Does "$\forall Z(C(X,Z) \cong C(Y,Z))$" imply $X\cong Y$? If $X, Y$ are topological spaces, let $C(X,Y)$ denote the collection of continuous maps $f: X\to Y$, endowed with the compact-open topology.
Assume that we are given topological spaces $X,Y$ such that for all spaces $Z$ we have $C(X,Z) \cong C(Y,Z)$. Does this imply that $X\cong Y$?
 A: Let $Z$ be the Sierpinski 2-point space.  Then the underlyying set of $C(X,Z)$ is naturally identified with the collection of open sets of $X$, and the specialization order from the compact-open topology is the inclusion order.  Thus we can recover the poset of open sets of $X$ from $C(X,Z)$.  So if you restrict your question to sober spaces, the answer is yes.
However, for arbitrary spaces the question seems more subtle.  For instance, let's consider the case where $X$ and $Y$ are indiscrete.  If $X$ is indiscrete, then for any $Z$, it is easy to see that $C(X,Z)$ looks like $Z$ except each maximal indiscrete subspace $A\subseteq Z$ has been replaced by $A^X$.  In particular, if $X$ and $Y$ are sets such that $|A^X|=|A^Y|$ for all sets $A$, then $C(X,Z)\cong C(Y,Z)$ for all $Z$, where $X$ and $Y$ are equipped with the indiscrete topology.  So the question reduces to the following: if $\kappa$ and $\lambda$ are cardinals such that $\mu^\kappa=\mu^\lambda$ for all cardinals $\mu$, must $\kappa=\lambda$?  The answer is yes: if $\kappa<\lambda$ and $\mu$ is a strong limit cardinal of cofinality $\kappa^+$, then $\mu^\kappa=\mu$ but $\mu^\lambda>\mu$.
So the answer is also yes if $X$ and $Y$ are both indiscrete.  However, it is noteworthy that this argument might require you to take a very large $Z$ to distinguish $X$ and $Y$.  For instance, when $|X|=\aleph_0$ and $|Y|=\aleph_1$, if $2^{\aleph_0}=2^{\aleph_1}=\aleph_2$ and GCH holds above $\aleph_1$, then $C(X,Z)\cong C(Y,Z)$ for all spaces $Z$ of cardinality less than $\aleph_{\omega_1}$.  So if the answer is yes in general, this is a sign that proving it probably isn't going to be very easy.
