Group of covering Transformation corresponding to proper discontinuous action of a group on a connected space is the group itself.

Question

Suppose $G$ acts properly discontinuously on a connected space $X$. Show that the group $G(X,p,X/G)$ of the covering transformation of $p:X \rightarrow X/G$ is $G$.

I have tried in this manner- Define the map $\phi: G \rightarrow G(X,p,X/G)$ by $\phi(g)=h_g$ where $h_g:X\rightarrow X$ is the homeomorphism $h_g(x)=g.x$. I have shown that $\phi$ is an injective homomorphism but I cannot show that $\phi$ is surjective.

Note that to show that it is a monomorphism we don't need the connectedness of $X$.

Answer 1

I'm assuming that you forgot to mention that the action was free (otherwise a finite $G$ acting trivially contradicts injectivity and actually the whole statement)

Hint : let $f$ be an automorphism of the covering, pick $x\in X$; then $f(x) = g\cdot x$ for some $g\in G$. By freeness of the action, this $g$ is unique. Show, using proper discontinuity of the action, that this $g$ is the same for all $y\in V$ for some open neighbourhood $V$ of $x$. Conclude using connectedness.