Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I found the following in section 8.14 of Munkres Topology, a first course.

"Now suppose we have a fixed path-connected space $B$ and a fixed base $b \in B$. Let $p: E \rightarrow B$ be a covering map, where $E$ is path connected. If we choose a point $e$ in $p^{-1}(b)$ as a base point for $E$, then we have an induced homomorphism $p_*: \pi_1(E,e) \rightarrow \pi(B,b)$."

How is this homorphism induced? I can get a surjection between the first fundamental group of a base space space at a point to the fiber over that point, but since it might not be injective I can't necessarily invert it.

share|cite|improve this question
Induced is not invertible. To define the homomorphism $p_*$ we need to use $p$ at some point, thus "induced" from $p$. – user27126 Apr 17 '13 at 23:07

If $\gamma : S^1\to E$ is a loop in $E$ (with base point $e$), then $p\circ\gamma : S^1\to B$ is a loop in $B$ with base point $b = p(e)$. Then we simply take the homotopy class of the projected loop: \begin{align*} p_* : \pi_1(E,e)&\to\pi_1(B,b).\\ \left[\gamma\right]&\mapsto\left[p\circ\gamma\right] \end{align*} Note that this may only be a homomorphism, not necessarily an isomorphism, so it need not be injective.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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