Fundamental group of a torus with points removed

Question 5.33 from "Topology and its Applications" by Baesner is to compute the fundamental group of the torus ($T^2$) with $n$ points removed. I can "see" in my mind that if we remove one point we get a bouquet of two circles. Less clear is what happens when we remove two (or more) points. Any hints?

Solution 1:

It's a bouquet of $n+1$ circles, so the free group on $n+1$ generators. Think of a rectangle with $n-1$ horizontal lines across it. Roll it up into a tube, so you have a line segment with $n+1$ circles attached to it. Then identify the top and bottom circles. So you have a circle with $n$ circles attached to it which is homotopic to a bouquet of $n+1$ circles.

Solution 2:

Hint: removing $n$ points will also give something that only consists of 1-dimensional things.

You can also use van Kampen's theorem to calculate the fundamental group directly. EDIT: Err, you can't, at least not in the way I thought. Let $T_n$ be a torus with $n$ holes. Then, van Kampen's theorem gives a short exact sequence of groups

$$ 1 \to \pi_1(S^1) \to \pi_1(T_n) \to \pi_1(T_{n-1}) \to 1$$

but that's not enough information to deduce $\pi_1(T_n)$.