It is well-known that if a topological space $Y$ is
locally compact (not necessarily Hausdorff),
then the map
$$
\operatorname{Hom}(X \times Y, Z) \to
\operatorname{Hom}(X, Z^Y)
$$
(here we use the compact-open topology for $Z^Y$)
is bijective for arbitrary topological spaces $X, Z$.
**Does the converse of this hold?**


If we impose no restriction on the topolgy for $Z^Y$,
$Y$ satisfies this condition (exponentiable in $\mathsf{Top}$)
if and only if $Y$ is [core-compact](https://ncatlab.org/nlab/show/exponential+law+for+spaces).
Hence $Y$ satisfies the condition if and only if
$Y$ is core-compact and the topology on $Z^Y$ given
[here](https://ncatlab.org/nlab/show/exponential+law+for+spaces)
coincides with the compact-open topology for every topological space $Z$.


### Edit


I use the definitions given [here](https://ncatlab.org/nlab/show/compact-open+topology)
for the local compactness and the compact-open topology.
Under this definition, the map is bijective for every $X, Z$.


For Hausdorff (actually sober is enough) spaces, core-compact implies
locally compact.



Exponential law w.r.t. compact-open topology
The answer is yes. Let $C(Y,Z)$ denote the compact-open topology on $Z^Y$ and let $E(Y,Z)$ denote the exponential topology on $Z^Y$. Let $S$ denote the Sierpinski space. In the following, we will identify $S^Y$ with the set of open subsets of $Y$ in the natural way.


**Claim:** If $Y$ is core-compact and $C(Y,S)$ refines $E(Y,S)$, then $Y$ is locally compact.


**Proof:** Suppose not. Then after passing to an open subspace if necessary, there is a point $y \in Y$ such that no open neighborhood of $y$ is contained in a quasicompact set. Because $Y$ is core-compact, there is an open neighborhood $U \ni y$ with $U << Y$. The set $\{V \in S^Y \mid U << V\}$ is an open neighborhood of $Y$ in $E(Y,S)$. Because $C(Y,S)$ refines $E(Y,S)$, there is a quasicompact set $A \subseteq Y$ such that $A \subseteq V \Rightarrow U << V$ for all $V \in S^Y$. We may assume without loss of generality that $A$ is an intersection of open sets (for if $A$ is quasicompact, so is the intersection of all open sets containing $A$). In particular, $A \subseteq V \Rightarrow U \subseteq V$, so $U \subseteq A$. But then $A$ is a quasicompact set containing an open neighborhood of $y$, contradicting the choice of $y$.


