I just started studying covering spaces and I have been really stuck on showing the following:
Let $G$ be a finite group acting freely on a Hausdorff space $E$. Show that the quotient $q : E \rightarrow E /G :=B$ is a covering map.
Any help or solution will be highly appreciated.
Let us define the equivalence relation on $E$ such that $x\sim y$ if and only if $x$ and $y$ are within the same orbit under the action of $G$ on $E$. Then the resulting canonical projection $$q:E \rightarrow E/G$$ onto the quotient topology is surjective and continuous, so it remains to prove that it is covering.
Consider a point $[x] \in E/G$. Its preimage is the set $G.x$, which is finite by assumption. For any open set $U$, let $g.U$ be its $g$-translate.
Let us now assemble some standard facts.
Lemma 1: There exists an open neighborhood $U_x$ of $x$ such that all translates $g.U_x$ are pairwise disjoint.
Proof: Since our topology is Hausdorff, we may, for each element $g.x$, choose open sets $U_g$ each containing $g.x$ and which are pairwise disjoint. Let us take $$U_x = \bigcap_{g\in G}g^{-1}.U_g.$$ Clearly $x\in U_x$. Also note that $g.U_x \subseteq U_g$ so that the translates $g.U_x$ are also pairwise disjoint. $\square$
Lemma 2: The canonical projection $q$ is an open map.
Proof: Let $U$ be an open set in $E$. Then each of the translates $g.U$ are also open due to the continuity of the action, and hence $\bigcup_{g\in G}g.U$ is open. But this says that $q^{-1}(q(U))$ is open, so that $q(U)$ is open. $\square$.
Lemma 3: Consider $q|_{g.U_x}$, the restriction of $q$ onto $g.U_x$. Then $q|_{g.U_x}$ is a homeomorphism between $g.U_x$ and $q(U_x)$.
Proof: Since $q$ is a continuous open map, it remains to show that $q|_{g.U_x}$ is a bijection. Clearly $q(g.U_x) = q(U_x)$, so that $q|_{g.U_x}$ is surjective.
If $q(y)=q(z)$ for $y,z\in g.U_x$ then $x=g'.y$ for some $g'\in G$. But then we have $y \in g'g.U_x\cap g.U_x$, where by Lemma 1 the intersection is empty except when $g'g = g$. So we must have $g'=e$ and hence $a=b$. Thus $q|_{g.U_x}$ is injective. $\square$
Assembling all the ingredients, we can finally prove that $q$ is a covering map. By Lemma 2, the set $q(U_x)$ is open in $E/G$. Its preimage is the set $$q^{-1}(q(U_x)) = \bigcup_{g\in G}g.U_x,$$ which is a disjoint union of open sets in $E$. By Lemma 3, each $g.U_x$ is mapped homeomorphically onto $q(U_x)$. Hence $q$ is a covering map.