I was reading Hatcher's Algebraic topology book,in page 66 prop 1.36,there is a theorem ,which says that :
Suppose $X$ is path-connected, locally path-connected, and semilocally
simply-connected. Then for every subgroup $H \subset \pi_{1}\left(X,
x_{0}\right)$ there is a covering space $p: X_{H} \rightarrow X$ such
that $p_{*}\left(\pi_{1}\left(X_{H}, \tilde{x}_{0}\right)\right)=H$
for a suitably chosen basepoint $\tilde{x}_{0} \in X_{H}$
I have some difficulty in figuring out why constructed $p$ is a covering map.
Assuming we have already construced the case that $H$ is trivial subgroup,then we have the universal covering $\tilde{X} = \{[f]\mid f(0) = x_0 \}$ be the path homotopy class with starting point at some $x_0$.Then the proof for the general $H$ is given as follows:
Proof:
Step1.(define the space $X_H$ as quotient of the univeral covering $\tilde{X}$)
For points $[\gamma],\left[\gamma^{\prime}\right]$ in the simply-connected covering space $\tilde{X}$ constructed above, define $[\gamma] \sim\left[\gamma^{\prime}\right]$ to mean $\gamma(1)=\gamma^{\prime}(1)$ and $\left[\gamma \cdot \overline{\gamma^{\prime}}\right] \in H .$ It is easy to see that
this is an equivalence relation . Let $X_{H}$ be the quotient space of $\tilde{X}$ obtained by identifying $[\gamma]$ with $\left[\gamma^{\prime}\right]$ if $[\gamma] \sim\left[\gamma^{\prime}\right] .$
We have quotient topology on space $X_H$ and the covering $p_H:X_H\to X$ is given by $p:\tilde{X}\to X$ $p_H$ is continuous since $p$ is continuous.
Step2.(Prove it's a covering map)
Note that if $\gamma(1)=\gamma^{\prime}(1)$, then $[\gamma] \sim\left[\gamma^{\prime}\right]$
iff $[\gamma\cdot\eta] \sim\left[\gamma^{\prime} \cdot \eta\right]$. This means that if any two points in basic neighborhoods $U_{[\gamma]}$ and $U_{\left[\gamma^{\prime}\right]}$ are identified in $X_{H}$ then the whole neighborhoods are identified. Hence the natural projection $X_{H} \rightarrow X$ induced by $[\gamma] \mapsto \gamma(1)$ is a covering space.
I don't really know how to prove it's covering i.e. find a piece of slice that homeomorphic to some neiborhood of $X$,I think it should use the unversal covering is a covering map,I can't figure it out.
Best Answer
You know that projections $pU_{[\gamma]}=U$ are trivializing neighborhoods for the universal covering, while $U_{[\gamma]}$ are distinct pieces of its preimage. What is happening in prop. 1.36 is you take those pieces $U_{[\gamma]}$ and glue some of them together. As Hatcher says there, if two points of such pieces are identified in $X_H$, then the whole pieces $U_{[\gamma]}$ are identified, and this all happends "above" the trivializing neighborhood $pU_{[\gamma]}=U$. So, before you had that $p^{-1}pU_{[\gamma]} = \coprod_{\gamma'}U_{[\gamma']}$, and now you have the same story but some of those pieces are glued together. So, the preimage of $pU_{[\gamma]}$ is again a disjoint union of some pieces $U_{[\gamma']}$, but for a smaller index-set of $\gamma'$.