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

Question

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?

Answer 1

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 $H_1 (\Sigma \times [0,1])\cong H_1(\Sigma)$.