Let $G=\pi(X,x)$ be the fundamental group of a compact orientable surface of genus $g\ge 2$. It is well known that a presentation of $G$ is $$G=\langle x_1,y_1,\dots,x_g,y_g \ | \ [x_1,y_1]\cdots [x_g,y_g]=1\rangle$$ (where $[x,y]=xyx^{-1}y^{-1}$ is the commutator).

Denote by $F$ be the free group with $2g$ generators $x_1,y_1,\dots,x_g,y_g$ and by $R$ be the normal closure of the relation $r=[x_1,y_1]\cdots [x_g,y_g]$, so $G=F/R$.

It is clear that $r\in [F,F]$.

Question: Is there an elementary proof that $r=[x_1,y_1]\cdots [x_g,y_g]\not\in [F,[F,F]]$?

This result appears when one considers the Stallings exact sequence associated to $$1\to [G,G]\to G\to G^{ab}\to 1$$ to get $$ H_2(G,\mathbb{Z})\to H_2(G^{ab},\mathbb{Z})\to [G,G]/[G,[G,G]]\to H_1(G,\mathbb{Z})\to H_1(G^{ab},\mathbb{Z}) \to 0$$

Since $H_1(G,\mathbb{Z})\cong H_1(G^{ab},\mathbb{Z})\cong G^{ab}$ we obtain a short exact sequence $$ H_2(G,\mathbb{Z})\to H_2(G^{ab},\mathbb{Z})\to [G,G]/[G,[G,G]]\to 0$$ which should be injective at the left (see at the end some argument why).

Now, Hopf's formula gives $$H_2(F/R,\mathbb{Z})=(R\cap [F,F])/[R,F]=R/[R,F]$$ since $R\subset [F,F]$, hence $H_2(F/R,\mathbb{Z})$ is cyclic and the generator is given by (the class of) $r$. So the map $\psi:H_2(G,\mathbb{Z})\to H_2(G^{ab},\mathbb{Z})$ is either injective or zero. But $$H_2(G^{ab},\mathbb{Z})\cong [F,F]/[F,[F,F]]$$ since $G^{ab}\cong F/[F,F]$, and so the map $\psi$ is given by the natural map $$\psi:R/[R,F]\to [F,F]/[F,[F,F]]$$ coming from the inclusion $R\hookrightarrow [F,F]$, hence $\psi$ is injective if and only if $r\not\in [F,[F,F]]$.

One possibility is to use another description of the map $\psi$ as $$H_2(G,\mathbb{Z})\to \bigwedge H_1(G^{ab},\mathbb{Z})$$ that should correspond to the dual of the cup product in cohomology via Poincaré duality (i.e. dual universal coefficient theorem) (but I am not sure if to consider this approach as really elementary).