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 $Z$.


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.

  • $\begingroup$ Maybe it's worth noting that according to the same nlab page you link to, the exponential topology on $Z^Y$ coincides with the compact-open topology if $Y$ and $Z$ are Hausdorff. But also according to the nlab page, for Hausdorff spaces, core-compactness coincides with local compactness. So the answer is yes if you restrict to the category of Hausdorff spaces. $\endgroup$ – Tim Campion Aug 4 '18 at 20:18
  • $\begingroup$ Actually, if $Y$ is not Hausdorff, precisely which definition of the compact-open topology are you using? $\endgroup$ – Tim Campion Aug 4 '18 at 23:14
  • $\begingroup$ @TimCampion I added their definitions. $\endgroup$ – B. W. Aug 5 '18 at 2:39
  • $\begingroup$ Okay, it appears the definition agrees with the one I used below. $\endgroup$ – Tim Campion Aug 5 '18 at 5:04

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 $y$.

  • $\begingroup$ Finite spaces are locally compact by definition. $\endgroup$ – B. W. Aug 5 '18 at 5:28
  • $\begingroup$ Wow, I guess I misread -- I thought we were talking about the locally compact Hausdorff condition. Do you have a reference for the first sentence of your question in the non-Hausdorff case? $\endgroup$ – Tim Campion Aug 5 '18 at 5:32
  • $\begingroup$ I think my new answer addresses your actual question. Sorry for the confusion! $\endgroup$ – Tim Campion Aug 5 '18 at 6:51
  • $\begingroup$ Thanks! I don't know a reference for the first sentence of my question, but its proof is the same as the Hausdorff case. $\endgroup$ – B. W. Aug 5 '18 at 7:13

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