Question on the symmetric product $\mathrm{Sym}^g\Sigma$ Let $\Sigma$ be a genus $g$ closed Riemann surface.
Recall that the symmetric product $\mathrm{Sym}^g\Sigma$ is the quotient of the $g$-fold product $\Sigma \times \dots \times \Sigma$ under the action of the symmetric group $\mathfrak{S}_g$ on $g$ letters.
In Lemma 2.6 of Osvath and Szabo's first paper 
http://arxiv.org/PS_cache/math/pdf/0101/0101206v4.pdf
on Heegaard Floer homology, they prove that 
$$H_1(\Sigma)\cong H_1(\mathrm{Sym}^g\Sigma).$$
Unfortunately I do not understand their proof. In particular, I do not understand their construction of the map $H_1(\mathrm{Sym}^g\Sigma) \rightarrow H_1(\Sigma)$, nor why this map inverts the (obvious map) $H_1(\Sigma) \rightarrow H_1(\mathrm{Sym}^g\Sigma)$. Could anyone explain this please?  
 A: 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 \operatorname{Sym}^g(\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. $\blacksquare$
