Let $p:(E,e_0) \rightarrow(X,x_0)$ be a covering projection. Show that $p \sharp: \pi_{1}(E,e_0) \rightarrow \pi_{1}(X,x_0)$ is a monomorphism.

I was wondering here do I need to prove this is a homomorphism?

As I'm confused as I know you have to prove that it is an injection. Which is just by considering what is in the kernel of $\pi_{1}(E,e_0) \rightarrow \pi_{1}(X,x_0)$, something that is homotopic to the constant map, then you just lift the homotopy to the covering space to get something homotopic to the constant map in $(E,e_0)$.

However, is this enough or do you need to show homomorphism? How would you show a homomorphism? would you need to do concatenation of loops?

  • $\begingroup$ In most books on algebraic topology it is proved that any continuous map $f$ induces a homomorphism on $\pi_1$s. It's actually not so hard: if $\gamma_i$ are loops in the domain then $f \circ (\gamma_1 * \gamma_2) = (f \circ \gamma_1) * (f \circ \gamma_2)$. $\endgroup$ – Dylan Moreland May 21 '12 at 0:49
  • $\begingroup$ It is not enough to show that the kernel is trivial. The statement "$\varphi$ is injective $\iff$ $\ker(\varphi) = 0$" requires that $\varphi$ is a homomorphism. For a simple counterexample, the map $x \mapsto x^2$ on $(\mathbb{R}, +)$ has trivial "kernel" but is certainly not injective. $\endgroup$ – Henry T. Horton May 21 '12 at 0:56
  • $\begingroup$ And I suppose you should also prove that the map is well defined first. $\endgroup$ – Dylan Moreland May 21 '12 at 1:07

If you are not sure about it being a homomorphism, what you should try and prove is that for any map of topological spaces $p:X \to Y$ the induced map $p_*:\pi_1(X,x) \to \pi_1(Y,p(x))$ is a homomorphism. What is this map?

Let $[f]$ be a class in $\pi_1(X,x)$. Then the induced map is $p_*([f]) = [p \circ f]$ where $p \circ f$ is the composite $f:I \to X$ with $p:X \to Y$. Then check that this map is a homomorphism. Recall if we have paths $f:x \to y$ and $g:y \to z$ then $g \cdot f$ is the map obtained by 'going twice as fast' and that this passes through equivalence class via $[g]\cdot[f] = [g \cdot f]$.

The fact that the induced map of a covering space is injective is a consequence of the homotopy lifting property as you say.


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.