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If $M$ is a connected topological manifold with dimension $n\ge 3$, let $q\in M$. How can I show that $\pi_1(M)=\pi_1(M-\{q\})$?

As Neal said, since M is locally connected, M is path connected, then we can use the Seifert-Van Kampen theorem, I have problems to find the open path connected sets $U$ and $V$ such that we can use Seifert-Van Kampen theorem. I'm thinking about $U=W-\{p\}$, where $W$ is the neighborhood of $p$ which is homeomorphic to $\mathbb R^n$ am I correct? and about the another open set $V$?

I need help here

Thanks

  • 1
    Can you prove this result for $\Bbb R^n$ ($n\ge 3$)?2012-11-26
  • 0
    @ChrisEagle yes, of course2012-11-26
  • 0
    Then this is an easy application of van Kampen.2012-11-26
  • 0
    I didn't understand, in order to use Van Kampen theorem, M has to be path connected.2012-11-26
  • 4
    But $M$ is path-connected!2012-11-26
  • 0
    For local path connected space, the connectedness is equivalent to path-connectedness.2012-11-26
  • 0
    Locally path-connected $\Rightarrow$ (connected $\Leftrightarrow$ path-connected)2012-11-26
  • 0
    @Neal yes, I forgot the manifold is locally path connected, thank you2012-11-26
  • 0
    Using the manifold property of $M$, consider a small enough n-ball $B_{\epsilon}$ in $M$ around q. What do you know about the subspace $(M\setminus\{q\})\cap B_{\epsilon}$? How about the fundamental group of this space?2012-11-26
  • 0
    @DanielR thank you, it helped a lot.2012-11-27
  • 0
    why downvoted??2012-11-27

1 Answers 1

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Following the hints in the commentaries in the question, choose the following open sets:

$U=M-\{q\}$

$V=B_\epsilon(q)$, where $B_\epsilon(q)$ is a open ball centered in $q$.

Note these open sets are path connected and the intersection $U\cap V=B_\epsilon(q)-\{q\}$ is path-connected.

Since $n\ge 3$, $B_\epsilon(q)-\{q\}$ is simply connected, and note $\pi_1(V)=1$ (trivial group).

Thus using Seifert-Van Kampen Theorem we have:

$\pi_1(M)=\pi_1(M-\{q\})$