# Compact group actions and automatic properness

I am currently re-reading a course on basic algebraic topology, and I am focussing on the parts that I feel I had very little understanding of. There is one exercise in the chapter devoted to groups acting on topological spaces (preceding the chapter on covering spaces) that I have spent several frustrating hours on, but I just can't crack.

I will state the question below, but PLEASE, DO NOT POST A SOLUTION OR A HINT. I am only interested in knowing wether the statement in question is true. I have looked on the internet for a reference about this fact, but didn't get anywhere. If you know a book that gives proves it, or a document online where this is discussed, I would like to get the reference, in case I continue to fail at giving a proof of this the next week.

I recall some terminology first: let $X,Y$ be two topological spaces, and $f:X\rightarrow Y$ a map that isn't supposed to be continuous. The author defines such a map to be $\mathrm{PROPER}$ whenever the following two properties are satisfied : $f$ is a closed map, and $\forall y\in Y, ~f^{-1}(\lbrace y\rbrace)$ is a compact subspace of $X$. It then follows that for all compact subsets $K$ of $Y,~f^{-1}(K)$ is a compact subset of $X$. Also, if $X$ is Hausdorff, and $Y$ is a locally compact Hausdorff, then properness is equivalent to this property.

Let $X$ be a topological space, and $G$ a topological group. Suppose there is a continuous left group action $\rho: G\times X\rightarrow X,~(g,x)\mapsto g\cdot x$. Let $\theta:G\times X\rightarrow X\times X, ~(g,x)\mapsto (x,g\cdot x)$. The author defines the group action to be $\mathrm{proper}$ if $\theta$ is a proper map.

Here is the question: "Show the action of a compact Hausdorff group $G$ on a Hausdorff space $X$ is always proper".

As I said, I have struggled with this for days (since friday). IS THAT STATEMENT TRUE? There are no further hypothesis, $X$ is not supposed to be locally compact, and the action is completely arbitrary (continuous of course, but not supposed free, or other things).

• I think it is a very unusual, and hence confusing, idea not to include "continuous" in the definition of "proper". Anyway, Theo certainly makes this continuity assumption in the fine text (mentioned in his answer) that he wrote for MathOverflow. – Georges Elencwajg May 23 '11 at 10:19

• @Olivier: I think it's rather straightforward and no trickery is required. I'm reluctant to say anything more, since you don't want any hints... However, you can try to show first that the projection map $G \times X \to X$ and the action map $\rho$ are proper (since $G$ is compact). Their "combination" $\theta$ is then quite easily seen to be proper, as $X$ is Hausdorff. – t.b. May 23 '11 at 15:08