Prove a compact Hausdorff space with a group structure is a topological group

Question

Here is an exercise from the textbook of topology by Armstrong.

Let $G$ be a compact Hausdorff space which has the structure of a group. Show that $G$ is a topological group if the multiplication function $m:G\times G \rightarrow G$ is continuous.

I tried to use the conditions to prove the inverse function is also continuous, but I couldn't find out how to connect the inverse with the multiplication...

Any help will be appreciated.

Answer 1

Consider the map $(a,b)\to (a,ab)$. Prove that it's a homeomorphism from $G\times G$ to $G\times G$ and consider its inverse.

Answer 2

This has an elegant proof using universal nets (or ultrafilters, or convergence of maps with respect to ultrafilters which would be my favorite way to do it). To show the inverse map is continuous, it suffices to show it preserves convergence of universal nets. So suppose $(g_i)$ is a universal net converging to some $g\in G$. The net $(g_i^{-1})$ is then also a universal net, and so since $G$ is compact it must converge to some $h\in G$. Now the net $(g_i,g_i^{-1})$ converges to $(g,h)$ in $G\times G$, so by continuity of $m$, $(g_ig_i^{-1})$ must converge to $gh$. But $g_ig_i^{-1}=1$ for all $i$, and so since $G$ is Hausdorff this constant net can only converge to $1$. Thus $gh=1$, and $h=g^{-1}$. That is, $(g_i^{-1})$ converges to $g^{-1}$, as desired.

More generally, a similar argument shows that if $Y$ is compact and $f:X\to Y$ is a map whose graph is closed in $X\times Y$, then $f$ is continuous. In this case, the graph of the inverse map is closed in $G\times G$ since it is $m^{-1}(\{1\})$.