7

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.

Community
  • 1
The math god
  • 759
  • 1
    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 00:39

1 Answers1

6

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.

Robert Thingum
  • 1,918
  • it’s an old question, but how is the hausdorff condition actually needed? to me it seems like you only need $Y$ to be locally compact in the neighbourhood base sense: then given a compact set $K ⊆ Y$ and an open set $U' ⊆ Y$ (with $U' = g^{-1}(U)$ in the proof), there are compact neighbourhoods around each point in $K$ within $U'$. as $K$ is compact, it is covered by (the interiors of) finitely many such compact neighbourhoods, whose finite union $C$ is hence again compact, so $C$ is a compact neighbourhood of $K$ in $U'$ (yielding $B(C, U) ∘ B(K, C^∘) ⊆ B(K,U)$). – windfish Jun 05 '22 at 11:54
  • 1
    @windfish I assume that by "locally compact in the neighbourhood base sense" you mean that every point $x$ admits a neighbourhood basis made up of compact sets? If so, then yes I believe you are correct. – Robert Thingum Jun 06 '22 at 09:45