Suppose that $(X,d_X)$ and $(Y,d_Y)$ are complete metric spaces and let $\phi:X\rightarrow Y$ be a homeomorphism. Let $C(Z,Z)$ be the set of continuous functions from $Z$ to itself (where Z is a standing for either $X$ or for $Y$) and endow it with the compact-open topology.
Define the induced map between function spaces by \begin{align} \Phi:&C(X,X)\rightarrow C(Y,Y)\\ & f\mapsto \phi\circ f\circ \phi^{-1}. \end{align}
Is $\Phi$ then also a homeomorphism?
Update: Tentative Argument
Since $\phi$ is a homeomorphism, then:
- $K$ is compact in $X$ if and only if $\phi(K)$ is compact in $Y$,
- $\tilde{K}$ is compact in $Y$ if and only if $\phi^{-1}(\tilde{K})$ is compact in $X$,
- $U$ is open in $X$ if and only if $\phi(U)$ is open in $Y$,
- $\tilde{U}$ is open in $Y$ if and only if $\phi^{-1}(\tilde{U})$ is open in $X$,
- $\phi$ and $\phi^{-1}$ are bijections so they preserve inclusions of sets,
Therefore, for every $f\in C(X,X)$ such that $f(K)\subseteq U$, where $K,U$ are as above, we have that $ f(K)\subseteq U $ if and only if there is a compact of the form $\tilde{K}=\phi^{-1}(K)\subseteq Y$ such that $$ f\circ \phi^{-1}(K)\subseteq U \Leftrightarrow \phi\circ f\circ \phi^{-1}(K)\subseteq \phi(U), $$ and by the above remarks every open subset $V$ of $Y$ can be written in the form $\phi(U)$ for some open subset $U$ of $X$.
Therefore, there is a bijections between the subases of the compact-open topologies; whence $\Phi$ is a homeomorphism.
It is true with any assumptions on $X,Y$. Let us write $Y^X$ for the set of continuous functions $f : X \to Y$ which is endowed with the compact-open topology. This topology depends on the interpretation of "compact". If we understand "compact" to include "Hausdorff", then the resulting compact-open topology is in general coarser than that based on not necessarily Hausdorff compact subsets $K \subset X$. However, if $X$ is Hausdorff, both variants agree.
Given maps $u : Y \to Z, v : T \to X$, it is well-known (and easy to verify) that $$u_* : Y^X \to Z^X, g_*(f) = u \circ f$$ is continuous and $$v^* : Y^X \to Y^T, v^*(f) = f \circ v$$ is continuous provided images $v(C)$ of compact $C \subset T$ are compact in the given interpretation of "compact". Thus, in the interpretation not including Hausdorff it is always true, but in the other interpretation it may fail. Anyway, if $v$ is a homeomorphism, then it is true also in the second case.
Now let $\phi : X \to Y$ be a homeomorphism. Then $\phi_* : X^X \to Y^X$ and $(\phi^{-1})_* : Y^X \to X^X$ are inverse homeomorphism since $(\phi^{-1})_* \circ \phi_* = id$ and $\phi_* \circ (\phi^{-1})_* = id$. Similarly $(\phi^{-1})^* : Y^X \to Y^Y$ is a homeomorphism. This shows that $\Phi = (\phi^{-1})^* \circ \phi_*$ is a homeomorphism.
Remark: As observed by freakish, we know that $\Phi = (\phi^{-1})^* \circ \phi_*$ is continuous. Similarly $\Psi = \phi^* \circ (\phi^{-1})_*$ is continuous and obviously $\Psi \circ \Phi = id$ and $\Phi \circ \Psi = id$.