# Find fundamental group of space $X=(S^1\times S^1)/(S^1\times \{s_0\})$

Find fundamental group of space $X=(S^1\times S^1)/(S^1\times \{s_0\})$

I am not sure how to do it. It seems obvious that $\pi_1(X)=\mathbb{Z}$ because $X$ is torus where one class of nontrivial loops is contracted to the point.

There is also fact I found which may be used here to show that $X$ is homotopically equivalent to $S^2\vee S^1$ which would surely make above result obvious. However I am not sure how to use this fact which says:

For $i=1,2$ let $Z_i=X \sqcup_{h_i}X$ be topological space resulting from gluing $\overline{B^n}$ to $X$ alongside map $h_i:S^{n-1}\to X$. If $h_1$ and $h_0$ are homotopic, then $Z_0$ and $Z_1$ are homotopically equivalent.

• You may either use the fact that $X$ is homotopic to $S^2 \vee S^1$ as you say, which is best proved pictorially (yes most would consider this a perfectly rigorous proof), or one may prove this by using the Seifert-van Kampen theorem using a covering of the space by a small neighbourhood around the contracted circle and its complement. – Dan Rust Jul 2 '18 at 12:02
• Here's one point where the groupoid-version of Van Kampen's theorem is very useful, because there's a very interesting contractible subset in this space – Max Jul 2 '18 at 14:55

Let $X$ be your space, and note that it can be realized as a quotient of $I^2$. In particular, the usual structure of the torus is a quotient map $q:I^2 \to I^2/\sim$ by taking $(x,0) \sim (x,1)$ and $(0,y) \sim (1,y)$.
There is another quotient now $p:I^2 \to I^2/\sim$ by taking $(x_1,0) \sim (x_2,0)$ and $(x_1,1) \sim (x_2,1)$. Your space is done by doing $q$ first and then $p$ (or technically the map induced by $p$.)
However, the situation is clearer if you do $p$ first and then $q$.
Pinching two sides of the square down to a point gives a structure of $2$ zero cells, and two $1$-cells joining them, and filling this with a $2$-cell. The induced bap by $q$ glues these two $1$ cells together giving a sphere, but we still have to identify the $0$-cells, so this is a sphere with two points identified.
There is an excellent illustration that a sphere with two points identified is homotopically just $S^2 \vee S^1$ here.