Show that the action of $G$ over $X\cong G/K$ is proper for $K$ compact

Question

During a proof on Brown's book (details below) I encountered the following claim

Let $G$ be a topological group that acts transitively on a manifold $X$, and let $K$ be the isotropy group. Assume that $X$ is homeomorphic to $G/K$. Assume further that $K$ is compact in $G$. Then the action of $G$ over $X$ is proper.

I'm asked to show that the map $$ G \times X \to X \times X $$ $$ (g,x) \mapsto (g.x,x)$$ is proper. The condition of properness is easily verified for points, but I'm looking for a clever way to extend the reasoning for any compact subset of $X$.

If someone is interested, I'm filling the details of ex $6$ at page $39$ of Brown's Group Cohomology

Any hint?

Answer 1

With a little bit of point-set topology it's sufficient to show that preimages of points are compact in order to prove properness of a map.

Let $f: X \to Y$ be a continuous closed map (as in our case which is very easy to show), such that $f^{-1}(y)$ is compact (in $X$) for all $y \in Y$. Let $K$ be a compact subset of $Y$. We will show that $f^{-1}(K)$ is compact. Let $\{ U_{\lambda} \mid \lambda \in \Lambda \}$ be an open cover of $f^{-1}(K)$. Then for all $k \in K$ this is also an open cover of $f^{-1}(k)$. Since the latter is assumed to be compact, it has a finite sub-cover. In other words, for all $k\in K$ there is a finite set $\gamma_k \subset \Lambda$ such that $f^{-1}(k) \subset \cup_{\lambda \in \gamma_k} U_{\lambda}$. The set $X \setminus \cup_{\lambda \in \gamma_k} U_{\lambda}$ is closed. Its image is closed in Y, because f is a closed map. Hence the set $V_k = Y \setminus f(X \setminus \cup_{\lambda \in \gamma_k} U_{\lambda})$ is open in Y. It is easy to check that $V_k$ contains the point $k$. Now $K \subset \cup_{k \in K} V_k$ and because K is assumed to be compact, there are finitely many points $k_1,\dots , k_s$ such that $K \subset \cup_{i =1}^s V_{k_i}$. Furthermore the set $\Gamma = \cup_{i =1}^s \gamma_{k_i} $ is a finite union of finite sets, thus $\Gamma$ is finite. Now it follows that $f^{-1}(K) \subset f^{-1}(\cup_{i=1}^s V_{k_i}) \subset \cup_{\lambda \in \Gamma} U_{\lambda}$ and we have found a finite subcover of $f^{-1}(K)$, which completes the proof of this claim.

In order to prove what you've asked in your question, notice that the preimage $\phi_G^{-1}(x,x)$ is $K\times \{x\}$ by construction, because $K$ is the isotropy group of all points (due to transitiveness of the action). It's clearly compact in $G\times X$ by hypothesis and therefore the action is proper.