algebraic-topologyfundamental-groups
Let $X = S^1 \times S^1 – \Delta$ , where $\Delta = \{ (x,y) \in S^1 \times S^1 | x = y \}$. Determine the fundamental group of $X$.
I know $\pi_1 (S^1 \times S^1) = \pi_1 (S^1) \times \pi_1 (S^1) = \mathbb{Z} \times \mathbb{Z}$. So taking away the diagonal should in some way limit the amount of paths that I can make. But I am not sure how to visualize this one. I usually like to use deformation retracts to solve these problems.
Any help is appreciated!
Thanks
The $n$-holed torus has as fundamental group the group presented as
$$\langle a_1, b_1, \ldots, a_n, b_n \mid [a_1,b_1]\cdots[a_n,b_n] = 0\rangle$$
where $[a, b] = aba^{-1}b^{-1}$.
As an example, consider this octagon:
Identify all corners, then identify the edges as labeled, and you get a 2-holed torus. The sequence $a_1b_1a_1^{-1}b_1^{-1}a_2b_2a_2^{-1}b_2^{-1}$ of edge labelings immediately gives you the generating relation for the fundamental group.
Let $S^2$ be a sphere of radius $2$ centred at the origin, and let $L$ be the straight line segment connecting the points $(2, 0, 0)$ and $(-2, 0, 0)$. The space $R^3 - A$ deformation retracts onto $S^2 \cup L$, whose fundamental group is $\mathbb Z$.
Hatcher, page 46, example 1.23 contains pictures of these kinds of deformation retractions for a number of examples, including $\mathbb R^3 - S^1$, which is very similar to the present example.
Best Answer
There is an automorphism of the torus taking the $(1, 1)$ curve (the diagonal) to a $(1, 0)$ curve (a meridian). Removing the meridian leaves you with a cylinder...