Munkres page 289 #7 Show that if $Y$ is locally compact Hausdorff, then composition of maps
$p: C(X,Y)\times C(Y,Z)\rightarrow C(X,Z)$ is continuous, provided the compact-open topology is used throughout.
Prove: let $C$ be compact subspace of $X$ and $U$ be open subset of $Z$. Then $S(C,U)$ is an open subset of $C(X,Z)$. Let $(h,g)\in p^{-1}(S(C,U))$ where $h\in C(X,Y)$ and $g\in C(Y,Z)$ such that $g\circ h=f$ for some $f\in S(C,U)$. Then $h(C)$ is compact and is contained in the open subset $g^{-1}(U)$ of $Y$. 
We want to find an open set $V\subset Y$ such that  $h(C)\subset V$ and $g(\bar{V})\subset U$. Then $S(C,V)\times S(\bar{V}, U)$ open neighbourhood of $h\times g$ is contained in $p^{-1}( S(C,U))$ hence $p^{-1}( S(C,U))$ is open in $C(X,Y)\times C(Y,Z)$. Therefore, $p$ is continuous.
Let $y\in g^{-1}(U)/h ( C ) $. Since $Y$ is Hausdorff, there exist open subset $W_1$ and $W_2$ of $Y$ such that $y\in W_1$ and $h( C )\in W_2$ and $W_1\cap W_2=\emptyset$. Let $V=W_2\cap g^{-1}(U)$. Then $V$ containing $h( C )$ is properly contained in $g^{-1}(U)$.
But I do not know how to prove that $\bar{V}\subset g^{-1}(U)$.
Does any one have any ideas?
 A: 
But I do not know how to prove that $\bar{V}\subset g^{-1}(U)$.

With $V = W_2 \cap g^{-1}(U)$, you will in general neither have $\overline{V}$ compact nor $\overline{V}\subset g^{-1}(U)$.
To obtain such a $V$, you need to use that $K = h(C)$ is compact and contained in $W = g^{-1}(U)$. Since $Y$ is a locally compact Hausdorff space, every neighbourhood of a compact set contains a compact neighbourhood of that set.
Namely, for every $y\in K$, there is a compact neighbourhood $K_y \subset W$ of $y$. Let $V_y = \overset{\large\circ}{K}_y$. Then
$$\mathfrak{V} = \{ V_y : y \in K\}$$
is an open cover of $K$, therefore it has a finite subcover $\{ V_{y_k} : 1 \leqslant k \leqslant n\}$ for some $y_1,\dotsc,y_n\in K$.
$$V = \bigcup_{k=1}^n V_{y_k}$$
is then an open neighbourhood of $K$, and
$$\overline{V} = \bigcup_{k=1}^n \overline{V_{y_k}} \subset \bigcup_{k=1}^n K_{y_k} \subset W.$$
A: Old and already answered question but locally compact Hausdorff implies regular. So if you already have a $g^{-1}(U)$ open neighborhood of $y$ in $Y$, then by regularity there is a neighborhood $V$ of $y$ such that $\overline{V}\subset g^{-1}(U)$.
