I'm trying to prove for a ($n \geq 3$)-dimensional path connected manifold M, if remove a point in M, then use Van Kampen theorem can somehow show the fundamental group of original $M$ and $M-{x}$, and $M-U_x$ where $U_x$ is the open neighbourhood of $x$ are the same.
I know the open neighbourhood $U_x$ is homeomorphic to open n-ball in $\mathbb{R}^n$, I can feel it is true as then intersect $U_x$ with $(M-{x})$ is homeomorphic to $S^{n-1}$. But I just cannot get my head around with all the details and how the logic actually goes.
Can anybody please try to help me in details? I really wish to understand the details in between as I do have rough intuition, it is details missing here.
Also, what is a good counter example for $n=2$?
Thanks a lot for helping.
Let $U=M-x$, $V=U_x$, then $M=U \cup V \cong R^n$, $U \cap V \cong S^{n-1}$ Then use Van Kampen's theorem. Actually since $\pi_1(S^{n-1})=0$, that normal subgroup $N$ must be $0$. So $\pi_1(M)=\pi_1(U)$.
Similarly $\pi(M)=\pi(M-U_x)$. Just let $V$ be a larger ball.
For n=2. The problem is $\pi_1(S^{1})=\mathbb{Z}$. Think about the torus. The fundamental group of torus minus one point is $\mathbb{Z} * \mathbb{Z}$.