# Computing the homology of genus $g$ surface, using Mayer-Vietoris

Let $\Sigma_g$ be the compact orientable surface of genus $g$. I'm trying to compute the homology groups of $\Sigma_g$ using Mayer-Vietoris sequence. To do this, I first calculated the homology groups of wedge of circles. Using induction, one can easily show that the wedge of $k$ circles have $H_0=\mathbb{Z},H_1=\mathbb{Z}^k$ and other $H_n$s zero.

Now to calculuate the homology of $\Sigma_g$, Let $U$ be $\Sigma_g$ with a small hole on it, and $V$ be a disk on the surface slightly larger than the hole. This pair $(U,V)$ satisfy the conditions of excision (M-V sequence), and one can see that $U$ is homotopically equivalent to the wedge of $2g$ circles. Since $U\cap V$ clearly deformation retracts to a circle, the M-V sequence

$$H_2(U)\oplus H_2(V)\rightarrow H_2(X)\rightarrow H_1(U\cap V)\rightarrow H_1(U)\oplus H_1(V)\rightarrow H_1(X)\rightarrow H_0(U\cap V)\rightarrow H_0(U)\oplus H_0(V)\rightarrow H_0(X)\rightarrow H_{-1}(U)\oplus H_{-1}(V)$$

becomes

$$0 \rightarrow H_2(X)\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}^{2g}\oplus 0 \rightarrow H_1(X)\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}\oplus\mathbb{Z}\rightarrow\mathbb{Z}\rightarrow 0.$$

This is the part where I'm stuck. How does one conclude from here that $H_2(X)=\mathbb{Z}$ and $H_1(X)=\mathbb{Z}^{2g}$? I've managed to prove that the map $\mathbb{Z}^{2g}\oplus 0 \rightarrow H_1(X)$ must be an epimorphism, but I don't see why this has to be an isomorphism; if it is an isomorphism, the result follows immediately.

• why $U$ is homotopically equivalent to the wedge of $2g$ circles? Jul 17, 2019 at 4:04
$$H_1(U\cap V)$$ is generated by the attaching map of the 2-cell which includes each generator twice, once with $$+$$ sign and once with $$-$$ sign. Therefore it is homologous to zero. Hence the map $$\mathbb{Z}\to \mathbb{Z}^{2g}$$ is the zero map. Hence $$H_2(X)=\mathbb{Z}$$ and $$H_1(X)=\mathbb{Z}^{2g}$$.
• Are you saying that the generator of $H_1(U \cap V)$ is homologous to zero? Wouldn't this mean that $H_1(U \cap V)$ is trivial? But we know that this is $\mathbb{Z}$. What am I missing? Mar 6, 2019 at 20:56