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Edit: I already received a good answer to my second question. I'd be interested in a hint about the first one, as well. Thanks in advance!

I'm interested in compact Riemann surfaces and their homology. In this question, Kundor proposes a nice drawing of the connected sum of tori, saying that it is clearer than the traditional drawing of regular $4g$-gon whose sides are to be identified.
At first I was convinced that this was a smarter way to draw the $4g$-gon (just a continuous deformation), but then I realized that this is not the case (it works for genus 2, though).

So I was wondering if there exists a way to make the transition from the $4g$-gon to the "connected sum of squares", that makes the construction of the genus $g$ surface a lot easier to visualize.

Also, I had in mind a theorem, saying that the homology groups of the connected sum of two spaces are the direct sum of the homology groups of the single spaces (well, apart from $H_0$ and $H_n$).

Only, I wasn't able to find a reference: I was convinced I saw the result in Nakahara, Geometry, Topology and Physics, but I couldn't find it anymore. After some googling, I found these two sources (link1, link2), but nothing conclusive on any book I consulted.

Does this theorem have a name? Do you know a book where it is stated/proved?

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Computing the homology of a connected sum is a matter of applying Mayer--Vietoris.

We have (let's say closed, connected, orientable) manifolds $M_1$ and $M_2$. In each of them we choose an $n$-ball $B_i$, and a slightly smaller $n$-ball $B_i'$ contained in $B_i$ (here $i = 1,2$).

For a moment, let's omit the subscripts and just let $M$ be a closed $n$-manifold. Let $B$ be an $n$-ball in $M$, let $B'$ be a slightly smaller ball contained in $B$, so that $B \setminus B'$ is a spherical shell, while $M = M\setminus B' \cup B.$ Then by Mayer--Vietotris, one computes that $H_n(M\setminus B') = 0$, while $H_i(M \setminus B') \cong H_i(M)$ for $i < n$.

Now reintroduce the subscripts: the connected sum $M_1 \# M_2$ of $M_1$ and $M_2$ is obtained by gluing $M_1 \setminus B'_1$ and $M_2 \setminus B_2'$ along their (homeomorphic) spherical shells $B_1 \setminus B_1'$ and $B_2 \setminus B_2'$.

Applying Mayer--Vietoris again, we find that $H_n(M_1 \# M_2) = H_0(M_1 \# M_2) = \mathbb Z$, and that $H_i(M_1 \# M_2) = H_i(M_1) \oplus H_i(M_2)$ if $0 < i < n$.

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Thanks, this is interesting. Have you got any advice on my first question, please? – Andrea Orta Jul 27 '13 at 13:31
@Matt E: Is the orientability assumption needed for your argument in the connected sum? in other words, if you were to assume that $M_1$ and $M_2$ are non-orientable, can you still glue them easily along the boundary and apply Mayer-Vietoris? that seems to be not the case, but i don't see where your argument fails. – adrido Jan 31 '14 at 12:27
@adrido: Dear adrido, Implicit in my argument is that the boundary map $H_n(M) \to H_{n-1}(B\setminus B') = H_{n-1}(S^{n-1})$ is an isomorphism. This uses the the description of the fundamental class in the top cohomology of a closed connected orientable manifold. If $M$ were not orientable, then this map certainly wouldn't be an isomorphism, and what we would find instead is that $H_{n-1}(M\setminus B')$ is different from $H_{n-1}(M)$. E.g. think about removing a disk from the real projective plane to get a Mobius band. Regards, – Matt E Jan 31 '14 at 19:49
Dear @Matt: this makes sense. Thank you! but I do not know how to compute $H_{n-1}(B\backslash B')$ in the case of non-orientable manifolds. My main problem is that I don't know how to picture the intersection. Can you please tell me how this is done? – adrido Feb 3 '14 at 1:33
@adrido: Dear adrido, $B$ is still a ball, and $B'$ is still a smaller ball, so $B\setminus B'$ is still a spherical shell (i.e. an $n-1$-sphere times an interval). The ambient manifold $M$ doesn't play any role in this computation. Regards, – Matt E Feb 3 '14 at 2:50

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