TITLE: A question on the fundamental group of a compact orientable surface of genus >1
QUESTION [9 upvotes]: 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).

REPLY [3 votes]: 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.