# de Rham cohomology of doubly punctured torus

Let $$T^2=S^1\times S^1$$. I'd like to know all de Rham cohomology groups of $$M=T^2-\{a,b\}$$ but I couldn't find a result. So I want to compute it and I'm thinking of using Mayer Vietoris sequence. I need two open sets whose union covers $$M$$. I'm having difficulty choosing these open sets. Any help is appreciated.

• The most easiest argument is probably to show that $M$ is homotopy equivalent to the wedge sum of three circles. Do you want an explicit calculation of the differential forms, or is an argument using homotopy equivalences enough for you? – klirk Dec 7 '18 at 23:37
• How did you know it contracts to the wedge sum of 3 circles? I can't visualize it. An argument in homotopy equivalence works. Thanks. – tangentbundle Dec 7 '18 at 23:39

Its easy to visualize when looking at the fundamental domain of the torus:

$$U$$ is a neighbourhood of $$x$$, $$V$$ is a neighbourhood of $$y$$. Removing $$x$$ and $$y$$ from the torus is homotopy equivalent to removing the whole neighbourhoods $$U$$ and $$V$$. Further we can choose $$U,V$$ so large, that they fill the entire triangle they lie in.

So the fundamental domain is homotopy equivalent to the union of the boundary of the square with the diagonal. But some parts of the boundary are identified with each other. Doing the identifications, we obtain the wedge sum of three circles (their common point is $$A$$). One circle corresponds to the left side = right side of the bundary, one circle corresponds to the top side = bottom side and one circle corresponds to the diagonal.

More precisely, first identifying top with bottom, we get two distinct edges connecting the bottom/top left vertex with the bottom/top right vertex (corresponding to the bottom/top edge and the diagonal) and two circles (corresponding to the left and right edges).
Then we identify the left and right edges. The two circles we already had get identified, and as the bottom/top left edge gets identified with the bottom/top right edge, the edges which previously connected these two vetices become circles.
So in the end we get three circles which are connected in their common point $$A$$, the bottom/top now left/right vertex.

I hope the detailled description did not make things more confusing. I probably should've also drawn this.

• Thanks for the elaborate explanation. – tangentbundle Dec 8 '18 at 0:19