Let $X_{n}$ be the (unordered) configuration space of $n$ distinct points in $\mathbb{P}_{\mathbb{C}}^{1}$. The fundamental group of $X_{n}$ is the braid group on $n$ strands on the Riemann sphere, which I denote by $\pi_{1}X_{n}$, and which is generated by $\sigma_{1}, ... , \sigma_{n - 1}$, where $\sigma_{i}$ wraps the $i$th strand over the $(i + 1)$th strand. Consider the family over $X_{2g + 2}$ of hyperelliptic curves of genus $g$ such that for each element of $X_{2g + 2}$, the fiber is the hyperelliptic curve branched over $\mathbb{P}_{\mathbb{C}}^{1}$ at the corresponding set of $2g + 2$ points. After choosing a fiber $C$, this induces a monodromy representation $R : \pi_{1}X_{n} \to \mathrm{Sp}(H_{1}(C, \mathbb{Z}) \cong \mathrm{Sp}_{2g}(\mathbb{Z})$.

I need to show that the element $(\sigma_{1} ... \sigma_{2g + 1})^{2g + 2}$ lies in the kernel of $R$. I know I could do this through a tedious, elementary proof involving induction on rows and columns of $2g \times 2g$ matrices, but I feel that there should be a more enlightening proof using algebraic topology.

For instance, in *Braids, Links, and Mapping Class Groups*, Birman describes a map taking $\pi_{1}X_{n}$ to the mapping class group $M(0, 2g + 2)$ whose kernel is exactly the center of $\pi_{1}X_{n}$, which in fact is generated by the element $(\sigma_{1} ... \sigma_{2g + 1})^{2g + 2}$. How do I show that the monodromy action $R : \pi_{1}X_{n} \to \mathrm{Sp}_{2g}(\mathbb{Z})$ factors through this map $\pi_{1}X_{n} \to M(0, 2g + 2)$? A number of people (for instance, Mumford in *Tata Lectures on Theta II*) seem to assume this; can anyone provide a source that proves it, or advise me on what would be a sufficiently rigorous topological argument?