A question on the fundamental group of a compact orientable surface of genus >1

Probably the easiest way to see that the map $\psi\colon H_2(G) \rightarrow H_2(G^{\text{ab}})$ is injective is as follows. Since we're dealing with a surface group, the surface $\Sigma_g$ itself is an Eilenberg-MacLane space. Let $\{a_1,b_1,\ldots,a_g,b_g\}$ be the usual collection of oriented simple closed curves that one draws whose homology classes form a basis for $H_1(\Sigma_g)$. Thus $a_i$ intersects $b_i$ once, and otherwise the curves are disjoint. Let $f\colon G^{\text{ab}} \rightarrow \mathbb{Z}^2$ be the map whose kernel is spanned by $\{[a_2],[b_2],\ldots,[a_g],[b_g]\}$ and which takes $[a_1]$ and $[b_1]$ to the usual basis for $\mathbb{Z}^2$. To prove that $\psi$ is injective, it is enough to prove that the composition $$\phi\colon H_2(G) \stackrel{\psi}{\longrightarrow} H_2(G^{\text{ab}}) \stackrel{f_{\ast}}{\longrightarrow} H_2(\mathbb{Z}^2)$$ is injective. But $\phi$ is easy to understand geometrically: the surface $\Sigma_g$ is an Eilenberg-MacLane space for $G$, a torus $T$ is an Eilenberg-MacLane space for $\mathbb{Z}^2$, and $\phi$ is induced by the map $\Phi\colon \Sigma_g \rightarrow T$ that collapses a genus $(g-1)$-subsurface with one boundary component to a point. This subsurface contains $a_2,b_2,\ldots,a_g,b_g$. The point here is that it is obvious that $\Phi_{\ast}$ takes the fundamental class of $\Sigma_g$ to the fundamental class of $T$, and thus induces an isomorphism on $H_2$.

You can soup up this argument to show that $\psi$ takes the fundamental class of $\Sigma_g$ to the element $a_1 \wedge b_1 + \cdots + a_g \wedge b_g$ of $H_2(\mathbb{Z}^{2g}) \cong \wedge^2 \mathbb{Z}^{2g}$. For more details, see Theorem 2.7 of the lecture notes from my Park City course on the Torelli group, which are available here.

The dual to the map $\psi\colon H_2(G,\mathbb{Z}) \to H_2(G^{\operatorname{ab}},\mathbb{Z})$ is the cup-product map $\cup\colon H^1(G,\mathbb{Z})\wedge H^1(G,\mathbb{Z}) \to H^2(G,\mathbb{Z})$; see e.g. Lemma 1.10 in arXiv:math/9812087. Clearly, the latter map is surjective; hence, the former map must be injective.