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Let $X,Y,Z$ be topological spaces. It is well-known that if $F:X\times Y\to Z$ is a continuous map, we can define a map $$\overline{F}:X\to C(Y,Z) \\\overline{F}(x)(y)=F(x,y)$$ where $C(Y,Z)$ is the topological space of all continuous maps from $Y$ to $Z$ equipped with the compact-open topology, and this induced map is continuous.

If $Y$ is locally compact Hausdorff, the converse holds, i.e. if a function $\overline{F}:X\to C(Y,Z)$ is continuous, and related to $F$ by the equation above, then $F$ is also continuous. This converse seems a bit unsatisfying, so I was wondering:

Does the converse also hold under some weaker assumptions? What is the best known theorem in this regard?

Also, I suspect the problem might be with the choice of topology on $C(Y,Z)$. As nice as the compact-open topology might be, I imagine there might be some other (more or less natural) choice of topology availible for which the continuity of $F$ and continuity of $\overline{F}$ would be equivalent.

Does there exist such a topology?

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Have a look at this MO answer and the refernces therein – Juan S May 21 '12 at 1:30
Also Wikipedia has an entry that claims that $Y$ need by locally compact Hausdorff, but no reference: – Juan S May 21 '12 at 1:32
up vote 3 down vote accepted

Looking at the paper referenced in the Mathoverflow question, the first line of the abstract states:

It is well-known that a Hausdorff space is exponentiable if and only if it is locally compact, and that in this case the exponential topology is the compact-open topology.

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Thanks, the article you link answers pretty much everything I wanted to know (and more). – Dejan Govc May 21 '12 at 2:18

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