Find the fundamental group of the product of two circles with the diagonal removed.

Question

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

Answer 1

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...

Answer 2

To give a more geometric way of viewing this. Recall that $S^1\times S^1$ is homeomorphic to the torus $T^2$ which we can view as a square $[0,1]\times[0,1]$ with opposite edges glued together in the usual way. Take this square and cut out the diagonal $\Delta=\{x,y\mid x=y\}$. Now glue the left edge to the right edge to leave us with a parallelogram with the top and bottom edges glued together in the same orientation. Well, by a bijective linear transformation of the plane (a shear), we can homeomorphically map this parallelogram onto a square with one pair of opposite edges glued together. This is (one of the) usual model for the cylinder which is homotopy equivalent to the circle and so $\pi_1(T^2\setminus\Delta)\cong\pi_1(S^1)\cong \mathbb{Z}$.