\Theta(x_n,t_0b_n)\\| > \\frac{1}{2} \\text{ for all $m$.}\\]\nChoose $n_1 < n_2 < \\cdots$ so that \n\\begin{equation}\\label{e6.4}\\|\\Theta(x_{n_m},s_mb_{n_m}) - \\Theta(x_{n_m},t_0b_{n_m})\\| > \\frac{1}{2} \\text{ for all $m$.}\\end{equation}\nClearly, $(x_n)$, and hence $(x_{n_m})$, is a separated sequence by choice. Since $(t_0b_{n_m})$ is bounded, there exist $g_1\\in U_*(X,F)$ so that \n$g_1(x_{n_m}) = t_0b_{n_m}$ for all $m$.\nIf there exists $\\delta > 0$ and an infinite set $M$ so that $B(x_{n_m},\\delta) = \\{x_{n_m}\\}$ for all $m\\in M$, then each $\\{x_{n_m}\\}$ is a $\\delta$-set in $X$, contradicting Lemma \\ref{l6.11}.\nTherefore, there is a sequence $(x'_m)$ in $X$ so that $0 < d(x_{n_m},x'_m)\\to 0$.\nNote that \n$\\|s_mb_{n_m} - t_0b_{n_m}\\| \\to 0$. \nHence there exists $h \\in U_*(X,F)$ so that $h(x_{n_m}) = s_mb_{n_m} - t_0b_{n_m}$ and $h(x'_m) = 0$ for all sufficiently large $m$.\nSet $g_2 = g_1 +h$.\nSince $g_1(x'_m) = g_2(x'_m)$, $S^{-1}g_1(x'_m) = S^{-1}g_2(x'_m)$ for all sufficiently large $m$. Thus\n\\begin{align*}\n\\|\\Theta(x_{n_m},&\\, t_0b_{n_m}) - \\Theta(x_{n_m},s_mb_{n_m})\\| = \\|S^{-1}g_1(x_{n_m}) - S^{-1}g_2(x_{n_m})\\| \\\\\n&\\leq \\|S^{-1}g_1(x_{n_m}) - S^{-1}g_1(x'_m)\\| + \\|S^{-1}g_2(x'_{m}) - S^{-1}g_2(x_{n_m})\\|.\n\\end{align*}\nAs $S^{-1}g_1, S^{-1}g_2$ are uniformly continuous functions and $d(x_{n_m},x'_m) \\to 0$, both terms on the right of the inequality tend to $0$. So we have reached a contradiction with (\\ref{e6.4}).\nThis completes the proof of the claim and hence of the lemma.\n\\end{proof}\n\nLemmas \\ref{l6.11} and \\ref{l6.12} prove Theorem \\ref{t6.10}.\nIf $T:U(X,E)\\to U_*(Y,F)$ is a biseparating map, then $X$ is Bourbaki bounded by Theorem \\ref{t6.10}.\nHence $U(X,E) = U_*(X,E)$ \\cite{O'F}.\nThus the characterization Theorem \\ref{t6.7.2} applies.\n\n\\section{Spaces of Lipschitz functions}\\label{s8}\n\nWe focus on biseparating maps on Lipschitz spaces in this section. Again, we restrict consideration to complete metric spaces $X$, $Y$ and Banach spaces $E, F$.\nIn contrast to previous cases, we will see that there is no difference between spaces of bounded and unbounded Lipschitz functions. In fact, it is even sufficient to consider bounded metric spaces $X$ and $Y$.\nIndeed, if $(X,d)$ is a metric space, let $X_1$ be the set $X$ with the bounded metric $d\\wedge 1$.\nThen clearly $\\operatorname{Lip}_*(X,E) = \\operatorname{Lip}(X_1,E)$.\nTo see that $\\operatorname{Lip}(X,E)$ is equivalent to some $\\operatorname{Lip}(Z,E)$ for a bounded metric space $Z$ via a linear biseparating map, we employ essentially the same argument from \\cite[Proposition 5.2]{LT}, which has its roots in \\cite{W}.\nFix a distinguished point $e$ in $X$ and define a function $\\xi:X\\to {\\mathbb R}$ by $\\xi(x) = d(x,e)\\vee 1$.\nDenote the Lipschitz constant of a function $f$ by $L(f)$. Let $d':X\\times X\\to {\\mathbb R}$ be given by\n\\[ d'(p,q) = \\sup_{\\stackrel{f\\in \\operatorname{Lip}(X)}{L(f),|f(e)| \\leq 1}}\\bigl|\\frac{f(p)}{\\xi(p)} - \\frac{f(q)}{\\xi(q)}\\bigr|,\\]\nwhere $\\operatorname{Lip}(X)$ is the space of all real-valued Lipschitz functions on $X$.\n\n\n\\begin{lem}\\cite[Proposition 5.1]{LT}\\label{p6.1}\n\\begin{enumerate}\n\\item $d'$ is a metric on $X$ that is bounded above by $4$.\n\\item \n\\[ \\frac{d(p,q)}{\\xi(p)\\vee \\xi(q)}\\leq d'(p,q) \\leq \\frac{3d(p,q)}{\\xi(p)\\vee \\xi(q)}\\]\nfor all $p,q\\in X$.\n\\item If $\\xi(p) \\leq \\xi(q)$, then\n\\[ d'(p,q) \\leq d'(p,q)\\xi(p) \\leq 3d(p,q).\\]\n\\item If $X$ is complete with respect to the metric $d$, then it is complete with respect to the metric $d'$.\n\\end{enumerate}\n\\end{lem}\n\nLet $Z$ be the set $X$ with the metric $d'$.\n\n\\begin{prop}\\label{p6.2}\n$f\\in \\operatorname{Lip}(X,E)$ if and only if $\\frac{f}{\\xi} \\in \\operatorname{Lip}(Z,E)$.\nIn particular, $T:\\operatorname{Lip}(X,E)\\to \\operatorname{Lip}(Z,E)$, $Tf = \\frac{f}{\\xi}$, is a linear biseparating map.\n\\end{prop}\n\n\\begin{proof}\nThe second assertion follows easily from the first.\nSuppose that $f\\in \\operatorname{Lip}(X,E)$. Set $c = L(f)\\vee \\|f(e)\\| \\vee 1$.\nFor any $x^* \\in E^*$, $\\|x^*\\| \\leq 1$, $g = (x^*\\circ f)\/c\\in \\operatorname{Lip}(X)$ and $L(g), |g(e)| \\leq 1$.\nBy definition of $d'$,\n\\[ cd'(p,q) \\geq c\\bigl|\\frac{g(p)}{\\xi(p)} - \\frac{g(q)}{\\xi(q)}\\bigr| = \\bigl|x^*\\bigl(\\frac{f}{\\xi}(p) - \\frac{f}{\\xi}(q)\\bigr)\\bigr|.\n\\]\nTaking supremum over $\\|x^*\\|\\leq 1$ shows that $\\frac{f}{\\xi}\\in \\operatorname{Lip}(Z,E)$ with Lipschitz constant at most $c$.\n\nConversely, suppose that $g= \\frac{f}{\\xi}\\in \\operatorname{Lip}(Z,E)$. Let $p,q$ be distinct points in $X$ so that $\\xi(p) \\leq \\xi(q)$. Denote the Lipschitz constant of $g$ with respect to the metric $d'$ by $L'(g)$.\nThen\n\\[ \\|g(q)\\| \\leq \\|g(q) - g(e)\\| + \\|g(e)\\| \\leq L'(g)d'(q,e) + \\|g(e)\\| \\leq 4L'(g) + \\|g(e)\\|\\]\nsince $d' \\leq 4$.\nHence\n\\begin{align*}\n\\|f(p) - f(q)\\| & \\leq \\|g(p) - g(q)\\|\\xi(p) + \\|g(q)\\|(\\xi(q) - \\xi(p))\\\\\n&\\leq L'(g)d'(p,q)\\xi(p) + (4L'(g)+ \\|g(e)\\|)d(p,q)\\\\\n&\\leq (7L'(g)+\\|g(e)\\|)d(p,q) \\text{ by Lemma \\ref{p6.1}(3)}.\n\\end{align*}\nThus $f\\in \\operatorname{Lip}(X,E)$.\n\\end{proof}\n\n\\subsection{$\\varphi$ is a Lipschitz homeomorphism} \n\nIn view of the above, throughout the rest of this section, $X$ and $Y$ will be assumed to be bounded complete metric spaces. Let $T:\\operatorname{Lip}(X,E)\\to \\operatorname{Lip}(Y,F)$ be a biseparating map so that $T0 =0$.\nOnce again, we have a representation (Proposition \\ref{p4.2})\n\\begin{equation}\\label{e7.1}Tf(y) = \\Phi(y,f(\\varphi^{-1}(y))) \\text{ for all $y\\in Y$ and all $f\\in \\operatorname{Lip}(X,E)$},\\end{equation}\nwhere $\\varphi:X\\to Y$ is a homeomorphism and $\\Phi:Y\\times E\\to F$ \nis a function such that $\\Phi(y,\\cdot):E\\to F$ is a bijection for all $y\\in Y$.\nDenote the inverse of $\\Phi(y,\\cdot)$ by $\\Psi(x,\\cdot)$, where $\\varphi(x) = y$.\n\n\\begin{prop}\\label{p7.3}\nSuppose that $(x_n), (x'_n)$ are sequences in $X$. Let $y_n = \\varphi(x_n), y_n' = \\varphi(x_n')$ for all $n$.\nIf $(y_n)$ is a separated sequence, then there exists $C<\\infty$ so that $d(y_n,y_n') \\leq Cd(x_n,x_n')$ for all $n$.\n\\end{prop}\n\n\\begin{proof}\nAssume that the proposition fails. There are sequences as in the statement of the proposition so that $d(y_n,y_n')\/d(x_n,x_n') \\to \\infty$. Since $Y$ is bounded, $d(x_n,x_n') \\to 0$.\nIf $(y_n')$ has a convergent subsequence, then $(x_n')$, and hence $(x_n)$ has a convergent subsequence, which in turn implies that $(y_n)$ has a convergent subsequence, contrary to the choice of $(y_n)$.\nThus, by taking a subsequence if necessary, we may assume that both $(y_n)$ and $(y_n')$ are separated sequences. Fix a nonzero vector $a\\in E$ and let $g = T(1\\otimes a)$.\n\n\\medskip\n\n\\noindent\\underline{Case 1}. $d(y_n,y_n') \\not\\to 0$.\n\nIn this case, by taking a subsequence, we may assume that $(y_n)\\cup (y_n')$ is a separated set.\nSince $g\\in \\operatorname{Lip}(Y,F)$, $(g(y_n))$ is a bounded sequence in $F$.\nHence there exists $h\\in \\operatorname{Lip}(Y,F)$ so that $h(y_n) = g(y_n)$ and $h(y_n') = 0$ for all large $n$.\nThen $T^{-1}h(x_n) = a$ and $T^{-1}(x_n') = 0$ for all large $n$.\nSince $T^{-1}h$ is Lipschitz and $d(x_n,x_n')\\to 0$, we have a contradiction.\n\n\\medskip\n\n\\noindent\\underline{Case 2}. $d(y_n,y_n')\\to 0$.\n\nIf $(\\frac{\\|g(y_n)\\|}{d(y_n,y_n')})$ is bounded, then there is a function $h\\in \\operatorname{Lip}(Y,F)$ so that \n$h(y_n) = g(y_n)$ and $h(y_n') = 0$ for all large $n$.\nThus $T^{-1}h(x_n) = a$ and $T^{-1}h(x_n') = 0$. This is impossible since $T^{-1}h$ is Lipschitz and $d(x_n,x_n') \\to 0$.\n\nFrom the unboundedness of $(\\frac{\\|g(y_n)\\|}{d(y_n,y_n')})$, we may assume without loss of generality that for each $n$, there exists $s_n$ so that $d(y_n,y'_n)\\leq s_n< 2d(y_n,y_n')$ and that $k_n = \\frac{\\|g(y_n)\\|}{s_n}\\in {\\mathbb N}$.\nSince $\\Phi(y_n,a) = g(y_n)$, $\\Psi(x_n,g(y_n)) = a$. Thus\n\\[ \\sum^{k_n}_{j=1}[\\Psi(x_n,\\frac{jg(y_n)}{k_n}) - \\Psi(x_n,\\frac{(j-1)g(y_n)}{k_n})] = \\Psi(x_n,g(y_n)) = a.\n\\]\nHence there exists $j_0\\in \\{1,\\dots, k_n\\}$ so that \n\\[ \\|\\Psi(x_n,\\frac{j_0g(y_n)}{k_n}) - \\Psi(x_n,\\frac{(j_0-1)g(y_n)}{k_n})\\| \\geq \\frac{\\|a\\|}{k_n}.\\]\nTherefore, there exists $i_0\\in \\{j_0-1,j_0\\}$ so that \n\\begin{equation}\\label{e7.2.0} \\|\\Psi(x_n, \\frac{i_0g(y_n)}{k_n}) - \\Psi(x'_n, \\frac{j_0g(y_n')}{k_n})\\| \\geq \\frac{\\|a\\|}{2k_n}.\\end{equation}\nNow\n\\begin{align}\\label{e7.2}\n\\|\\frac{i_0g(y_n)}{k_n} &- \\frac{j_0g(y_n')}{k_n}\\| \\leq \\frac{\\|g(y_n)\\|}{k_n} + \\frac{j_0}{k_n}\\|g(y_n)-g(y_n')\\| \\\\ \\notag\n&\\leq s_n + \\frac{j_0L(g)}{k_n}d(y_n,y_n') \\leq (2+L(g))d(y_n,y_n'),\n\\end{align}\nwhere $L(g)$ is the Lipschitz constant of $g$.\nSince $(y_n)$ is separated and $(\\frac{i_0g(y_n)}{k_n})$ is a bounded sequence in $F$, there exists $h_1\\in \\operatorname{Lip}(Y,F)$ so that $h_1(y_n) = \\frac{i_0g(y_n)}{k_n}$ for all $n$.\nLet $L(h_1)$ be the Lipschitz constant of $h_1$. By (\\ref{e7.2}),\n\\begin{align*} \\|\\frac{j_0g(y_n)}{k_n} - h_1(y_n')\\|& \\leq \\|\\frac{j_0g(y_n)}{k_n} - h_1(y_n)\\| + L(h_1)d(y_n,y_n')\n\\\\&\\leq (2+L(g)+L(h_1))d(y_n,y_n').\n\\end{align*}\n Therefore, \none can construct a function $h_2\\in \\operatorname{Lip}(Y,F)$ so that \n\\[ h_2(y_n) = 0 \\text{ and } h_2(y_n') = \\frac{j_0g(y_n')}{k_n}- h_1(y_n') \\text{ for all large $n$}.\\]\nLet $f= T^{-1}(h_1+h_2)$.\nThen $Tf(y_n) = h_1(y_n)$ and hence $f(x_n) =\\Psi(x_n, \\frac{i_0g(y_n)}{k_n})$ for all large $n$.\nSimilarly, $f(x_n') = \\Psi(x_n, \\frac{j_0g(y_n)}{k_n})$ for all large $n$.\nNote that $g$ is a bounded function. Set $\\|g\\|_\\infty = \\sup_{y\\in Y}\\|g(y)\\|$. By (\\ref{e7.2.0}, \n\\[ \\|f(x_n) -f(x_n')\\| \\geq \\frac{\\|a\\|}{2k_n} \\geq \\frac{\\|a\\|s_n}{2\\|g\\|_\\infty} \\geq \\frac{\\|a\\|}{2\\|g\\|_\\infty}d(y_n,y_n')\\] for all large $n$.\nSince $f$ is Lipschitz, it follows that $d(y_n,y_n')\/d(x_n,x'_n)\\not\\to \\infty$.\nThis contradiction completes the proof of the proposition.\n\\end{proof}\n\nIf $(x_0,e_0)\\in X\\times E$, $C<\\infty$ and $r >0$, let\n\\begin{equation}\\label{e7.4} \\Delta(x_0,e_0,C,r) = \\{(x,e)\\in X\\times E: d(x,x_0) \\leq r, \\|e-e_0\\| \\leq Cd(x,x_0)\\}.\\end{equation}\nThen set $\\Delta(x_0,e_0,C) = \\bigcup_{r > 0} \\Delta(x_0,e_0,C,r)$.\nIt is not surprising that understanding the map $T$ depends on analyzing the sets $\\Delta(x_0,e_0,C,r)$ and $\\Delta(x_0,e_0,C)$. For a very special instance, see \\cite[Section 7.2]{AZ}.\nDefine sets in $Y\\times F$ in a similar manner.\nLet $M:X\\times E\\to Y\\times F$ be the function\n\\[M(x,u) = (\\varphi(x),\\Phi(\\varphi(x),u)).\\]\nThen $M$ is a bijection. Moreover, if $f \\in \\operatorname{Lip}(X,E)$, then $M(x_0, f(x_0)) = (\\varphi(x_0), Tf(\\varphi(x_0)))$.\n\n\n\n\nSuppose that $\\varphi:X\\to Y$ is not Lipschitz. There are sequences $(x_n), (x'_n)$ in $X$, $x_n\\neq x_n'$ for all $n$, so that taking $y_n = \\varphi(x_n), y_n' = \\varphi(x_n')$, we have $d(y_n,y_n')\/d(x_n,x_n') \\to \\infty$.\nSince $Y$ is bounded, $d(x_n,x_n') \\to 0$. By Proposition \\ref{p7.3}, $(y_n)$ cannot be a separated sequence.\nTaking a subsequence if necessary, we may assume that $(y_n)$ converges to some $y_0$.\nThen $(x_n)$ converges to $\\varphi^{-1}(y_0) = x_0$. The same must hold for $(x'_n)$. Therefore, $(y_n')$ also converges to $y_0$.\nWith further subsequences and relabeling the primed and unprimed terms if necessary, we may assume that $d(y_n,y_0) \\leq d(y_n',y_0)$ for all $n$.\nWith this assumption, $y_n'\\neq y_0$ for all $n$. For otherwise $y_n = y_0 = y_n'$, which implies that $x_n = x'_n$, contrary to their choice. Hence we may further assume that \n\\[ 2d(y'_{n+1},y_0) m d(x_{m}',x_{m}) \\text{ for all $m$.}\\]\nLet $d_m = d(y_m',y_m)$. \nDefine a function $g:Y\\to F$ by\n\\[ g(y) = \\begin{cases}\n v_0 + (1- \\frac{4d(y,y_m')}{d_m})(v_m'-v_0) &\\text{if $m\\in {\\mathbb N}$, $d(y,y_m') < \\frac{d_m}{4}$}\\\\\n v_0 &\\text{otherwise}.\n \\end{cases}\\]\nFrom the disjointness of the balls $B(y_m',\\frac{d_m}{4})$ and the inequality $\\|v_m'-v_0\\| \\leq d_m$ for all $m$,\nwe see that $g\\in \\operatorname{Lip}(Y,F)$.\n\nNext, we claim that $y_n \\notin B(y_m',\\frac{d_m}{4})$ for any $m,n\\in {\\mathbb N}$.\nIndeed, this is obvious if $m =n$.\nNote that \\[ d_m \\leq d( y_m',y_0) + d(y_0,y_m) \\leq 2d(y_m',y_0).\\]\nIf $m < n$, then \n\\[ d(y_n,y_0) \\leq d(y_n',y_0) < \\frac{d(y_m',y_0)}{2}.\n\\]\nHence\n\\[ d(y_n,y_m') \\geq d(y_m',y_0) - d(y_n,y_0)> \\frac{d(y_m',y_0)}{2} \\geq \\frac{d_m}{4}.\n\\]\nOn the other hand, if $m > n$ and $y_n\\neq y_0$, then\n\\[2d(y_m',y_0) \\leq 2d(y_{n+1}',y_0) < d(y_n,y_0).\n\\]\nHence\n\\[ d(y_n,y_m') \\geq d(y_n,y_0) - d(y_m',y_0) \\geq d(y_m',y_0) \\geq \\frac{d_m}{2}.\n\\]\nFinally, if $y_n = y_0$, then $d(y_n,y_m') = d(y_0,y_m') \\geq \\frac{d_m}{2}$.\nThus $y_n \\notin B(y_m',\\frac{d_m}{4})$ in all cases.\n\nObviously, $g(y_m') = v_m'= \\Phi(y_m',u_m')$. \nFrom the fact that $y_n \\notin B(y_m',\\frac{d_m}{4})$ for all $m,n$,\n $g(y_m) = v_0 = \\Phi(y_m,u_m)$ for all $n$.\nTherefore, $T^{-1}g(x_m) = u_m$ and $T^{-1}g(x'_m) = u_m'$ for all $m$.\nBut $\\|u_{m}'-u_{m}\\| > m d(x_{m}',x_{m})$, which contradicts the fact that $T^{-1}g$ is Lipschitz.\n\\end{proof}\n\nWe are now ready to prove the main result regarding the homeomorphism $\\varphi$.\n\n\n\\begin{thm}\\label{t7.5}\nLet $T:\\operatorname{Lip}(X,E)\\to \\operatorname{Lip}(Y,F)$ be a biseparating map, where $X, Y$ are complete bounded metric spaces and $E, F$ are Banach spaces. \nIn the notation of (\\ref{e7.1}), $\\varphi:X\\to Y$ is a Lipschitz homeomorphism.\n\\end{thm}\n\n\\begin{proof}\nIf $\\varphi$ is not a Lipschitz function, then we obtain sequences $(x_n), (x_n')$, $(y_n)$ and $(y_n')$ as in the discussion before Proposition \\ref{p7.5}.\nFor each $v\\in F$, determine $m = m(v) \\in {\\mathbb N}$ by Proposition \\ref{p7.5}.\nSet $F_k = \\{v\\in F: m(v) \\leq k\\}$ for each $k\\in {\\mathbb N}$.\nThen $F = \\bigcup_{k=1}^\\infty\\overline{F_k}$.\nBy the Baire Category Theorem, there are an open ball $O$ in $F$ and $k_0\\in {\\mathbb N}$ so that $O\\subseteq \\overline{F_{k_0}}$.\nPick distinct points $a,b\\in O\\cap F_{k_0}$. Since $d_n = d(y_n,y_n') \\to 0$, we may assume without loss of generality that $\\|a-b\\| > d_n$ for all $n$. For each $n$, choose $k_n\\in {\\mathbb N}$ so that \n\\[ \\frac{k_nd_n}{2} \\leq \\|a-b\\| < {k_nd_n}.\\]\nNote that $a + \\frac{j}{k_n}(b-a) \\in O$, $0\\leq j\\leq k_n$. By making small perturbations, one can find $w_{nj}\\in O\\cap F_{k_0}$, $0\\leq j\\leq k_n$, so that $w_{n0} = a$, $w_{nk_n} = b$ and $\\|w_{nj} - w_{n,j-1}\\|$ is sufficiently close to $\\frac{\\|a-b\\|}{k_n}$ so as to make it $< d_n$.\nNow\n\\begin{align*}\n\\|\\sum^{k_n}_{j=1}[\\Psi(x_n&,w_{nj}) - \\Psi(x_n,w_{n,j-1})]\\| = \\|\\Psi(x_n,v_1) -\\Psi(x_n,v_0)\\|\n\\\\ & = \\|T^{-1}(1\\otimes v_1)(x_n) - T^{-1}(1\\otimes v_0)(x_n)\\|\\\\\n& \\to \\|T^{-1}(1\\otimes v_1)(x_0) - T^{-1}(1\\otimes v_0)(x_0)\\|\\\\\n& = \\|\\Psi(x_0,b) -\\Psi(x_0,a)\\| = c.\n\\end{align*}\nSince $\\Psi(x_0,\\cdot)$ is a bijection, $c > 0$.\nFor all $n$, there exists $1\\leq j_n\\leq k_n$ so that \n\\[ \\|\\Psi(x_n,w_{n,j_n}) - \\Psi(x_n,w_{n,j_n-1})\\| > \\frac{c}{2k_n}.\\]\nNow choose $i_n\\in \\{j_n-1,j_n\\}$ so that, setting \n\\[ u_n = \\Psi(x_n,w_{n,j_n}) \\text{ and } u_n' = \\Psi(x_n',w_{n,i_n}),\\]\nwe have $\\|u_n-u_n'\\| > \\frac{c}{4k_n}$.\nNote that \n\\[ M^{-1}(y_n,w_{n,j_n}) = (\\varphi^{-1}(y_n),\\Psi(\\varphi^{-1}(y_n), w_{n,j_n}))\n=\n(x_n,u_n).\n\\]\nSimilarly, $M^{-1}(y_n',w_{n,i_n}) =(x_n',u_n')$.\nBy choice,\n\\[ \\|w_{n,i_n} - w_{n,j_n}\\| \\leq \\|w_{n,j_n-1} - w_{n,j_n}\\| < d_n = d(y_n',y_n).\\]\nHence $(y_n', w_{n,i_n}) \\in \\Delta(y_n,w_{n,j_n},1)$.\nSince $w_{n,j_n} \\in F_{k_0}$, $m=m(w_{n,j_n}) \\leq k_0$.\nBy definition of $m(w_{n,j_n})$, this implies that for all $n \\geq k_0$,\n\\[ (x_n',u'_n) \\in \\Delta(x_n,u_n,m)\\subseteq \\Delta(x_n,u_n,k_0),\n\\]\nwhich in turns yields that \n$\\|u_n'-u_n\\| \\leq k_0d(x_n',x_n).$\nTherefore,\n\\[ \\frac{c}{2\\|a-b\\|}\\cdot d_n \\leq \\frac{c}{4k_n} < \\|u_n - u_n'\\| \\leq k_0d(x_n',x_n).\\]\nSince this holds for all sufficiently large $n$, and $d_n = d(y_n,y_n')$, it contradicts the assumption that $d(y_n,y_n')\/d(x_n,x'_n) \\to \\infty$.\n\nThis completes the proof that $\\varphi$ is a Lipschitz function. By symmetry, so is $\\varphi^{-1}$. Hence $\\varphi$ is a Lipschitz homeomorphism.\n\\end{proof}\n\n\\subsection{Section problem for Lipschitz functions}\n\nLet $X$ be a complete bounded metric space and let $E,F$ be Banach spaces. Consider a given function $\\Xi:X\\times E\\to F$. Define $M:X\\times E \\to X\\times F$ by $M(x,e) = (x,\\Xi(x,e))$. Recall the sets $\\Delta(x,e,C,r)$ and $\\Delta(x,e,C)$ in $X\\times E$ as given by (\\ref{e7.4}). Similar definitions apply in $X\\times F$. Theorem \\ref{t7.7} solves the section problem for spaces of Lipschitz functions. For a very special case, refer to \\cite[Theorem 7.1]{AZ}.\n\n\\begin{lem}\\label{l7.6}\nSuppose that $x_0\\in X$, $u_0\\in E$ and $C<\\infty$.\nLet $(x^n_1), (x^n_2)$ in $X$, $(u^n_1), (u^n_2)$ in $E$ be sequences so that $x^n_1 \\neq x^n_2$ for all $n$,\n\\[ \\|u^n_1-u^n_2\\|\\leq Cd(x^n_1,x^n_2),\\\n\\|u^n_i-u_0\\| \\leq Cd(x^n_i,x_0) \\text{ $i =1,2$, $n\\in {\\mathbb N}$, and}\\]\n$\\lim_n d(x^n_i,x_0)=0$, $i =1,2$.\nThen there exists $f\\in \\operatorname{Lip}(X,E)$ so that $f(x^n_i) = u^n_i$, $i =1,2$, for infinitely many $n$.\n\\end{lem}\n\n\\begin{proof}\nSet $r^n_i = d(x^n_i,x_0)$. There is no loss of generality in assuming that $r^n_2 \\geq r^n_1$ for all $n$. After taking subsequences, we may divide the proof into the following cases.\n\n\\medskip\n\n\\noindent \\underline{Case 1}. $x^n_1 = x_0$, i.e., $r^n_1 = 0$ for all $n$.\n\n\\medskip\n\n\nNote that in this case $u^n_1 = u_0$ for all $n$. Since $x^n_2 \\neq x^n_1$ and $r^n_2=d(x^n_2,x_0) \\to 0$, we may further assume that $r^{n+1}_2 < \\frac{1}{3}r^n_2$ for all $n$, which implies that the balls $B(x^n_2,\\frac{r_n}{2})$ are pairwise disjoint.\nDefine $f:X\\to E$ by\n\\[ f(x) = \\begin{cases}\nu_0 + (1 - \\frac{2d(x,x^n_2)}{r_n})(u^n_2-u_0) &\\text{if $d(x,x^n_2) < \\frac{r_n}{2}$}\\\\\nu_0 &\\text{otherwise}.\n\\end{cases}\\]\nThen it can be checked that $f\\in \\operatorname{Lip}(X,E)$, $f(x^n_2) = u^n_2$, $f(x^n_1) = f(x_0) = u_0=u^n_1$ for all $n$.\n\n\\medskip\n\n\\noindent \\underline{Case 2}. $r^n_1 >0$ and there exists $c > 0$ so that $d(x^n_1,x^n_2) \\geq cr^n_2$ for all $n$.\n\n\\medskip\n\nWe may of course assume that $0 < c < 1$. The assumptions imply that the balls $B(x^n_1, \\frac{cr^n_1}{2})$ and $B(x^n_2, \\frac{cr^n_2}{2})$ are disjoint.\nSince $r^n_2\\to 0$, we may further assume that $B(x^{n+1}_1, \\frac{cr^{n+1}_1}{2})\\cup B(x^{n+1}_2, \\frac{cr^{n+1}_2}{2})$ is disjoint from $\\bigcup^n_{k=1}[B(x^{k}_1, \\frac{cr^{k}_1}{2})\\cup B(x^{k}_2, \\frac{cr^{k}_2}{2})]$ for any $n$.\nAs a result, the sets $B(x^n_1, \\frac{cr^n_1}{2})$, $B(x^n_2, \\frac{cr^n_2}{2})$, $n\\in {\\mathbb N}$, are all mutually disjoint.\nDefine $f:X\\to E$ by\n\\[ f(x) = \\begin{cases}\nu_0 + (1 - \\frac{2d(x,x^n_i)}{cr^n_i})(u^n_i-u_0) &\\text{if $d(x,x^n_i) < \\frac{cr^n_i}{2}$, $n \\in {\\mathbb N}$, $i=1,2$,}\\\\\nu_0 &\\text{otherwise}.\n\\end{cases}\\]\nThen it can be checked that $f\\in \\operatorname{Lip}(X,E)$, $f(x^n_i) = u^n_i$, $i=1, 2$, $n\\in {\\mathbb N}$.\n\n\n\n\\medskip\n\n\\noindent \\underline{Case 3}. $r^n_1 >0$ for all $n$ and $d(x^n_1,x^n_2)\/r^n_2 \\to 0$.\n\n\\medskip\n\nAs in Case 2, we may assume that the sets $B(x^n_2, \\frac{r^n_2}{2}), n\\in {\\mathbb N}$, are dsijoint.\nIn this instance, we may further assume that $B(x^n_1,d(x^n_1,x^n_2)) \\subseteq B(x^n_2, \\frac{r^n_2}{2})$ for all $n$.\nDefine $g:X\\to E$ by\n\\[ g(x) = \\begin{cases}\n (1 - \\frac{2d(x,x^n_2)}{r^n_2})(u^n_2-u_0) &\\text{if $d(x,x^n_2) < \\frac{r^n_2}{2}$}\\\\\n0 &\\text{otherwise}.\n\\end{cases}\\]\nSince $\\|u^n_2-u_0\\| \\leq Cd(x^n_2,x_0)= Cr^n_2$ for all $n$, $g\\in \\operatorname{Lip}(X,E)$ and has Lipschitz constant at most $2C$.\nClearly, $g(x^n_2) = u^n_2-u_0$ for all $n$.\nNow let \n$h:X\\to E$ be given by\n\\[ h(x) = \\begin{cases}\n (1 - \\frac{d(x^n_1,x)}{d(x^n_1,x^n_2)})(u^n_1-u_0-g(x^n_1)) &\\text{if $d(x^n_1,x) < d(x^n_1,x^n_2)$}\\\\\n0 &\\text{otherwise}.\n\\end{cases}\\]\nNote that\n\\[\\|u^n_1-u_0-g(x^n_1)\\| \\leq \\|u^n_1-u^n_2\\| + \\|g(x^n_2)-g(x^n_1)\\| \\leq 3Cd(x^n_1,x^n_2).\n\\]\nTaking into account the disjointness of the sets $B(x^n_1, d(x^n_1,x^n_2))$, it follows that $h\\in \\operatorname{Lip}(X,E)$.\nFurthermore, \n\\[ \n(g+h)(x^n_1) = u^n_1-u_0 \\text{ and } (g+h)(x^n_2) = g(x^n_2) = u^n_2-u_0.\\]\nFinally, the function $f(x) = g(x) +h(x) +u_0$ is the one we seek.\n\\end{proof}\n\n\n\\begin{thm}\\label{t7.7}\nLet $\\Xi:X\\times E\\to F$ be a given function. Define $Sf(x) = \\Xi(x,f(x))$ for any function $f:X\\to E$.\nSuppose that $Sf$ belongs to $\\operatorname{Lip}(X,F)$ for all $f\\in \\operatorname{Lip}(X,E)$. Then\n\\begin{enumerate}\n\\item If $(x_n)$ is a separated sequence in $X$, and $B$ is a bounded set in $E$, then there is a finite set $N \\subseteq {\\mathbb N}$ so that $\\bigcup_{n\\notin N}\\Xi(x_n,B)$ is bounded.\n\\item Suppose that $x_0\\in X$, $u_0\\in E$ and $C<\\infty$. There exist $r >0$ and $D<\\infty$ so that \n\\[ \\|\\Xi(x_1,u_1) - \\Xi(x_2,u_2)\\| \\leq Dd(x_1,x_2) \n\\]\nwhenever $\\|u_1-u_2\\| \\leq Cd(x_1,x_2)$, $\\|u_i-u_0\\| \\leq Cd(x_i,x_0)$ and $d(x_i,x_0) \\leq r$, $i=1,2$.\n\\item Let $(x_n)$ be a separated sequence in $X$ and $(u_n)$ be a bounded sequence in $E$.\nFor any $C<\\infty$, there exist $r>0$ and $D<\\infty$ so that \n\\[ \\|\\Xi(x_n',u'_n) - \\Xi(x_n,u_n)\\| \\leq Dd(x_n',x_n) \\text{ for all $n$}\\]\nwhenever $\\|u_n'-u_n\\| \\leq Cd(x_n',x_n)$ and $d(x_n',x_n) \\leq r$ for all $n$.\n\\end{enumerate}\nConversely, suppose that conditions (1), (2) and (3) hold. Then $Sf\\in \\operatorname{Lip}(X,F)$ for any $f\\in \\operatorname{Lip}(X,E)$.\n\\end{thm}\n\n\\begin{proof}\nSuppose that $Sf\\in \\operatorname{Lip}(X,F)$ for any $f\\in \\operatorname{Lip}(X,E)$. \nLet $(x_n)$ be a separated sequence in $X$ and $B$ be a bounded set in $E$. If $\\bigcup_{n\\notin N}\\Xi(x_n,B)$ is unbounded for any finite set $N\\subseteq {\\mathbb N}$, there exists a sequence $(u_n)\\subseteq B$ so that $(\\Xi(x_n,u_n))$ is unbounded. Since $(x_n)$ is separated and $(u_n)$ is bounded, there is a Lipschitz function $f:X\\to E$ so that $f(x_n) = u_n$ for all $n$.\nThen $Sf\\in \\operatorname{Lip}(X,F)$, $(\\Xi(x_n,u_n)) = (Sf(x_n))$ is bounded in $F$, a contradiction. This proves condition (1).\n\nSuppose that condition (2) fails. Then there are $(x^n_1), (x^n_2)$ in $X$, $(u^n_1), (u^n_2)$ in $E$ so that \n$\\|u^n_1-u^n_2\\| \\leq Cd(x^n_1,x^n_2)$, \n$\\|u^n_i-u_0\\| \\leq Cd(x^n_i,x_0)$ and $d(x^n_i,x_0) \\leq \\frac{1}{n}$, $i =1,2$, $n\\in {\\mathbb N}$, but $\\|\\Xi(x^n_1,u^n_1) - \\Xi(x^n_2,u^n_2)\\| > nd(x^n_1,x^n_2)$.\nIn particular, the last inequality implies that $x^n_1\\neq x^n_2$ for all $n$.\nApply Lemma \\ref{l7.6} to find a function $f\\in \\operatorname{Lip}(X,E)$ so that $f(x^n_i) = u^n_i$ for infinitely many $n$.\nLet $L$ be the Lipschitz constant of $Sf$.\nThen \n\\[\\|\\Xi(x^n_1,u^n_1) - \\Xi(x^n_2,u^n_2)\\| = \\|Sf(x^n_1) - Sf(x^n_2)\\| \n\\leq Ld(x^n_1,x^n_2)\\]\nfor all $n$, contrary to their choices.\n\nLet $(x_n)$ be a separated sequence in $X$ and let $(u_n)$ be a bounded sequence in $E$. Assume that condition (3) fails for a constant $C$.\nFor each $k\\in {\\mathbb N}$, there exist $n_k\\in {\\mathbb N}$, $x_k\\in X $ and $u_k'\\in E$ so that $\\|u_k'- u_{n_k}\\| \\leq Cd(x_k',x_{n_k})$ and $d(x_k',x_{n_k}) < \\frac{1}{k}$ but\n\\begin{equation}\\label{e7.4.1}\\|\\Xi(x_k',u_k') - \\Xi(x_{n_k},u_{n_k})\\| > kd(x_k',x_{n_k}).\\end{equation}\nIf $(n_k)$ has a constant subsequence, then, say, $x_{n_k} = x_0$ and $u_{n_k} = u_0$ for infinitely many $k$.\nIn this case, we have a contradiction to condition (2), which has been shown above.\nOtherwise, we may assume that $n_k \\uparrow \\infty$.\nSince $(x_{n_k})$ is separated and $(u_{n_k})$ is bounded, there exists $g\\in \\operatorname{Lip}(X,E)$ so that $g(x_{n_k}) = u_{n_k}$ for all $k$.\nLet $L$ be the Lipschitz constant of $g$. We have\n\\[ \\|u_k'- g(x_k')\\| \\leq \\|u_k'-u_{n_k}\\| + \\| u_{n_k} - g(x_k')\\| \\leq (C+L)d(x_{n_k},x_k').\\]\nAs $(x_{n_k})$ is separated and $d(x_k',x_{n_k})\\to 0$, we can find $h\\in \\operatorname{Lip}(X,E)$ so that $h(x_{n_k}) = 0$ and $h(x_k') = u_k'-g(x_k')$ for all large $k$. Let $f = g+h\\in \\operatorname{Lip}(X,E)$.\nThen $Sf\\in \\operatorname{Lip}(X,E)$ and \n\\[ Sf(x_{n_k}) = \\Xi(x_{n_k},u_{n_k}),\\ Sf(x_k') = \\Xi(x_k',u_k').\\]\nThus (\\ref{e7.4.1}) leads to a contradiction.\n\nConversely, suppose that conditions (1) - (3) hold.\nLet $f\\in \\operatorname{Lip}(X,E)$ with Lipschitz constant $C$.\nFirst, let us show that $Sf$ is a bounded function.\nIf not, there is a sequence $(z_n)\\in X$ so that $\\|Sf(z_n)\\| \\to \\infty$.\nBy condition (1), $(z_n)$ cannot have a separated subsequence.\nHence we may assume that $(z_n)$ converges to some $z_0\\in X$.\nThen $\\|f(z_n) - f(z_0)\\| \\leq Cd(z_n,z_0)$ and $d(z_n,z_0) \\to 0$.\nApplying condition (2) with $(x_0,u_0) = (x_1,u_1) = (z_0,f(z_0))$ and $(x_2,u_2) = (z_n,f(z_n))$, we obtain $D<\\infty$ so that \n\\[ \\|\\Xi(z_0,f(z_0)) - \\Xi(z_n,f(z_n))\\| \\leq Dd(z_0,z_n) \\text{ for all sufficiently large $n$}.\\]\nHence $(Sf(z_n)) = (\\Xi(z_n,f(z_n))$ is surely bounded, contrary to its choice.\n\n\nNow suppose that $Sf\\notin \\operatorname{Lip}(X,F)$. There are sequences $(x_n)$, $(x_n')$ in $X$ so that \n\\begin{equation}\\label{e7.5} \\|\\Xi(x_n,u_n) - \\Xi(x_n',u_n')\\| = \\|Sf(x_n) - Sf(x_n')\\|> nd(x_n,x_n') \\text{ for all $n$},\n\\end{equation}\nwhere $u_n = f(x_n)$ and $u_n' = f(x_n')$.\nSince $Sf$ is a bounded function, we must have $d(x_n,x_n') \\to 0$.\nBy using subsequences, we may assume that either $(x_n)$ converges to some $x_0$ or that $(x_n)$ is a separated sequence.\nIn the former case, since $d(x_n,x_0), d(x_n',x_0)\\to 0$, $\\|f(x_n) - f(x_n')\\|\\leq Cd(x_n,x_n')$,\n$\\|f(x_n)-f(x_0)\\| \\leq Cd(x_n,x_0)$ and $\\|f(x_n')-f(x_0)\\| \\leq Cd(x'_n,x_0)$, it follows from condition (2) that there exists $D<\\infty$ so that for all sufficiently large $n$,\n\\[ \\|\\Xi(x_n,u_n) - \\Xi(x_n',u_n')\\| = \\|\\Xi(x_n,f(x_n)) - \\Xi(x_n',f(x_n'))\\| \\leq Dd(x_n,x_n'),\\]\ncontrary to (\\ref{e7.5}).\nThe proof is similar in case $(x_n)$ is a separated sequence, using condition (3) instead.\n\\end{proof}\n\nThe next theorem is easily deduced from Theorem \\ref{t7.7}, keeping in mind that $\\varphi:X\\to Y$ is a Lipschitz homeomorphism.\n\n\\begin{thm}\\label{t7.8}\nLet $X,Y$ be complete bounded metric spaces and let $E, F$ be Banach spaces.\nSuppose that $T:\\operatorname{Lip}(X,E) \\to \\operatorname{Lip}(Y,F)$ is a biseparating map. \nThen there are a Lipschitz homeomorphism $\\varphi:X\\to Y$ and a function $\\Phi:Y\\times E\\to F$ so that \n\\begin{enumerate}\n\\item For each $y\\in Y$, $\\Phi(y,\\cdot):E\\to F$ is a bijection with inverse $\\Psi(x,\\cdot):F\\to E$, where $\\varphi(x) =y$.\n\\item $Tf(y) = \\Phi(y,f(\\varphi^{-1}(y)))$ and $T^{-1}g(x) = \\Psi(x,g(\\varphi(x)))$ for all $f\\in \\operatorname{Lip}(X,E)$, $g\\in \\operatorname{Lip}(Y,F)$ and $x\\in X$, $y\\in Y$.\n\\item Let $(x_n)$ be a separated sequence in $X$. For any bounded sets $B$ in $E$ and $B'\\in F$, there is a finite set $N \\subseteq {\\mathbb N}$ so that $\\bigcup_{n\\notin N}\\Phi(\\varphi(x_n),B)$ and $\\bigcup_{n\\notin N}\\Psi(x_n,B')$ are bounded.\n\n\\item Suppose that $x_0\\in X$, $u_0\\in E$, $v_0\\in F$ and $C<\\infty$. There exist $r >0$ and $D<\\infty$ so that \n\\[\n \\|\\Phi(\\varphi(x_1),u_1) - \\Phi(\\varphi(x_2),u_2)\\|, \\|\\Psi(x_1,v_1) - \\Psi(x_2,v_2)\\|\\leq Dd(x_1,x_2) \n\\]\nwhenever $\\|u_1-u_2\\|, \\|v_1-v_2\\| \\leq Cd(x_1,x_2)$, $\\|u_i-u_0\\|, \\|v_i-v_0\\| \\leq Cd(x_i,x_0)$ and $d(x_i,x_0) \\leq r$, $i=1,2$.\n\\item Let $(x_n)$ be a separated sequence in $X$ and $(u_n), (v_n)$ be bounded sequences in $E$ and $F$ respectively.\nFor any $C<\\infty$, there exist $r>0$ and $D<\\infty$ so that \n\\[ \\|\\Phi(\\varphi(x_n'),u'_n) - \\Phi(\\varphi(x_n),u_n)\\|,\\ \n\\|\\Psi(x_n',v'_n) - \\Psi(x_n,v_n)\\| \\leq Dd(x_n',x_n)\\]\nfor all $n$,\nwhenever $\\|u_n'-u_n\\|, \\|v_n'-v_n\\| \\leq Cd(x_n',x_n)$ and $d(x_n',x_n) \\leq r$\nfor all $n$.\n \\end{enumerate}\n Conversely, if $\\varphi$, $\\Phi$ satisfy conditions (1)-(5) and $T$ is defined by (2), then $T$ is a biseparating map from $\\operatorname{Lip}(X,E)$ onto $\\operatorname{Lip}(Y,F)$.\n\\end{thm}\n\n\\subsection{A property of Lipschitz sections}\n\nLet $X$ be a bounded metric space and let $E$ and $F$ be Banach spaces.\nTheorem \\ref{t7.7} characterizes the ``section maps'' $\\Xi:X\\times E\\to F$ so that $Sf(x) = \\Xi(x,f(x))$ is Lipschitz whenever $f\\in \\operatorname{Lip}(X,E)$.\nAn example in \\cite[p.~190]{AZ}, where $X= [0,1]$ with the H$\\ddot{\\text{o}}$lder metric $d(x,y)= |x-y|^\\alpha$, $0<\\alpha < 1$, and $E = F = {\\mathbb R}$, shows that for a given $x\\in X$, the function $\\Xi(x,\\cdot):E\\to F$ need not be continuous.\nNevertheless, in this subsection, we will show that if $x$ is an accumulation point of $X$, then there is a dense open set $O$ in $E$ so that $\\Xi(x,\\cdot)$ is continuous on $O$.\nLet $\\Xi:X\\times E\\to F$ be a ``Lipschitz section''. Taking $(x_1,u_1) = (x,u)$ and $(x_2,u_2) = (x_0,u_0)$ in Theorem \\ref{t7.7}(2) yields the next lemma.\n\n\n\\begin{lem}\\label{l7.10}\nLet $(x_0,u_0) \\in X\\times E$ and let $v_0 = \\Xi(x_0,u_0)$.\nFor any $C<\\infty$, there exists $n = n(x_0,u_0,C) \\in{\\mathbb N}$ so that if $(x,u)\\in X\\times E$,\n$\\|u-u_0\\| \\leq Cd(x,x_0)$ and $d(x,x_0) < \\frac{1}{n}$, then \n\\[ \\|v-v_0\\|\\leq nd(x,x_0), \\text{ where $v = \\Xi(x,u)$}.\\]\n\\end{lem}\n\n\n\\begin{thm}\\label{t7.11}\nLet $x_0$ be an accumulation point of $X$.\nThere is a dense open set $O$ in $E$ so that $\\Xi(x_0,\\cdot)$ is continuous on $O$.\n\\end{thm}\n\n\\begin{proof}\nIn the notation of Lemma \\ref{l7.10}, for each $n\\in {\\mathbb N}$, let \n\\[ A_n = \\{u_0\\in E: n(x_0,u_0,1) \\leq n\\}.\\]\nBy the lemma, $E= \\bigcup_n \\overline{A_n}$.\nSince $E$ is a complete metric space, $O = \\bigcup_n \\operatorname{int}\\overline{A_n}$ is a dense open set in $E$.\nTo complete the proof of the theorem, let us show that $\\Xi(x_0,\\cdot)$ is continuous on $O$.\nClearly, it suffices to show that $\\Xi(x_0,\\cdot)$ is continuous on each $\\operatorname{int}\\overline{A_n}$.\nFix $N\\in {\\mathbb N}$. Suppose that $(u_n)$ is a sequence in $\\operatorname{int}\\overline{A_N}$ converging to $u_0 \\in \\operatorname{int}\\overline{A_N}$.\n\n\\medskip\n\n\\noindent\\underline{Claim}. There is a sequence $(u_n')$ in $A_N$ so that \n\\[ \\|u_n'-u_n\\| , \\|\\Xi(x_0,u_n') - \\Xi(x_0,u_n)\\| \\to 0.\\]\n\n\\medskip\n\nConsider a given $n\\in {\\mathbb N}$. Since $\\Xi(x,u_n)$ is a Lipschitz function of $x$ and $x_0$ is an accumulation point, there exists $x\\in X$ so that $0< Nd(x,x_0) < \\frac{1}{n}$ and that $\\|\\Xi(x,u_n) -\\Xi(x_0,u_n)\\| < \\frac{1}{n}$.\nAs $u_n\\in \\overline{A_N}$, there exists $u_n'\\in A_N$ so that $\\|u_n'-u_n\\| \\leq d(x,x_0) < \\frac{1}{nN}$.\nNote that $n(x_0,u_n',1) \\leq N$. Hence the condition $\\|u_n-u_n'\\| \\leq d(x,x_0)< \\frac{1}{N}$ implies\n\\[ \\|\\Xi(x,u_n) - \\Xi(x_0,u_n')\\| \\leq Nd(x,x_0)< \\frac{1}{n}.\\]\nTherefore,\n\\begin{align*}\n\\|\\Xi(x_0,u_n)- &\\Xi(x_0,u_n')\\| \\\\&\\leq \\|\\Xi(x_0,u_n)- \\Xi(x,u_n)\\| + \\|\\Xi(x,u_n)- \\Xi(x_0,u_n')\\|\\\\\n& < \\frac{2}{n}.\n\\end{align*}\nThis completes the proof of the claim.\n\n\\medskip\n\nIn view of the claim, in order to prove the continuity of $\\Xi(x_0,\\cdot)$ at $u_0$, it suffices to show that $\n\\Xi(x_0,u_n') \\to \\Xi(x_0,u_0)$.\nLet $\\varepsilon > 0$ be given. As before, one can choose $x'$ so that $0< d(x',x_0) < \\frac{1\\wedge \\varepsilon}{N}$ and that \n$\\|\\Xi(x',u_0) - \\Xi(x_0,u_0)\\| < \\varepsilon$.\nFor all sufficiently large $n$, $\\|u_n'-u_0\\| < d(x',x_0)$.\nOnce again, $\\|u_0-u_n'\\| \\leq d(x',x_0) < \\frac{1}{N}$ implies\n\\[ \\|\\Xi(x',u_0) - \\Xi(x_0,u_n')\\| \\leq Nd(x',x_0) < \\varepsilon.\\]\nTherefore,\n\\begin{align*}\n\\|\\Xi(x_0,u_n')- &\\Xi(x_0,u_0)\\| \\\\&\\leq \\|\\Xi(x_0,u_n')- \\Xi(x',u_0)\\| + \\|\\Xi(x',u_0)- \\Xi(x_0,u_0)\\|\n < 2\\varepsilon\n\\end{align*}\nfor all sufficiently large $n$.\n\\end{proof}\n\n\n\n\n\n\\section{Comparisons}\\label{s9}\n\nWe close with some results comparing different types of spaces under nonlinear biseparating maps.\nThroughout this section, $X,Y$ will be complete metric spaces and $E$, $F$ will be Banach spaces.\n\n\\begin{prop}\\label{p8.1}\nLet $T:A(X,E)\\to \\operatorname{Lip}(Y,F)$ be a biseparating map, where $Y$ is bounded. If $A(X,E) = U(X,E)$, then $X$ is separated. If $A(X,E) = U_*(X,E)$, then both $X$ and $Y$ are separated.\n\\end{prop}\n\n\\begin{proof}\nNormalize $T$ by taking $T0 = 0$. Suppose that $A(X,E)$ is either $U(X,E)$ or $U_*(X,E)$. First assume, if possible, that there is a convergent sequence $(x_n)$ in $X$ consisting of distinct points.\nLet $x_0$ be its limit, which we may assume to be distinct from all $x_n$'s.\nSet $y_n = {\\varphi}(x_n)$, $n\\in {\\mathbb N}\\cup\\{0\\}$, and $r_n = d(y_n,y_0)$, $n\\in {\\mathbb N}$.\nSince $r_n \\to 0$, without loss of generality, we may further assume that $r_{n+1} < \\frac{r_n}{3}$ for all $n\\in {\\mathbb N}$.\nFix a nonzero vector $b\\in F$. For each $m\\in{\\mathbb N}$, define $g_m: Y\\to F$ by \n\\[ g_m(y) = \\begin{cases}\n\\bigl(1- \\frac{2d(y,y_n)}{r_n}\\bigr)mr_nb &\\text{if $d(y,y_n) < \\frac{r_n}{2}$, $n\\in {\\mathbb N}$,}\\\\\n0 &otherwise.\n\\end{cases}\\]\nThen $g_m\\in \\operatorname{Lip}(Y,F)$, ${g_m}(y_n) = mr_nb$ for all $n\\in {\\mathbb N}$ and ${g_m}(y_0) =0$.\nBy Proposition \\ref{p4.2}, ${T^{-1}g_m}(x_0) =0$.\nBy continuity of $g_m$, there is an increasing sequence $(n_m)$ so that $T^{-1}g_m(x_{n_m}) \\to 0$.\nThus, there is a function $f\\in U_*(X,E)\\subseteq A(X,E)$ so that $f(x_{n_m}) = T^{-1}g_m(x_{n_m})$ for all $m\\in {\\mathbb N}$ and $f(x_0) =0$.\nBy Proposition \\ref{p4.2}, \n\\[ {Tf}(y_{n_m}) = {g_m}(y_{n_m}) = mr_{n_m}b\\text{ and } {Tf}(y_0) = 0.\\]\nHowever, $Tf$ is Lipschitz on $Y$.\nWe have reached a contradiction since \n\\[ \\|{Tf}(y_{n_m}) - {Tf}(y_0)\\| = mr_{n_m}b = md(y_{n_m},y_0)b.\\]\nThis shows that $X$ does not contain any nontrivial convergent sequence.\n\nIf $X$ is not separated, there are points $x_n,x_n'\\in X$ so that $0