Compute $\pi_1 \left (S^1 \times S^1 \setminus \{(1,0,1,0)\}, (-1,0,-1,0) \right ).$
I know that it's torus minus a point. The idea is to use the polygonal representation of a torus i.e. a square with opposite edges identified in the same direction. To do that I took the square $I \times I$ where $I = [-1,1]$ and remove it's centre of mass $(0,0)$ and then do the necessary computation using Van Kampen's. But here the basepoint and the point to be removed are quite awkward looking. Although we can make a cutting of torus in such a way that the point $(-1,0,-1,0)$ becomes the centre of mass of it's polygonal representation and then do the same thing as before. But I want to make sure my ideas are fine so that when I write it down as a solution no ambiguity arises.
Could anyone give me some suggestions? Thanks for your time.
Think of the torus $T^2$ as the square $[0,1]^2$ with the edges identified: $(0,t)\sim (1,t)$ and $(s,0)\sim (s,1)$.
Also consider just the boundary of the square $\{0,1\}\times[0,1]\sqcup[0,1]\times\{0,1\}$ with the edges identified in the same way, call this space $B$ (for boundary).
I claim $B$ and $T^2\backslash\{p\}$ are homotopy equivalent, where $p=(0.5,0.5)$. Indeed, there is an obvious inclusion $B\hookrightarrow T^2\backslash \{p\}$ and the homotopy inverse is given by drawing a line from $p$ to $q\in [0,1]^2\backslash\{p\}$, which intersects the boundary of the square at a unique point (this is a so-caled deformation retraction).
Now it is easy to see that $B\simeq S^1\vee S^1$, which has fundamental group $F_2$, the free group on the two standard loops of $T^2$.