the fundamental group of a Riemann surface with n points removed Can anybody give me some suggestions about how to calculate the fundamental group of a Riemann surface with $n$ points removed? For example, what is the fundamental group of the genus $1$ torus with $4$ points removed? Thanks. 
 A: Let's start with a torus. Have you seen this picture for the cell complex structure on a torus? 

This tells you you identify the sides labeled $a$ with the sides labeled $b$ according to the given orientation. Then since your attaching a one cell that makes the loop going around the whole thing trivial this gives you a presentation of the fundamental group as generated by the loops $a$ and $b$ with the one relation given by gluing the cell which gives us $[a,b]=1$. Now what happens if we puncture the torus at one point? Well if we look at the picture on the right and we puncture a point, we see a that the 2-cell we attached in the middle deformation retracts to the boundary so we get a wedge sum of 2 loops one for $a$ and one for $b$ giving us a fundamental group $F_2$ the free group on $n$-generators. 
Now what happens when we puncture the torus $n$ times? By the same argument it will now deformation retract into the wedge sum of $n + 1$ circles instead of just $2$ giving us the free group $F_{n+1}$. 
Now let's raise the genus. For a higher genus surface we have a picture like this:
 
This picture is for genus 3 but we have a similar picture for any genus $g$. Again if we puncture just once it deformation retracts onto the boundary which after identification gives a wedge sum of $2g$ circles. 
Suppose we puncture $n$ times. We can write the surface of genus $g$ as a surface of genus $1$ with one puncture and a surface of genus $g - 1$ with on puncture glued along the puncture as pictured: 

Then putting all the $n$ punctures on the torus side, we get that the $n$ punctured genus $g$ surface is covered by a $1$ punctured genus $g-1$ surface $U$ and an $n + 1$ punctured genus $1$ surface $V$ whose intersection $U \cap V$ is just a circle. Then we can use van Kampans theorem. 
$\pi_1(U) = F_{2(g-1)}$, $\pi_1(V) = F_{n+2}$ and $\pi_1(U \cap V) = F_1$ so when we take the amalgamated free product $\pi_1(U) *_{\pi_1(U \cap V)} \pi_1(V)$ is generated by all the generators of $\pi_1(U)$ and $\pi_1(V)$ but where $\pi_1(U \cap V)$ just kills off on generator so we get a free group on $2(g-2) + (n + 2) - 1 = 2g + n - 1$ generators. 
Credit for the pictures goes to Hatcher's Algebraic Topology for the first two and google for the last. 
