Let $X=Sym^d(\Sigma_g)$ be the d-fold symmetric product of a genus-g surface, $d\ge 2$.
Is there / what is a (quick simple) way to see that $\pi_1(X)$ is abelian?
The link in the comments (Ozsvath-Szabo paper) gives me the answer, except I don't follow the last part of the proof:
Here is my understanding:
Since $\pi_1(X)\to H_1(X)$ is surjective, it suffices to show that the kernel is trivial. A curve $\gamma:S^1\to X$ in general position (i.e. missing the codimension-1 diagonal $D\subset X$ consisting of elements in $\Sigma^{\times d}$ where at least two entries coincide) corresponds to a $d$-fold cover $\hat{\gamma}:S^1\to \Sigma$, via pullback along the branched cover $\Sigma^{\times d}\to X$. A null-homologous $[\gamma]=0\in H_1(X)$ thus gives a null-homologous $\hat{\gamma}$, i.e. there is a map $j:F\to\Sigma$ (for some surface $F$ with boundary) with $j|_{\partial F}=\hat{\gamma}$.
Now somehow, using this surface $F$ we can induce a null-homotopy of $\gamma\in \pi_1(X)$. Can someone elaborate on this?
The fundamental group of a symmetric product d>1 for any simplicial complex X is isomorphic to the first homology group of X. This fact is true for many other "permutation products" as well, like cyclic products for instance (i.e. quotients by the cyclic group). Full proofs and discussions are in http://cms.math.ca/10.4153/CJM-2012-028-3 (also available on arXiv:1010.1507). The argument showing that the fundamental group is abelian is simple enough: any loop in Symd(X) based in a diagonal element lifts to a loop in the cartesian product Xd. For two given elements in the fundamental group of Symd(X), their lifts can be chosen to live in different copies of (pi_1X)^d, d>1, and hence they commute there. Their images would therefore have to commute.