I've recently begun thinking a little about Heegaard splittings and Heegaard diagrams for closed orientable 3 manifolds. I'd like to understand:
Given a Heegaard diagram for $Y$, can one 'read off' $H_{1}(Y)$ and $H_{2}(Y)$?
Ideally, 'read off' would mean something like apply an algorithm to a Heegaard diagram $(\Sigma, \{\alpha_{i}\}, \{\beta_{i}\})$ which describes the groups via generators and relations phrased in terms of the $\alpha_{i}$ and $\beta_{i}$.
Applying Mayer-Vietoris to the Heegaard decomposition $Y= M_{1} \cup_{\Sigma}M_{2}$, one finds it is sufficient to understand the map $(i_{1},i_{2})_{*} : H_{1}(\Sigma) \rightarrow H_{1}(M_{1}) \oplus H_{1}(M_{2})$ induced by the inclusions, as $H_{1}(Y) \cong coker((i_{1},i_{2})_{*})$ and $H_{2}(Y) \cong ker((i_{1},i_{2})_{*})$.
If I assume the $\alpha_{i}$ are chosen "nicely", I've been able to write down a matrix presenting this map. In the general case, I believe I can do the same so long as the following claim is true:
Given a Heegaard diagram $(\Sigma, \{\alpha_{i}\}, \{\beta_{i}\})$, there exist curves $\{a_{i}\}$ and $\{b_{i}\}$ in $\Sigma$ such that $a_{i} \cap \alpha_{j} = \{pt\} \delta_{ij}$, $b_{i} \cap \beta_{j} = \{pt\} \delta_{ij}$, and $\{a_{i}, b_{i}\}$ generate $H_{1}(\Sigma)$.
In all examples I've considered, it's easy to do this. Need this be true? Or is there an easy way to use M.V. to answer the original question?
Edit: I have seen this post, and now understand the presentation given for $H_{1}(Y)$, which is very nice. I'd still like to understand how to write this down directly from M.V., since that seems to me like a natural method of attack.
You find this in Section 3.5. of Jesse Johnson's notes: http://users.math.yale.edu/~jj327/notes.pdf