# properly discontinuous action of group of homeomorphisms on topological space

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)

• Is $X$ connected ?
– Bass
Nov 15, 2017 at 20:46
• @Bass Well, it is not mentioned specifically, but in one of the previous chapters it is said that from then on all spaces are supposed to be path connected and locally path connected. Nov 15, 2017 at 21:52

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.
• Thanks for reminding me that covering space automorphisms agreeing on a point agree everywhere. Using that, I think I could answer my question :) Nov 18, 2017 at 10:23

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 $g(U) \cap U = \emptyset$.