# why $H_{1}(\Sigma)\cong H_{1}(Sym^{g}(\Sigma))$ ?

In paper holomorphic disks and 3-manifold invariants, Ozsvath and Szabo connstruct two homeomoephisms $\mathcal {f} : H_{1}(\Sigma)\rightarrow H_{1}(Sym^{g}(\Sigma))$ and $\mathcal {g} : H_{1}(Sym^{g}(\Sigma))\rightarrow H_{1}(\Sigma)$. Then they says these two maps are inverses of each other.

Sorry for my weak ability, I can not see how this work. So can somebody explain it, like what f or g maps the generator to ? And why these two maps are inverses of each other? Thank you.

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You may want to give a precise reference, or explain the notation... –  Mariano Suárez-Alvarez Oct 15 '11 at 7:54
This question is answered here: math.stackexchange.com/questions/45923/… –  Will Merry Oct 15 '11 at 10:30
Mariano is absolutely right that the notation should be explained in the question, but from the link that Will gave, I gather that $\Sigma$ is a genus $g$ Riemann surface. So the result also follows by noting that $Sym^g\Sigma$ is birational to a $g$-dimensional complex torus by the Abel-Jacobi theorem, and that $H_1$ is birational invariant. –  Donu Arapura Oct 15 '11 at 15:16
en, suppose $C$ is a representing closed curve in $H_{1}(Sym^{g}(\Sigma))$, what is the $g(C)$? –  yanqing Oct 16 '11 at 6:55
Yanqing, the inverse to $f$ [$g$ is not a good letter to choose!] is outlined by Jim Conant in the answer linked to by Will. If $\gamma\colon S^1\to Sym^g(\Sigma)$ is a smooth loop, transverse to the diagonal (hence disjoint from it), the $g$ points in $\gamma(t)$, for varying $t$, trace out a 1-manifold $T$ which comes with two maps: a $g$-fold covering map $T \to S^1$ (this orients $T$) and a map $T\to \Sigma$. Take the fundamental class of $T$ in $H_1(\Sigma)$. –  Tim Perutz Oct 16 '11 at 21:00