properly discontinuous action of group of homeomorphisms on topological space

Question

I tried to solve the following question from Munkres (for reference: chapter 81, exercise 3 a, page 493)

Let $p:X \to B$ be a covering map and $G$ its group of covering transformations. Then $G$ acts properly discontinuous on $X$.

So the action is properly discontinuous if for every $x \in X$ there is an open neighborhood such that if $g(U) \cap U \neq \emptyset$ then $g$ is the identity homeomorphism. Considering $x \in X$, there is an evenly covered neighborhood of $p(x)$, say $V$. Hence we can write $p^{-1}(V)$ as disjoint union of open subsets $U_\beta$ in $X$ which are homeomorphic under $p$ to $V$. Consider $U$ to be the open neighborhood containing $x$. I tried to prove that for $g$ not the identity homeomorphism, we must have that $g(U) \cap U$ is empty, but I could not do this. Any hints would be appreciated.

EDIT I think I could answer my question using Bass' answer. Any remarks/hints on how to improve my answer would be appreciated (I was not sure if I had to post my answer as an actual answer, or if I should have edited it in this question. I chose the former, see my answer below)

Answer 1

In the event that $X$ is connected, you can follow the proof given by Hatcher on page 72 of his book (you can quickly check that your definition of a properly discontinuous action is indeed equivalent to his). The important points are the following (this is why we need connectedness):

1. The two points $\tilde{x_1}$ and $\tilde{x_2}$ belong to the same fiber $p^{-1}(x)$ because the action is by covering space automorphisms.

2. The fiber $p^{-1}(x)$ will intersect such a $U$ in only one point (because $X$ is connected).

3. Two (connected) covering space automorphisms that agree on one point agree everywhere.

Answer 2

Bass' answer reminded me that covering transformations agreeing on a point are equal maps. Using this, I can answer my question I guess:

Let $x \in X$, then $p(x)$ is a point in $B$, so there is an evenly covered neighborhood $V$ containing $p(x)$. This implies that $p^{-1}(V)$ is the disjoint union of open neighborhoods $U_\beta$ in $X$, let $U$ be the open neighborhood containing $x$. Note that $p$ restricted to $U$ is a homeomorphism of $U$ with $V$.

Let $g$ be a covering transformation, which is not the identity map of $X$. Suppose $y \in g(U) \cap U$, then there is an element $z \in U$ such that $g(z) = y$. By Bass' remark, we have that $y \neq z$. Since $g$ is a covering transformation, we have that $p = p \circ g$. This implies that $$p(z) = p(g(z)) = p(y)$$ but $p$ restricted to $U$ is a homeomorphism, hence injective and $z \neq y$, so this is not possible.

Therefore, we must have that $g(U) \cap U = \emptyset$.