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Let $S$ be a connected orientable topological compact surface of genus $g$. There are various ways to prove that the second singular homology group with integer coefficients $H_2(S, \mathbb{Z})$ is a free abelian group of rank $1$.

This question concerns how to concretely write down a generator using the definition of singular homology. That is, we need to find a singular $2$-chain with zero boundary.

The idea would be of course to use a triangulation of $S$ and choose suitable signs on each singular $2$-simplex so that the corresponding chain has no boundary. But when I try to make this precise I run into the problem that one oriented face of a triangle minus the same face with the reversed orientation is not the zero $1$-chain.

Is there some tricky way to choose the signs that makes things work (for example, representing $S$ as the quotient of a $4g$-gon and considering a triangulation out of its center) or one is forced to use that singular homology is isomorphic to simplicial or cellular homology?

I would be very grateful if someone knew how to do that, or a reference where it is properly explained.

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Firstly, if you start with one triangle then there is a unique sign for any neighbouring triangle such that the common boundaries add up to zero. Thus you can start at one triangle and distribute signs accordingly. Using a linear order on the vertices as in Hatchers book one can turn this into an algorithm. The point you are making, I think, is that one needs external knowledge to know that this is possible for the whole surface. If instead of ordering the vertices you use orientations, it immediately becomes clear that $H^2$ is $\mathbb Z$ for orientable surfaces, because you choose a global orientation, this induces orientations on each triangle, these induce orientations on each of their boundary 1-simplices and that the two orientations that a given 1-simplex gets, cancel each other out.

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  • $\begingroup$ Could you please elaborate on the second part of your answer? It is not clear to me why from a global orientation you get 1-simplexes cancelling out $\endgroup$
    – lalin
    Sep 16, 2022 at 3:50
  • $\begingroup$ Say you have a triangle with vertices a,b,c where the order gives the orientation. Neighbouring it, you have the triangle a,b,d. Now, what is the orientation that this neighbouring triangle gets? Draw a picture and you see, it is b,a,d. The common face is a,b, which from the first triangle gets the orientation a,b, friom the second the orientation b,a. The two cancle out. $\endgroup$
    – user473423
    Sep 16, 2022 at 6:00

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