I recently noticed a few things in some recent questions on MO:
1) the fundamental group of $S^2$ minus, say, 4 points, is $\langle a,b,c,d\ |\ abcd=1\rangle$.
2) The fundamental group of a torus minus a point is $\langle a,b,c\ |\ [a,b]c=1\rangle$.
I was just wondering if you have a manifold $M$, and you know a presentation of its fundamental group, can you quickly get a presentation of the fundamental group of $M$ minus $n$ points?
Of course, this could merely be a coincidence. These are both surfaces, and both their fundamental groups only have one relation. But perhaps there is more to it than that?
Any and all helpful comments appreciated,
Steve
Best Answer
The basic tool here is Van Kampen's theorem. Let $M$ be the manifold, and let $M' = M\setminus {x}.$ Let $D$ be a small ball around $x$. Then $M = M' \cup D,$ and $M \cap D = D\setminus {x}$ is homotopic to $S^{n-1}$ (here $n$ is the dimension of $M$).
So Van Kampen's theorem says that $\pi_1(M) = \pi_1(M')*_{\pi_1(S^{n-1})} \pi_1(D)$. If $n > 2,$ then $S^{n-1}$ is simply connected, and of course $\pi_1(D)$ is simply connected, and so this reduces to $\pi_1(M) = \pi_1(M')$; in other words, deleting points doesn't change $\pi_1$ in dimensions $> 2$. (You can probably convince yourself of this directly: imagine you have some loop contracting in a 3-dimensional space; then if you remove a point, you can always just perturb the contraction slightly so that it misses the point. Of course, this is not the case in two dimensions.)
If $n = 2$, then $\pi_1(S^1) = \mathbb Z$ is infinite cyclic, while $\pi_1(D)$ is trivial again. So we see that $\pi_1(M)$ is obtained from $\pi_1(M')$ by killing a loop. One can be more precise, of course (and I will restrict myself to the orientable case, so as to make life easier): if you have a genus $g$ surface, with $r>0$ punctures, and also $g + r > 1$, then $\pi_1$ is a free group on $2 g + r - 1$ generators. (Every puncture after the first adds another independent loop, namely the loop around that puncture.) If $g \geq 1,$ and $r = 0$, then $\pi_1$ is obtained via Van Kampen as above: you begin with a free group on $2g$ generators for $M'$, and then you kill off the loop around the puncture when you fill it in (so you get the standard presentation for the $\pi_1$ of a compact orientable surface of genus $g \geq 1$). (Conversely, going from the compact surface $M$ to the once-punctured surface $M'$ does not add any generators to $\pi_1$, but gets rid of a relation.) If $g = 0$ and $r \leq 1,$ then you have either a disk ($r = 1$) or a sphere ($r = 0$) and so $\pi_1$ is trivial.