# Computing the cohomology ring of $\Sigma_2$

This question is from an old exam. I was completely lost on it and not sure where to start- hoping for even a point in the right direction.

Let $\Sigma_2$ be the genus 2 surface and $C\subset\Sigma_2$ a separating circle. So that $\Sigma_2/C$ is the wedge of two tori. Study the quotient map and your knowledge of the cup product structure on a torus to compute the cohomology ring of $\Sigma_2$.

Now I'm pretty shaky on using the cup product. I do know the cohomology ring of the torus is isomorphic to the exterior algebra on two variables. I don't know where the quotient map comes into play here at all. I have a feeling that degree comes into play somehow, but am not sure.

Any help would be much appreciated. Thank you.

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Hint: $H^*(\Sigma_2/C) \cong \tilde{H}^*(\Sigma_2,C)$, and the latter fits into a long exact sequence. – Aaron Mazel-Gee Jul 28 '11 at 14:49
@Aaron I wasn't aware of that isomorphism (though I suppose that's a pretty useful one). I will have to do some reading. With that I think I have it though. Thank you :) – MJoszef Jul 28 '11 at 19:38
Sure. You can find this e.g. in Hatcher; the key word is "excision". (Excision is perhaps the strongest reason why (co)homology is often explicitly computable whereas homotopy isn't. Higher homotopy groups don't satisfy excision, except in special cases.) – Aaron Mazel-Gee Aug 5 '11 at 17:47