# Why does a covering map has the injective induced homomorphism?

Below is how I tried:

Let $p:(C,c_0)\rightarrow (X,x_0)$ be a covering map.

Let $[\gamma]\in \ker(p_*)$

Let $e_X,e_C$ be the constant loops at $x_0,c_0$ respectively.

Then $[e_X]=[p\circ \gamma]$.

Let $F$ be a path-homotopy between $e_X$ and $p\circ \gamma$.

Then, there exists a unique homotopy $G:I\times I\rightarrow C$ such that $p\circ G=F$ and $G(s,0)=c_0$, by homotopy path lifting theorem.

I have no idea how this applies that $p_*$ is injective.. Please help.. Why is $[\gamma]=[e_C]$?

• Moreover, someone please recommend me a basic algebraic topology text which covers "covering space". Hatcher is extremly terse for beginners I guess so.. – Rubertos Nov 25 '14 at 4:54
• Check out Munkres. I rather like his treatment of these topics (it's very, very clear). – Cameron Williams Nov 25 '14 at 4:55
• @CameronWilliams On which page of Munkres has this theorem? I can't find – Rubertos Nov 25 '14 at 5:36
• I think it's in section 53 or something? Just search for covering spaces. It'll be in that general vicinity. – Cameron Williams Nov 25 '14 at 5:38
• For an approach using covering morphisms of groupoids see my book "Topology and Groupoids", do a web search for info. This idea has been around since the first 1968 edition. The point is that a covering map is modelled algebraically by a covering morphism. – Ronnie Brown Nov 25 '14 at 10:03

Show that $G$ is a path homotopy between $\gamma$ and the constant loop $e_C$.
This should imply that $[\gamma] = [e_C]$ which means $\ker (p_*)$ is trivial, which means that $p_*$ is injective.