In John Lee's book on "Introduction to smooth manifolds", he proves the following result:
Let $G$ be a Lie group and $M$ be a smooth manifold. Suppose that $G$ acts smoothly, freely and properly on $M$. Then, the orbit space $M/G$ (equipped with the quotient topology) is a topological manifold of dimension equal to $\dim M – \dim G$. Moreover, $M/G$ as a unique smooth structure such that the quotient map $\pi: M \rightarrow M/G$ is smooth.
The definitions used are:
An action is free if $g \cdot p = p$ if and only if $g = e$.
An action is proper if the map $F: G \times M \rightarrow M \times M$ given by $F \left( g, p \right) = \left( g \cdot p, p \right)$ is proper (preimages of compact sets are compact).
In the proof of the result I mentioned above, the author proves that the orbit maps $\theta^{\left( p \right)} : G \rightarrow M$ are embeddings. To do so, we prove that $\theta^{\left( p \right)}$ is an injective proper immersion. Injectivity and immersion is easy to prove. However, I could not follow the properness of $\theta^{\left( p \right)}$. The author says that for any compact set $K \subseteq M$, we have $\left( \theta^{\left( p \right)} \right)^{-1} \left( K \right) \subseteq \left\lbrace g \in G | g \cdot K \cap K \neq \emptyset \right\rbrace$.
I have found a counterexample to this claim:
Consider $K = \left\lbrace g_0 \cdot p \right\rbrace$, for some $g_0 \neq e \in G$. Then, clearly $K$ is compact and $\left( \theta^{\left( p \right)} \right)^{-1} \left( K \right) = \left\lbrace g_0 \right\rbrace$. Now, we have $g_0 \cdot K = \left\lbrace g_0^2 \cdot p \right\rbrace$ which does not intersect $K$!
However, here, we do observe that $\left( \theta^{\left( p \right)} \right)^{-1} \left( K \right)$ is compact. If my counterexample is correct, how do we prove that $\theta^{\left( p \right)}$ is a proper map?
Best Answer
Note I'm looking at the second edition of Lee's Introduction to Smooth Manifolds.
Proposition 21.7 reads
"Suppose $\theta$ is a proper smooth action of a Lie group G on a smooth manifold $M$. For any point $p\in M$, the orbit map $\theta^{(p)}:G\rightarrow M$ is a proper map, and thus the orbit $G\cdot p=\theta^{(p)}(G)$ is closed in $M$. If in addition $G_p=\{e\}$, then $\theta^{(p)}$ is a smooth embedding, and the orbit is a properly embedded submanifold."
In the proof he claims that, for $p\in M$, \begin{equation*} (\theta^{(p)})^{-1}(K)\subset G_{K\cup\{p\}}=\{g\in G:(g\cdot K\cup\{p\})\cap(K\cup\{p\})\}. \end{equation*}
In particular, since $K\cup\{p\}$ is compact, and $G_{K\cup\{p\}}$ is compact by properness (Proposition 21.5), we have that $(\theta^{(p)})^{-1}(K)$ is a closed subset of a compact space and hence compact.
It seems either there is a typo in the edition you are looking at or perhaps you missed the union with the point. In any event, your counterexample is no longer true. In your counterexample, we have that \begin{equation*} (\theta^{(p)})^{-1}(K)\subset G_{K\cup\{p\}}, \end{equation*} which is what was claimed. Therefore the space is compact, as desired.