Why are there always pairwise intersections in a Heegaard splitting? Let $M=A\cup B$ be a Heegaard splitting, such that $\{\alpha_i\}_{i=1}^g$ is a set of boundaries for meridian disks of $A$, and $\{\beta_i\}_{i=1}^g$ is a set of boundaries for meridian disks of $B$ (Recall that meridian disks are those properly embedded in the manifold, and such that the union of their boundaries doesn't separate the boundary of said manifold).
Why is it true that we can always find a permutation $\tau$ of $\{1,...,g\}$ such that $\alpha_i$ and $\beta_{\tau(i)}$ intersect? Moreover, if I'm not mistaken, once we've found such pairwise intersections for a subset of the $\alpha$'s, we can complete it to the rest of the $\alpha$'s (so that in the end each $\alpha$ intersects some $\beta$, and vice versa). Why is this possible?
Edit: It appears that this is not true in general, however it is true, for example, in the case of the rational homology sphere. I would like to understand why in this particular case the statement is true.
Thanks!
 A: As I noted in the comments above, what you originally asked is not true - the standard genus one splitting of $S^1 \times S^2$ does not have an intersection between its $\alpha$ and $\beta$ curve. However, it does hold for rational homology spheres. 
We take a Heegard splitting for $M$ with $\alpha$ and $\beta$ curves $(\alpha_i)_{i=1}^g$, $(\beta_j)_{j=1}^g$. We can consider this Heegard splitting as a handle decomposition of $M$, where the $\alpha$ curves are belt disks if 1-handles and the $\beta$ curves are are the attaching disks for the 2-handles. This means that $H_1(M)$ is presented by the matrix $A=(\alpha_i \cdot \beta_j)_{1\leq i,j\leq g}$, where $\alpha_i \cdot \beta_j$, is the algebraic intersection of the $\alpha_i$ and $\beta_j$.
In particular, if $M$ is a rational homology sphere, then $|H_1(M)|=|\det(A)|> 0$.
Consider the expression coming from the definition of the determinant (see wikipedia link below):
$$|\det(A)|= |\sum_{\sigma \in S_g} \epsilon(\sigma) \prod_{1\leq i\leq g}\alpha_i \cdot \beta_{\sigma(i)}|,$$
where $\epsilon(\sigma)=\pm 1$ is the sign of the permutation $\sigma$. Now since $\det A \ne 0$, this shows there exists $\sigma \in S_g$ such that $\prod_{1\leq i\leq g}\alpha_i \cdot \beta_{\sigma(i)}\ne 0$. So for this $\sigma$, $\alpha_i \cdot \beta_{\sigma(i)}\ne 0$ for $1\leq i \leq g$. Since algebraic intersection non-zero implies geometric intersection non-zero, this is the permutation you want.
http://en.wikipedia.org/wiki/Determinant#n.C2.A0.C3.97.C2.A0n_matrices 
