So I'm just learning about the compact-open topology and am trying to show that for a compact, Hausdorff space ,$X$, the group of homeomorphisms of $X$, $H(X)$, is a topological group with the compact open topology. This topology has a subbasis of sets $\{f\in H(X):f(C)\subseteq V\}=S(C,V)$ for compact $C$, open $V$. This is my first attempt at getting to know this topology, so I'd appreciate some help with the part of a proof I have so far, possibly a better way to go about proving this, and any other help understanding this topology.

First, let $c:H(X)\times H(X)\rightarrow H(X)$ be composition. For $f,g\in H(X)$, let $S(C,V)$ be a subbasis set with $f\circ g\in S(C,V)$. So $f(g(C)\subseteq V$ which means $f\in S(g(C),V)$ and $g\in S(C,f^{-1}(V)$. The product of these open sets works since if $h_1(C)\subseteq f^{-1}(V)$ and $h_2(g(C))\subseteq V$, then $h_2\circ h_1(C)\subseteq V$.

Then let $i:H(X)\rightarrow H(X)$ be inversion. Take a subbasis set $O=\{g:g(C)\subseteq U\}$ and consider $i^{-1}(O)=\{g^{-1}:g(C)\subseteq U\}$. If $h^{-1}\in i^{-1}(O)$, then one thing we have is that $h(C)\subset U$, but I'm not totally sure what to do with this. This part seems like it should be easier, but I am just not seeing it.


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    $\begingroup$ Can I ask how do you show that $h_2h_1(C) \subset V$, it is not so clear why this is true. $\endgroup$ – Keith Feb 19 '16 at 0:53

Here you don't need much topology, it boils down to pure manipulation of sets and bijective functions, and the two following facts: $(*)$ closed subsets of compact spaces are compact, and $(**)$ compact subsets of Hausdorff spaces are closed. Just take complements and use the fact that $h$ is by definition a bijection, so $$\begin{array}{RCL} h\in S(C,U) & \Longleftrightarrow & h(C)\subset U\\ & \Longleftrightarrow & X\setminus U\subset \underbrace{X\setminus h(C)}_{=h(X\setminus C)}\\ & \Longleftrightarrow & h^{-1}(X\setminus U)\subset X\setminus C \\ & \Longleftrightarrow & h^{-1}\in S(X\setminus U,X\setminus C) \end{array}$$ (which is a subbasic open neighborhood) i.e. $$\mathrm{inv}^{-1}( S(C,U))=\mathrm{inv}(S(C,U))=S(X\setminus U,X\setminus C)$$ which proves the inversion map $\mathrm{inv}$ to be continuous.

  • $\begingroup$ Thanks! I suspected it wouldn't be so bad, but I just couldn't see where to go. I figured those would be the facts about $X$ being compact Hausdorff we would have to use. $\endgroup$ – Francis Adams Jul 20 '12 at 12:14

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