Cohomology groups of $\Sigma_{2}^1$

Question

I am trying to find the cohomology groups of $\Sigma_{2}^1$, that is an orientable surface with genus two and boundary one. However, I am really finding it difficult to understand what $\Sigma_{2}^1$ is. Is a punctured 2 torus $\Sigma_{2}^1$? Can I think of $\Sigma_{2}^1$ as a punctured $\Sigma_{2}$? Any help is appreciated!

Answer 1

The boundary of a compact manifold with boundary is again a compact manifold. In particular, the boundary of a compact surface with boundary is a compact one-dimensional manifold (i.e. a disjoint union of circles). The $1$ in $\Sigma^1_2$ denotes the number of boundary components (i.e. circles). That is, $\Sigma_2^1$ is a genus two surface with an open disk removed. This is homotopy equivalent to a punctured genus two surface, but not homeomorphic to it ($\Sigma_2^1$ is compact but a punctured genus two surface is not).

In order to compute the cohomology of $\Sigma_2^1$, note that $\Sigma_2 = \Sigma_2^1\cup D$ where $D$ is an open disk. Now you can use Mayer-Vietoris. Alternatively, you can use the homotopy invariance of cohomology and note that $\Sigma_2^1$ deformation retracts onto a bouquet of circles (I'll leave it to you to determine how many).