Covering through group action and corresponding deck transformations

Question

I'm having a bit of trouble with the following exercise:

Let $G$ be a group acting properly discontinuous and continuous on a topological space $E$. Then $p:E\to G\backslash E$ is a covering. Let $N_G(H)$ be the normalizer of $H$ in $G$. Show that:
i) For all $H\subset G,p_H:H\backslash E\to G\backslash E$ is a covering.
ii) The action of $g\in G$ on E induces a deck transformation of $p_H$ iff $g\in N_G(H)$.
iii) The action in ii) induces a homomorphism $\varphi:N_G(H)\to \Delta(p_H)$ with kernel $H$, but which in general is not surjective.
iv) If $E$ is connected, $\varphi$ is surjective.

My attempt:
i) Since $H\subset G$, the action of $H$ on $E$ is properly discontinuous and continuous aswell, thus $E\to H\backslash E$ is a covering aswell. Now $H\backslash E$ is a quotient space of $E$, so applying its universal property to the covering $p$ gives us the covering $p_H$.
ii) That the action induces a homeomorphism is clear, since the action is continuous and the inverse map is given by acting with the inverse element. So it seems the normalizer property is needed to show that $p_H$ is invariant under that action. This is where I'm stuck here.
iii) Let $D_g$ be the deck transformation corresponding to $g\in N_G(H)$. Then $\varphi(gg')(e)=D_{gg'}(e)=(gg').e=g.(g'.e)=D_g(D_{g'}(e))=(D_g\circ D_{g'})(e)=(\varphi(g)\circ\varphi(g'))(e)$, thus it is a homomorphism. It is clear that $ker(\varphi)=H$ since $h.e=e\forall h\in H, e\in H\backslash E$, thus the deck transformation is the identity. I was not able to find an example for the surjectivity.
iv) I have no idea how to approach this.

I would really appreciate any criticism and/or tips.

Answer 1

Proof. Let's say $\phi: E \to E$ is the homeomorphism $\phi(e)=g.e$. If $\pi_H: E\to H \backslash E$ is the quotient map, then the composition $\pi_H \circ \phi: E \to H\backslash E$ induces a unique map $\overline \phi: H \backslash E \to H \backslash E$, provided that \begin{equation} \pi_H \circ \phi(e)= \pi_H \circ \phi(h.e) \quad \forall h \in H, e \in E. \tag{$*$} \end{equation} If $(*)$ holds and $\phi$ descends to a homeomorphism of $H\backslash E$, then we have \begin{equation}\tag{$**$} p_H \circ \overline \phi(H_e)=p_H(g.H_e)=G.(g.H_e)=G.H_e=p_H(H_e). \end{equation} Thus $\overline \phi$ will be a deck transformation.

We now show that $(*)$ holds if and only if $g$ lies in $N_G(H)$. First observe that $(*)$ is equivalent to $H.(g.(h.e))=H.(g.e)$ (for all $h$ and $e$). Now if $g \in N_G(H)$, i.e. $g H=Hg$, then \begin{align} H.(g.(h.e))&= Hg.(h.e)=gH.(h.e)=gH.e=Hg.e=H.(g.e), \end{align} as desired. On the other hand, suppose $H.(g.(h.e))=H.(g.e)$ for all $h$. Fixing $h$ now, there must exist $h' \in H$ such that $h'.(g.(h.e))=g.e$, hence $(g^{-1}h'gh).e=e$. Since $e \in E$ can only be fixed by $1 \in G$, we have $g^{-1} h' g h=1$ and thus $$g h g^{-1} = (h')^{-1}.$$ Since $ghg^{-1}$ is an element of $H$ and $h$ was arbitrary, we must have $g \in N_G(H)$. $\square$

Answer 2

The example for $\text{Question (iii)} $:

Consider the space E consisted of a disjoint sequence of $S^{1}$ lying along the x-axis, each $S^{1}$ is centered at an integer point on x-axis and radius is $1/2$ . The group action G is Z, and taking H to be the trivial subgroup $\{e\}$. Consider a deck transformation from H\E to H\E by interchanging the circles that are symmetric w.r.t the origin, and this transformation cannot be represented by any element of G, hence this is a counterexample.