So in Benson Farb's text on Mapping Class Groups he discusses how we can construct a map from $\mathcal{M}(S_g)$, the mapping class group, to $Sp(2g;\mathbb{Z})$. In doing so he discusses how the isotopy classes preserves the symplectic inner product once we have affixed a basis, but I don't see how he deduces this. Can somebody explain in a bit more detail regarding this. Thanks. This is the introduction from chapter 6 in Benson Farb's text for reference. Below are pictures of the symplectic form and the representation he constructs.
A symplectic structure is a (nondegenerate) bilinear form $\omega: H_1 \to \Bbb Z$. There is no need to pick a basis for $H_1$ to define the symplectic structure on it; when the surface has a specified orientation to begin with, it's given by $\omega(a,b) = \# a \cap b$, the algebraic intersection number of (representatives) of the homology classes $a$ and $b$. (This is what Farb and Margalit were saying at the beginning of the section.) At this point it is completely clear that $\omega$ is preserved by the mapping class group, because the algebraic intersection number of curves is preserved by a diffeomorphism. (Do you see why?)
The choice of explicit basis was to get a representation $MCG(\Sigma_g) \to Sp(2g,\Bbb Z)$ (if you want an action on a vector space to give you a representation, you need to pick a basis).