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 $[\gamma]=[e_C]$?

  • 1
    $\begingroup$ Moreover, someone please recommend me a basic algebraic topology text which covers "covering space". Hatcher is extremly terse for beginners I guess so.. $\endgroup$ – Rubertos Nov 25 '14 at 4:54
  • 2
    $\begingroup$ Check out Munkres. I rather like his treatment of these topics (it's very, very clear). $\endgroup$ – Cameron Williams Nov 25 '14 at 4:55
  • $\begingroup$ @CameronWilliams On which page of Munkres has this theorem? I can't find $\endgroup$ – Rubertos Nov 25 '14 at 5:36
  • $\begingroup$ I think it's in section 53 or something? Just search for covering spaces. It'll be in that general vicinity. $\endgroup$ – Cameron Williams Nov 25 '14 at 5:38
  • 1
    $\begingroup$ 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. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.