Deck transformation acting properly discontinuously assumed covering space is path-connected

Question

Let $p:E\rightarrow X$ be a covering space, $x\in X$ fixed point, $E$ path-connected and $\Delta(p)$ – Deck transformation group of $p$, that is $\Delta(p) = \{f\in \text{Homeo}(E):pf=p\}$. Let $\pi_1(X,x)$ be a fundamental group, and it acts on $E$ from right in the following way: $(e, [w]) \mapsto \bar{w}(1)$, where $w:[0,1]\rightarrow X$ is a representative path of $[w]$ and $\bar{w}:[0,1]\rightarrow E$ is the (unique) path lift. You might be familiar with this construction if you recall Hatcher's.

The problem is:

Given all above, prove that $\Delta(p)$ acts properly discontinuously on $E$.

Properly discontinuous action is a group action $G\times X \rightarrow X$ such that $\forall x\in X \,\,\exists \text{ neighborhood} \,U \,\text{of}\, x : \forall g\neq 1 \in G$ holds $ g(U)\cap U = \varnothing$.

So roughly speaking it means that a deck transformation has no fixed points, or "areas" (neighborhoods) in this context, unless its identity.

1. In the first part of the exercise I proved that $f(e\cdot[w]) = f(e)\cdot[w]$.
2. Actually it's not necessary to prove this for the whole $E$, I guess it will be enough to show this only for preimage $p^{-1}(x)$, whence $\pi_1(X,x)$ was introduced.
3. What appears an obstacle for me is the whole deal with neighborhoods. For instance, if we had found a non-identity homeomorphism from $\Delta(p)$ with a fixed point $f(x)=x$ (which for me appears a natural interpretation of properly discontinuous map, but it's more general in the definition), then we'd be done because of 1., path-connectedness of $E$ and application of path lifting theorem (this map would be an identity).

If we try to prove this by contradiction then we only get a point whose each neighborhood and image of the neighborhood under some image $f$ share at least one point. Because of 1. it applies for all points, but it neither proves that $f$ is identity nor contradicts with the assumptions. So this is where I'm stuck.

I guess path lifting theorem should be used at some point. Just a guess.

Help appreciated. Thanks in advance!

Answer 1

Pick $x\in E$ and consider an evenly covered neighbourhood $U$ of $p(e)$. Let $U_a$ be the copy of $U$ which $x$ belongs to. Then pick $g \neq 1$ in $\Delta(p)$: if $g(z)=u$ for $z,u \in U_a$ then $p(z)=p\circ g(z)=p(u) \implies z=u$. Note that the thesis still holds if you remove the homeomorphism hypothesis.