# $T^2$ with $n$ points removed - Question from Basener

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 2 (and more) points.

Any hints?

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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 $\pi_1(T_n)$.

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Thanks. I had thought that removing $n$ points gives a bouquet of $n+1$ circles - I am guessing this is what you are referring to? I am interested in your comment regarding van Kampen's theorem - I had previously come across the punctured torus when using VK to calculate the fundamental group of the torus –  Juan S Feb 20 '11 at 22:23
@Qwirk: Oops. I thought that van Kampen's theorem can be used backwards to calculate the fundamental group of the torus with $n$ holes from the fundamental group of the torus with $n-1$ holes, but that's actually not the case. You get an extension of groups $1 \to \pi_1(S^1) \to \pi_1(T_n) \to \pi_1(T_{n-1}) \to 1$, but unfortunately that's not enough information to determine $\pi_1(T_n)$. –  Greg Graviton Feb 21 '11 at 8:18

its 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.

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