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Let $(\tilde{X},p)$ be a universal covering space of $X$. We know that if $G$ acts properly disctinously on $\tilde{X}$, then $\tilde{X}$ is a covering space of $\tilde{X}/G$ and $\pi_1(\tilde{X}/G)=G$; in particular, $X= \tilde{X}/ \pi_1(X)$.

Let $G$ be a subgroup of $\pi_1(X)$. There is a canonical map $q : \tilde{X}/G \to \tilde{X}/\pi_1(X)=X$ corresponding to the injection $G \to \pi_1(X)$.

My question: Is $(\tilde{X}/G,q)$ a covering space of $X$?

By my preliminary remark, $(\tilde{X},\pi)$ is a covering space of $\tilde{X}/G$ with the canonical surjection $\pi : \tilde{X} \to \tilde{X}/G$. So we have the following commutative diagram:

$$\begin{array}{ccc} \tilde{X}/G & \rightarrow^{q} & X \\ \uparrow{\pi} & \nearrow{p} & \\ \tilde{X} & & \end{array}$$

My main problem is that if $U \subset X$ then the connected components of $q^{-1}(U)$ don't have necessarly the same behaviour.

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If $X$ is path-connected, locally path-connected, and semilocally simply-connected, then there is a Galois correspondence between (homeomorphism classes of) covering spaces of $X$ and (conjugacy classes of) subgroups of $\pi_1(X)$ via the construction you mention. See, for example, Theorem 2.13 here. – Zhen Lin Aug 23 '12 at 16:06
up vote 1 down vote accepted

Using the link given by Zhen Lin, I answer my question:

Let $x \in X$ and $\tilde{x} \in p^{-1}(x)$. Because $\pi_1(X)$ acts properly discontinously on $\tilde{X}$, there exists an open neighborhood of $\tilde{x}$ such that for all $g \in \pi_1(X) \backslash \{e\}$, $g \cdot V \cap V= \emptyset$. Let $U$ denote the open neighborhood $p(V)$ of $x$ ($p$ is open).

Let $W$ be a connected component of $q^{-1}(U)$. We have $p^{-1}(U)= \coprod\limits_{g \in \pi_1(X)} g \cdot V$ so $q^{-1}(U)= \pi (p^{-1}(U))= \coprod\limits_{ g \in \pi_1(X)/H} g \cdot \pi(V)$ ($\pi$ is onto). So $W$ has the form $g \cdot \pi(V)$, $g \in \pi_1(X)/H$.

Notice that $q$ induces a continous open application from $W$ onto $U$. Moreover, if $x=g_1 \cdot \pi(v_1),y=g_2 \cdot \pi(v_2) \in W$ are such that $q(x)=q(y)$, then $p(v_1)=p(v_2)$ so there exists $h \in \pi_1(X)$ such that $v_2= h \cdot v_1$ with $h \in \pi_1(X)$. Thus, $v_2 \in h \cdot V \cap V$ so $h=e$ and $x=y$.

Consequently, $q$ induces a homeomorphism from $W$ onto $U$, proving that $(\tilde{X}/H,q)$ is a covering space of $X$.

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