Show composition mapping is continuous with compact-open topology

Let $$X$$ be a compact Hausdorff space, and $$H(X)$$ be the set of homeomorphisms from $$X$$ to $$X$$, with the compact-open topology.

Prove that the mapping $$h:H(X)\times H(X)\rightarrow H(X)$$, $$h(f,g)=f\circ g$$ is continuous.

Note, if $$C(X,X)$$ is the set of all continuous mappings from $$X$$ to $$X$$, the compact-open topology on $$C(X,X)$$ is generated by subsets of the form $$B(K,U)=\{f:f(K)\subset U\}$$ where $$K$$ is compact in $$X$$ and $$U$$ is open in $$X$$.

I honestly have no clue how to work with the compact open topology and would appreciate any hints.

Let's take $$U$$ open in $$H(X)$$. I want to show that $$h^{-1}(U)$$ is open in $$H(X)\times H(X)$$.

I believe I am overthinking this, and apologize for the lack of work, I am just really confused how to show this. Any help would be much appreciated.

• Two things. First, the function whose continuity you are proving is $h$, not $f$. Second, for proving continuity, you don't always have to prove that the inverse image of every open set is open; you can instead prove that the inverse image of every basis element is open. – Lee Mosher Nov 20 '18 at 0:39

We will use the characterization of continuity that says that a map $$f:X\rightarrow Y$$ is continuous if for all $$x\in X$$ and open $$U\subseteq Y$$ such that $$f(x)\in U$$ we have that there is an open $$V\subseteq X$$ such that $$x\in V$$ and $$f(V)\subseteq U$$.

Let $$B(K,U)\subseteq H(X)$$ be given where $$K\subseteq X$$ is closed (compact) and $$U\subseteq X$$ is open. Then, if $$g\circ f\in B(K,U)$$ we have that $$gf(K)\subseteq U$$. Because $$g$$ is continuous we have that $$g^{-1}(U)$$ is open in $$X$$ can contains $$f(K)$$. Because $$X$$ is normal (recall that compact Hausdorff spaces are normal) there is an open $$V\subseteq X$$ such that $$f(K)\subseteq V\subseteq\overline{V}\subseteq g^{-1}(U)$$. Because $$X$$ is compact we have that $$\overline{V}$$ is compact. Moreover it is clear that $$g(\overline{V})\subseteq U$$. We then claim the following:

$$(B(\overline{V},U)\circ B(K,V))\subseteq B(K,U)$$

To see this we simply let $$(k,l)\in B(\overline{V},U)\times B(K,V)$$. Then, by definition $$l(K)\subseteq V$$ and $$k(\overline{V})\subseteq U$$. Then we can easily see that $$(k\circ l)(K)\subseteq U$$. It is also clear that $$g\circ f\in B(\overline{V},U)\times B(K,V)$$. Therefore composition is continuous.

Note: This result generalizes quite easily to the following situation as seen in an exercise of Munkres. Let $$Y$$ be locally compact Hausdorff, and $$X$$ and $$Z$$ general spaces. Also let $$\mathcal{C}(X,Y),\,\mathcal{C}(Y,Z),$$ and $$\mathcal{C}(X,Z)$$ denote the spaces of continuous functions from the respective spaces with the compact open topology. Then the composition map

$$\mathcal{C}(X,Y)\times\mathcal{C}(Y,Z)\rightarrow\mathcal{C}(X,Z)$$

is continuous.