In this post, my main question is: How to calculate the $n$-th singular homology group $H_n(X)$ of the compact surfaces $X$?
My attempt 1: We know the classical theorem of the classification of compact sufaces:
Theorem: Let $X$ be a compact 2-dimensional manifold. If $X$ is orientable, then either $X \cong \mathbb{S}^2$, or $X \cong M_1 \# \cdots \# M_g$, where each $M_i$ is $\mathbb{T} := \mathbb{S}^1 \times \mathbb{S}^1$; if $X$ is nonorientable, then $X \cong M_1 \# \cdots \# M_g$, where each $M_i$ is $\mathbb{RP}^2$. Here, $\cong$ means homeomorphic and $\#$ means taking connected sums.
By the classification theorem, I'm trying to reduce the case to the basic "components" of the surfaces, namely $\mathbb{S}^2$, $\mathbb{T}$ and $\mathbb{RP}^2$. So there are mainly two steps:
Step 1: Calculate the homology groups of $\mathbb{S}^2$, $\mathbb{T}$ and $\mathbb{RP}^2$. I have already obtained these. So here is no problem.
Step 2: Calculate the homology group of the connected sums. I get stuck here. 🙁
So the general question may be: Let $X$ and $Y$ be two topological spaces, whose connected sum $X \# Y$ can be defined. If we know the homology groups of $X$ and $Y$, then how to calculate $H_n(X \# Y)$? I'm hoping someone to help me on this. My idea is to use the Mayer-Vietoris sequence, but the definition of connect sums are quite messy and hence I do not know how to implement it in details.
My attempt 2: In Munkres's book Topology, he calculated some 1st-homology groups in Sect. 75. By de Rham theorem, the homology groups (with real coefficients) are isomorphic to the de Rham cohomology groups of $X$, and since $X$ has dimension 2, it seems that we can worry less on the homology groups of higher $n$. However, using de Rham theorem is likely to be a little "cheaty" and I'm not sure that this is a correct way.
My attempts 3: We can use the other classification theorem, which used cutting and pasting:
Theorem: Every compact 2-dimensional manifold $X$ is homeomorphic to either $M$ or $M^{\prime}$ defined as follows:
Oriented case: Let $W_h$ be a $4h$-gon in the plane with edges labeled
$$
\alpha_1, \beta_1, \alpha_1^{-1}, \beta_1^{-1}, \ldots, \alpha_h, \beta_h , \alpha_h^{-1}, \beta_h^{-1},
$$
and let $M_h$ be the quotient space of $W_h$, in which the edges are identified according to the labels.
Non-oriented case: Let $W^{\prime}_n$ be a $2n$-gon on the plane with edges labeled
$$
\alpha_1, \alpha_1, \ldots, \alpha_n, \alpha_n,
$$
and let $M^{\prime}_n$ be the quotient space of $W^{\prime}_n$, in which the edges are identified according to the labels.
However, since I haven't learnt CW-complexes or simplicial complexes systematically, and the original question appears before introducing such tools, I don't want to use the constructions above to calculate.
Sorry for such a long post, and thank you all for commenting or answering!! 🙂
Best Answer
When you take a connect sum of surfaces $X$ and $Y$, you delete a disk (call it $D$ on both surfaces) from each surface and identify the resulting boundary circles. To use Mayer-Vietoris, consider an open set $U$ that consists of $X$ together with a small portion of $Y$; if you are willing to accept the existence of a metric $d$ on $X \# Y$, you can fix $\epsilon > 0$ sufficiently small and say that this open set is the union of 1) $X$ with the disk $D$ deleted; and 2) the set of points $y$ which lie in $Y \setminus D$ such that $d(y,D) < \epsilon$. If you want to avoid metrics you can say something like "union of sufficiently small open sets centered along the boundary circle." Let $V$ be the open set constructed by the same procedure with $X$ and $Y$ swapped.
I will do the example $X = Y = T$ the torus, so $X \# Y$ is a genus $2$ surface $\Sigma_2$. The case where one of the surfaces is a projective plane is similar, and then you just induct on the number of summands. Mayer-Vietoris reads $$ 0 \to H_2(S^1) \to H_2(U)\oplus H_2(V) \to H_2(\Sigma_2) \to H_1(S^1) \to H_1(U)\oplus H_1(V) \to H_1(\Sigma_2) \to H_0(S^1) \to H_0(U)\oplus H_0(V) \to H_0(\Sigma_2) \to 0 $$
which, since $U$ and $V$ deformation retract onto the wedge of two circles (the $1$-skeleton of the standard CW structure on the torus), evaluates to
$$ 0 \to 0 \to 0\oplus 0 \to H_2(\Sigma_2) \to \mathbb Z \to \mathbb Z^4 \to H_1(\Sigma_2) \to \mathbb Z \to \mathbb Z\oplus \mathbb Z \to \mathbb Z \to 0 $$
using the known $H_*(S^1)$ and the fact that $H_0(\text{connected space}) \cong \mathbb Z$. Since the $H_0$ portion of the sequence is exact on its own, we can reduce this to
$$ 0 \to H_2(\Sigma_2) \to \mathbb Z \to \mathbb Z^4 \to H_1(\Sigma_2) \to 0. $$
Finally, you can check that the generator of $H_1(U\cap V) \cong H_1(S^1) \cong \mathbb Z$ goes to $0$ in $H_1(U)$ and $H_1(V)$, so the middle map is the $0$ map and this final sequence splits up into isomorphisms $H_2(\Sigma_2) \cong \mathbb Z$ and $H_1(\Sigma_2) \cong \mathbb Z^4$.