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, \mathcal{V}[A,B]\subseteq\mathcal{C}[W,T]=\left\{f\in\mathcal{C}[W,T]:f(A)\subseteq B,\text{ }A\text{ compact and }B\text{ op}e\text{n}\right\}

Let X,Y,Z be topological spaces and Y locally compact Hausdorff. Then, if \mathcal{C}[X,Y],\mathcal{C}[Y,Z] and \mathcal{C}[X,Z] 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.
Proof: We need a quick little lemma (you can ignore me if you know/believe it).

Lemma: Let S be a locally compact Hausdorff space and G\subseteq S compact and N any neighborhood of it. Then, there exists some neighborhood O of G such that \overline{O} is compact and G\subseteq O\subseteq\overline{O}\subseteq N.

Proof: Now, since every locally compact Hausdorff space is regular (just consider that \iota:S\hookrightarrow S_{\infty} is an embedding and S_{\infty} is compact Hausdorrf thus normal thus regular) we have that for each g\in G there exists a neighborhood E_g of it such that g\in E_{g}\subseteq\overline{E_g}\subseteq N. Doing this for each g\in G forms an open subcover and by the compactness of S it must have a finite subcover E_{g_1},\cdots,E_{g_n} and so since for each g_k we have that E_{g_k}\subseteq \overline{E_{g_k}}\subseteq N we have that

G\subseteq E_{g_1}\cup\cdots\cup E_{g_n}\subseteq \overline{E_{g_1}}\cup\cdots\cup\overline{E_{g_n}}  =\overline{E_{g_1}\cup\cdots\cup E_{g_n}}\subseteq N

and so G has a neighborhood (call it \Sigma) such that G\subseteq\Sigma\subseteq\overline{\Sigma}\subsete  q N. But, by local compactness we have that for each g\in G there exists a neighborhood B_g such that \overline{B}_g is compact. Using the same concept as last time we may cover G with finitely many of these B_g's, say B_{g_1},\cdots,B_{g_m} and so

G\subseteq B_{g_1}\cup\cdots\cup B_{g_m}

and since the finite union of compact subspaces is compact and \overline{B_{g_1}}\cup\cdots\cup \overline{B_{g_m}}=\overline{B_{g_1}\cup\cdots\cup B_{g_m}} we may conclude that G has a neighborhood \Omega whose closure is compact. So, \Sigma\cap\Omega is a neighborhood of G whose closure is a closed subspace of \overline{\Omega} and thus compact and such that

G\subseteq\Sigma\cap\Omega\subseteq\overline{\Sigm  a\cap\Omega}\subseteq\overline{\Sigma}\subseteq N

The conclusion follows. \blacksquare

It suffices to check that for any subbasic open neighborhood of an element of the image of F there exists a neighborhood of the inverse image of that element in the domain which maps entirely into it. So, let f\circ g\in\mathcal{C}[X,Z] be arbitrary and \mathcal{V}[C,U] any neighborhood of it. Then, f(g(C))\subseteq U and thus g(C)\subseteq f^{-1}(U). But, notice that g(C) being the continuous image of a compact space is compact and that both g(C),f^{-1}(U) live in Y and so by the lemma there exists some neighborhood O of g(C) whose closure is compact and g(C)\subseteq O\subseteq \overline{O}\subseteq g^{-1}(U). So, clearly \mathcal{V}[C,O] and \mathcal{V}[\overline{O},U] are neighborhoods of f,g in \mathcal{C}[X,Y],\mathcal{C}[Y,Z] respectively and I finally claim that F\left(\mathcal{V}[C,O]\times\mathcal{V}[\overline{O},U]\right)\subseteq \mathcal{V}[C,U] but this is trivial since if (\varphi,\psi)\in\mathcal{V}[C,O]\times\mathcal{V}[\overline{O},U] then \varphi(C)\subseteq O and \psi(\overline{O})\subseteq U and so \psi(\varphi(C))\subseteq\psi(O)\subseteq\psi(\ove  rline{O})\subseteq U and so \psi\circ\varphi\in\mathcal{V}[C,U] and since f\circ g and \mathcal{V}[C,U] were arbitrary the conclusion follows.