Presentation of the fundamental group of a manifold minus some points 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
 A: 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.
