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. Hence $Y$ satisfies the condition if and only if $Y$ is core-compact and the topology on $Z^Y$ given here coincides with the compact-open topology for every topological space $Z$.

### Edit

I use the definitions given here 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.

yesif you restrict to the category of Hausdorff spaces. $\endgroup$ – Tim Campion♦ Aug 4 '18 at 20:18