Top cohomology of a non-orientable smooth surface with boundary.

Question

I would like to know what the singular relative cohomology $H^2(M,\partial M;\mathbb{Z})$ of a smooth connected surface with boundary $M$ is. In the orientable case I did the following: The zero-th singular homology $H_0(M;\mathbb{Z})$ is zero since the surface is connected. Then, by Lefschetz duality, $H^2(M,\partial M;\mathbb{Z})$ is also zero. In the non-orientable case, I tried considering the orientable double cover $\tilde{M}$. By the previous argument we get $H^2(\tilde{M},\partial \tilde{M};\mathbb{Z})=0$. However I don't know how to relate this to the cohomology of the surface. Can you help me?

Answer 1

The relative long exact sequence gives us the exactness of $H^1(M) \to H^1(\partial M) \to H^2(M,\partial M) \to H^2(M)$. The last term is zero ($M$ deformation retracts onto a 1-complex; to see this, think about the picture of closed manifolds as polygons mod edge identifications). Call $M'$ the manifold you get when you glue in discs to each of the boundary circles; we have a Mayer-Vietoris sequence $H^1(M) \to H^1(\partial M) \to H^2(M') \to H^2(M) = 0;$ and for a non-orientable manifold, $H^2(M') = \Bbb Z/2$ by the universal coefficient theorem. So $H^2(M,\partial M) = \Bbb Z/2$.