$\pi_1$ and $H_1$ of Symmetric Product of surfaces

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?

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?

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This question may be of use: math.stackexchange.com/questions/45923/… – Dan Rust May 18 '13 at 1:09
One proof is found here in Lemma 2.6: arxiv.org/abs/math/0101206. For more details on the cohomology, look at the following paper of MacDonald: sciencedirect.com/science/article/pii/0040938362900198 – Henry T. Horton May 18 '13 at 1:15

@ Sadok Kallel: The book Topology and Groupoids has Chapter 11 on "Orbit spaces, orbit groupoids". The main result is to give circumstances under which the morphism of fundamental groupoids $\pi_1(X) \to \pi_1(X/G)$ presents the latter as an orbit groupoid. As a consequence, 11.5.4 calculates the fundamental group of a symmetric square.