# Prove that $\frac{H_1(\Sigma)}{[\alpha_1], \ldots, [\alpha_g], [\beta_1], \ldots, [\beta_g]} \cong H_1(Y)$

algebraic-topologyhomology-cohomologylow-dimensional-topologymanifoldsmayer-vietoris-sequence

I am looking at Definition 2.11 in this paper: https://arxiv.org/pdf/math/0101206.pdf. In particular, given a Heegaard diagram $$(\Sigma, \alpha, \beta)$$ for a 3-manifold $$Y$$, how to prove the isomorphism
$$\frac{H_1(\Sigma)}{[\alpha_1], \ldots, [\alpha_g], [\beta_1], \ldots, [\beta_g]} \cong H_1(Y)$$
using Mayer-Vietoris?

For me, the easiest way to see this identity is to view $$Y$$ as being obtained by taking $$\Sigma \times [0,1]$$ attaching fattened disks (i.e 3-dimensional 2-handles) to $$\Sigma \times \{0\}$$ and $$\Sigma \times \{1\}$$ using the $$\alpha_i$$ and $$\beta_j$$ respectively. This gives a 3-manifold $$\widehat{Y}$$ with two spherical boundary components (exercise: see why the Heegaard splitting conditions imply this boundary), which we cap off with two balls to obtain $$Y$$. Since we are just computing $$H_1$$ of $$Y$$ we can ignore the balls and just compute $$H_1(\widehat{Y})\cong H_1(Y)$$. We can do this using Mayer-Vietoris on the decomposition $$\widehat{Y}=\Sigma \times [0,1] \cup \sqcup_{i=1}^{2g} D^2 \times D^1$$. Looking at the $$H_1$$ terms, we obtain a sequence of the form: $$H_1( \sqcup_{i=1}^{2g} \partial D^2 \times D^1) \rightarrow H_1 (\Sigma \times [0,1]) \rightarrow H_1(\widehat{Y}) \rightarrow 0$$ This gives the necessary identity, since the images of $$\partial D^2 \times D^1$$ are exactly the classes given by the $$\alpha_i$$ and $$\beta_i$$ in $$H_1 (\Sigma \times [0,1])\cong H_1(\Sigma)$$.