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?

# 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

#### Related Solutions

There is one answer to this question here. But it's useful to understand the style of argument of Ozváth and Szabó. Here's my attempt:

There's an obvious map $i_*: H_1(\Sigma)\to H_1(\operatorname{Sym}^g(\Sigma))$ induced by the map $i: \Sigma \hookrightarrow \operatorname{Sym}^g(\Sigma)$ given by $i(x)=[x,x_0,\ldots,x_0]$ for any fixed choice $x_0 \in \Sigma$.

**Claim 1.** *A map $S^1 \to \operatorname{Sym}^g(\Sigma)$ which misses the diagonal gives rise to a unique map from a $g$-fold cover of $S^1$ to $\Sigma$.*

*Proof.* Given a continuous map $f: S^1 \to \operatorname{Sym}^g(\Sigma)$, define
$$\tilde S = \{ (s,x)\in S^1 \times \Sigma : x \text{ is an entry of the $g$-tuple } f(s)\in \operatorname{Sym}^g(\Sigma)\}.$$
Consider the first and second coordinate projections $p_1: \tilde S \to S^1$ and $p_2: \tilde S \to \Sigma$. Since $f$ can be perturbed to avoid the big diagonal in $\operatorname{Sym}^g(\Sigma)$, each fiber $p_1^{-1}(s)$ consists of $g$ distinct points, and its easy to see that $p_1: \tilde S \to S^1$ is actually a $g$-fold cover. Then $p_2: \tilde S \to \Sigma$ is a continuous map from a $g$-fold cover of $S^1$, i.e. $\coprod S^1$, to $\Sigma$. $\blacksquare$

This gives a homomorphism of chain groups \begin{align*} \Phi: C_1(\operatorname{Sym}^g(\Sigma))&\to C_1(\Sigma) \\(f:S^1 \to \Sigma) &\mapsto (p_2: \tilde S \to \Sigma).\end{align*} Now for well-definedness at the level of homology:

**Claim 2.** *A cobordism $Z$ between two 1-chains $a,b \in C_1(\operatorname{Sym}^g(\Sigma))$ gives rise to a branched cover $\tilde Z$ of $Z$ with $\partial \tilde Z \mapsto \Phi(a)-\Phi(b)$.*

*Proof.* Suppose $F: Z \to \operatorname{Sym}^g(\Sigma)$ satisfies $F\big|_{\partial Z}=a-b$. (Note that $\Phi\left(F\big|_{\partial Z}\right)=\Phi(a-b)$.) We can assume $F$ hits the big diagonal in a discrete set of points, so now the space $\tilde Z$ (defined analogously to $\tilde S$) is a *branched* $g$-fold cover $P_1: \tilde Z \to Z$. Moreover, it restricts to an actual $g$-fold cover $ P_1\big|_{\partial \tilde Z}: \partial \tilde Z \to \partial Z$, just as above. Now $P_2: \tilde Z \to \Sigma$ satisfies
$$P_2 \big|_{\partial \tilde Z}= \Phi\left( F\big|_{\partial Z}\right)=\Phi(a-b)=\Phi(a)-\Phi(b),$$
so $\Phi(a)-\Phi(b)$ is nullhomologous. $\blacksquare$

It follows that $\Phi$ is well-defined on homology. And here's the final part:

**Claim 3.** *$\Phi$ and $i_*$ are inverses of each other.*

*Proof.* First we consider what $\Phi \circ i_*$ does to maps $\gamma: S^1 \to \Sigma$ that miss $x_0$, since these generate $H_1(\Sigma)$:
Since $i \circ \gamma : S^1 \to \operatorname{Sym}^g(\Sigma)$ takes $s$ to $[\gamma(s),x_0,\ldots,x_0]$, the corresponding cover $\tilde S$ is homeomorphic to $S^1 \coprod S^1$, with one connected component consisting of points $(s,\gamma(s))$ and one component of points $(s,x_0)$. Then $p_2$ maps one component via $\gamma$ and the other via the constant map $s \mapsto x_0$, so $\Phi (i_*[\gamma])=[\gamma]$.

For the other direction, first observe that a loop $\gamma: [0,1] \to \operatorname{Sym}^g(\Sigma)$ which is based at $[x_0,\ldots,x_0]$ but misses the diagonal for all $t \in (0,1)$ is equivalent to a collection of $g$ unordered loops $\gamma_j:[0,1] \to \Sigma$ based at $x_0$ satisfying $\gamma_j(t)\neq \gamma_k(t)$ for all $t\in( 0,1)$ and $j \neq k$. Fix an ordering on these loops, which is equivalent to fixing a lift $\tilde \gamma: [0,1] \to \Sigma^g$ of $\gamma$. Then, by the same argument that shows $\pi_1(\Sigma^g)$ is isomorphic to $\pi_1(\Sigma)^g$, we can homotope $\tilde \gamma$ to trace out the curves $\gamma_j$ one by one, i.e. $$\tilde \gamma' (t) = \begin{cases} \big(\gamma_1\left(gt \right),\gamma_2(0),\ldots,\gamma_g(0)\big) & 0 \leq t \leq 1/g \\ \qquad \vdots & \quad \vdots \\ \big(\gamma_1(1),\gamma_2(1),\ldots,\gamma_g(gt-g+1)\big) & \frac{g-1}{g} \leq t \leq 1. \end{cases}$$ Now project back down to $\operatorname{Sym}^g(\Sigma)$ where ordering doesn't matter. Since $\gamma_j(0)=\gamma_j(1)=x_0$, we get a loop which is of the form $\gamma'(t)=[\gamma_j(gt-j+1),x_0,\ldots,x_0]$ for $(j-1)/g \leq t \leq j/g$. This shows that all loops in $\operatorname{Sym}^g(\Sigma)$ can be realized in the form $s\mapsto [\gamma_1(s),x_0,\ldots,x_0]$. (Note that this shows $i_*$ is surjective, which is enough for our claim. But let's keep going.) So any loop $\gamma: S^1 \to \Sigma$ can be realized as $i \circ \gamma_1$ for some $\gamma_1 :S^1 \to \Sigma$. Then we again have $\tilde S\cong S^1 \coprod S^1$ with $p_2: \tilde S \to \Sigma$ restricting to $\gamma_1$ on one of the components and a constant map on the other component. Then $i_* \circ \Phi ([\gamma])= i_*[ \gamma_1]=[i \circ \gamma_1]=[\gamma]$, as desired. $\blacksquare$

You need to know how the $2$-cell $F$ is attached to the $1$-skeleton $\bigvee_{k=1}^hS_k^1$, namely by the word $a_1^2a_2^2\dotsc a_h^2$ (this word describes the homotopy class of the attaching map in the fundamental group $\pi_1(\bigvee_{k=1}^hS_k^1)=F(a_1,\dotsc,a_k)$, the free group on $a_1,\dotsc,a_k$, where $a_i$ denotes the loop traversing $S_i^1$ once counter-clockwise for each $i=1,\dotsc,h$). If you follow this attaching map by projecting onto the $i$-th $1$-cell $S_i^1$, you end with the map $S^1\rightarrow S_i^1$ represented by the word $a_i^2$, i.e. the map that traverses the circle $S_i^1$ twice in counter-clockwise direction. This map has degree $2$.

## Best Answer

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