This problem has been bothering me. I have come up with a solution but I'm not sure if it's correct. I'm not sure I actually believe anyone will actually read this (I know it's long), but it's worth a try.

Also,

Problem:

Proof: We need a quick little lemma (you can ignore me if you know/believe it).Let be topological spaces and locally compact Hausdorff. Then, if and are given the compact-open topologies the the map

g,f)\mapsto f\circ g" alt="F:\mathcal{C}[X,Y]\times\mathcal{C}[Y,Z]\to\mathcal{C}[X,Z]g,f)\mapsto f\circ g" />

is continuous.

Lemma:Let be a locally compact Hausdorff space and compact and any neighborhood of it. Then, there exists some neighborhood of such that is compact and .

Proof:Now, since every locally compact Hausdorff space is regular (just consider that is an embedding and is compact Hausdorrf thus normal thus regular) we have that for each there exists a neighborhood of it such that . Doing this for each forms an open subcover and by the compactness of it must have a finite subcover and so since for each we have that we have that

and so has a neighborhood (call it ) such that . But, by local compactness we have that for each there exists a neighborhood such that is compact. Using the same concept as last time we may cover with finitely many of these 's, say and so

and since the finite union of compact subspaces is compact and we may conclude that has a neighborhood whose closure is compact. So, is a neighborhood of whose closure is a closed subspace of and thus compact and such that

The conclusion follows.

It suffices to check that for any subbasic open neighborhood of an element of the image of there exists a neighborhood of the inverse image of that element in the domain which maps entirely into it. So, let be arbitrary and any neighborhood of it. Then, and thus . But, notice that being the continuous image of a compact space is compact and that both live in and so by the lemma there exists some neighborhood of whose closure is compact and . So, clearly and are neighborhoods of in respectively and I finally claim that but this is trivial since if then and and so and so and since and were arbitrary the conclusion follows.