# What is the fundamental group of the torus with two segments attached?

I'm trying to calculate the fundamental group of the following space:

I've been thinking that I should apply Seifert - Van Kampen theorem but I haven't been able to choose some nice open sets $U$ and $V$ because most of the times I end up with something whose intersection is not path connected or doesn't give me anything interesting to calculate the group of the whole space. Which sets should I consider to prove this using this theorem? Is there an easier way to calculate $\pi_1(X)$ in this case?

I don't know covering spaces theory yet so I can't use those results.

P.S. I think that $\pi_1(X)= \mathbb{Z}$ using that if $A \subset X$ is contractible then $X$ and $X / A$ have the same homotopy type and considering as contractible subsets both segments. Unfortunately, I can't use this result in my proof.

• If you move the segments around inside the body, you can show that the space is homotopy equivalent to $T^2\vee S^1 \vee S^1$. Commented Nov 23, 2015 at 18:46
• Following @DanRust idea it will be very easy to apply Van Kampen since the theorem works very well with wedge sums. In that case the group would be: $\mathbb{Z}^2 *\mathbb{Z}*\mathbb{Z}$. Just note that you can split the computations in two steps: 1) The fundamental group of $Y=T^2 \bigvee S^1$: take as open subsets: A= torus and a bit of S^1 so it is contractible to the torus B= S^1 and a bit of the torus so it is contractible to S^1 The intersection of A and B is contractible All subsets are path-connected 2) The fundamental group of $Y \bigvee S^1$. Just repeat the previous idea Commented Nov 23, 2015 at 18:57
• No, the torus has fundamental group $\mathbb{Z}\times \mathbb{Z}$, and the circle has fundamental group $\mathbb{Z}$. The fundamental group of a wedge of finitely many spaces is the free product of the fundamental groups of the individual spaces, so we get $\pi_1=\mathbb{Z}^2\ast\mathbb{Z}\ast\mathbb{Z}$. Commented Nov 23, 2015 at 18:59
• You are absolutely right @DanRust. I made a stupid mistake. Commented Nov 23, 2015 at 19:02
• @DanRust Your comment has been really helpful, that idea has helped me to solve not only this problem but also some other related ones. If you post it as an answer I'll be glad to accept it.
– S -
Commented Nov 25, 2015 at 13:52

## 2 Answers

If you move the ends of the segments around inside the body, you can show that the space is homotopy equivalent to $T^2\vee S^1\vee S^1$.

If you wanted to make this rigorous, just find a contractible subspace $A$ of the torus which contains the end points of each segment (a tree in the torus with the ends of the line segments at its leaves will do) and then quotient out this subspace to form $X/A$. As $A$ is contractible this does not change the homotopy type, and as $A$ contains the end points of each segment, it's not hard to see that the quotient will be homeomorphic to $T^2\vee S^1\vee S^1$.

Finally, the torus has fundamental group $\mathbb{Z}×\mathbb{Z}$, and the circle has fundamental group $\mathbb{Z}$. The fundamental group of a wedge of finitely many spaces is the free product of the fundamental groups of the individual spaces, so we get $\pi_1\cong \mathbb{Z}^2 \ast \mathbb{Z} \ast \mathbb{Z}$.

First just consider the torus with one segment attached. Pick a nice loop going through the segment. Now pick a regular neighborhood of that loop as one open set and the complement of the loop as the other set. Boom, you got yourself two sets satisfying Van Kampen's theorem.

Observing that the intersection is topologically trivial you get immediately a free product. Now do this again for attaching the other segment.