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Consider $\mathbb{P}_\mathbb{R}^2$, i.e. the $\mathbb{R}$-projective plane. I have two questions.

My first question is, what is an explicit description of the isomorphism between $H_*^{\text{cellular}}(\mathbb{P}_\mathbb{R}^2)$ and $H_*^{\text{simplicial}}(\mathbb{P}_\mathbb{R}^2)$?

My second question is, what is an explicit description of the abelianization map from $\pi_1(\mathbb{P}_\mathbb{R}^2)$ to $H_1(\mathbb{P}_\mathbb{R}^2)$?

Thanks for your time!

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In general, if you want the map from cellular homology to singular homology, just notice that if you have a homology class represented by a cell then that cell's closure can be regarded as a singular simplex, so send the cell to itself.

If you want to talk about simplicial homology here, probably the easiest way of thinking about things is to show that both cellular and simplicial homology are isomorphic to singular homology. That way you don't have to worry about comparing the simplicial structure you've chosen with a cellular structure you've chosen. And the situation above works for both cases.

Since the fundamental group of the projective plane is already abelian, the abelianization map is the identity.

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  • $\begingroup$ Hi @DanielMcLaury, thanks for the answer! I'm aware of your remarks on homology in general, but I'm interested in a very explicit description in the case of $\mathbb{P}_\mathbb{R}^2$ of the isomorphism between cellular and simplicial homology here. $\endgroup$ – user387384 Nov 8 '16 at 18:03
  • $\begingroup$ Well, of course that depends on explicitly saying which of the infinitely many possible cellular structures and simplicial structures you've chosen to endow $\mathbb{P}^2$ with. For instance, you could take both of them to be the same, at which point the map on homology is just induced by the identity map on chains. $\endgroup$ – Daniel McLaury Nov 8 '16 at 18:06

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