So I was thinking about Nielsen-Schreier formula for the generators of a free group subgroup and i tried to apply it. Something must be wrong but I can't find the mistake.
So I have $F(a,b)$ free group generated by two element and an surjective homomorphism $F(a,b)\to F(b)$ that acts by killing the "a" letter from a word (i.e. $ \phi(aba^{-1}b)=bb=b^2)$. So we should have $\frac{F(a,b)}{\mathrm{ker}(\phi)} \cong F(b)$ free group on one element.
I want to find the generators of the kernel by applying Nielsen-Schreier formula, which states that $\mathrm{ker}(\phi)$ is freely generated by this set $$B = \{ u x (\overline{u x})^{-1} \mid u \in U, x \in X \}$$ where $U$ is a Schreier transversal for $\mathrm{ker}(\phi)$ and $X$ is $\{a,b\}$.
I figured that $\mathrm{ker}(\phi)$ obviously contains $F(a)$ and that consists of all the words that has the same amount of $b$ and $b^{-1}$ or, in other words, of all the words $w=a^{n_1}b^{m1}a^{n2}b^{m2}\cdots $ so that $\sum_{i=1}^n m_i=0$.
In the next step I tried to compute the quotient $\frac{F(a,b)}{\mathrm{ker}(\phi)}=\{1,bH,b^{-1}H,b^2H,b^3H,\ldots\}$. I'm pretty sure that's it because two words are in relation only if, summing the power of all the $b$ and $b^{-1}$ in the word the result is the same. (for example $w=b^{-1}abaab$ with a $b$ total power of $-1+1+1=1$ is in relationship with $v=aba$, in fact $wv^{-1} \in \mathrm{ker}(\phi)$.
So the Schreier Transversal is $U=\{1,b,b^2,\ldots\}$. The problem is that applying now the formula to this Schreier Transversal I got that the generator of $\mathrm{ker}(\phi)$ is $\{a\}$, which makes no sense.
Where's the mistake?
I was miscalculating the base of $ker \phi$ ($xu(\overline{xu})^{-1}$ instead of $ux(\overline{ux})^{-1}$). Applyng the right formula we have that $B=\{b^iab^{-i}, i \in Z\}$