# How can I prove $\pi_1(M)=\pi_1(M-\{q\})$?

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

• Can you prove this result for $\Bbb R^n$ ($n\ge 3$)? Nov 26, 2012 at 11:53
• @ChrisEagle yes, of course Nov 26, 2012 at 11:55
• Then this is an easy application of van Kampen. Nov 26, 2012 at 11:57
• I didn't understand, in order to use Van Kampen theorem, M has to be path connected. Nov 26, 2012 at 11:59
• But $M$ is path-connected! Nov 26, 2012 at 12:00

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:

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