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](https://groupprops.subwiki.org/wiki/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](https://groupprops.subwiki.org/wiki/Hopf%27s_formula_for_Schur_multiplier) 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).



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](https://www3.nd.edu/~andyp/papers/TorelliLectures.pdf).


