## Compact-Open Topology

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

Problem:
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

$F:\mathcal{C}[X,Y]\times\mathcal{C}[Y,Z]\to\mathcal{C}[X,Z]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.