1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Lee Mosher Nov 20 '18 at 0:39
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.