Representation of nullhomologous loop on compact surface as a product of commutators. Why this sentence is true?:
Assume that $M$ is compact surface and $f: S^1 \to M$ is nullhomologous and without selfintersections. Letting $g$ be the genus and $b$ the number of boundary components of $M$, it follows that there is a generating set $S=\{\alpha_1, \beta_1, ..., \alpha_g, \beta_g, x_1, ..., x_b\}$ such that:
$$ \pi_1(M, *) = \langle a_1, b_1, \ldots , a_g, b_g, x_1, \ldots , x_b \mid [a_1, b_1]\cdots [a_g, b_g]= x_1\cdots x_b \rangle . $$ 
and such that:
$f$ is homotopic to $[\alpha_1, \beta_1] \cdots [\alpha_{g'}, \beta_{g'}]$ for some $g'<g$.
I know Hurewicz teorem, so I know that $f$ belongs to comutant of fundamental group, but why representation of $f$ is such as above? In particular why in this representation doesn't $x_i$ appear?
Could anybody help me or show me the book,where I can read about it in a algebraic topology way (I'm not familiar with differential topology)?
[Edit]: I forgot to write, that $f$ is without selfintersections 

What you are trying to prove is false: The least number of products of commutators needed to represent the given element x of the derived subgroup is called the "commutator length" of x. The theorem is that in any torsion free noncyclic hyperbolic group (like surface group of genus at least 2) there are elements of arbitrary high (finite) commutator length. In fact, take any nontrivial element y of the derived subgroup, then the commutator length of $y^n$ will diverge linearly to infinity. You can find a proof and much more in the book by D.Calegari "scl" (which is an acronym for "stable commutator length").
