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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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  • $\begingroup$ You may want to give a precise reference, or explain the notation... $\endgroup$
    – Mariano Suárez-Álvarez
    Oct 15, 2011 at 7:54
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    $\begingroup$ This question is answered here: math.stackexchange.com/questions/45923/… $\endgroup$
    – user8594
    Oct 15, 2011 at 10:30
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    $\begingroup$ 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. $\endgroup$
    – Donu Arapura
    Oct 15, 2011 at 15:16
  • $\begingroup$ en, suppose $C$ is a representing closed curve in $H_{1}(Sym^{g}(\Sigma))$, what is the $g(C)$? $\endgroup$
    – yanqing
    Oct 16, 2011 at 6:55
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    $\begingroup$ 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 $H_1(\Sigma)$. $\endgroup$
    – Tim Perutz
    Oct 16, 2011 at 21:00

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