Morphism of free groups that induces isomorphism on abelianizations I came up with the following question, that I'm not able to prove or disprove.
Let $\phi: F_I \to F_J$ a morphism between the free groups generated by the sets $I$ and $J$. This induces a morphism between the abelianizations $\phi_\mathrm{ab}: \oplus_I \mathbb{Z} \to \oplus_J \mathbb{Z}$. Suppose that $\phi_\mathrm{ab}$ is an isomorphism. Does this imply that $\phi$ is an isomorphism as well?
I am able to see that

*

*$|I| = |J|$

*$\phi$ is injective

But I'm not able to prove surjectivity nor to find a counterexample.
 A: The alternating group $A_5$ is generated by $a = (12345)$ and $b = (12)(34)$. These give a transitive action of the free group $F_2$ on $ \{ 1, 2, 3, 4, 5 \}$. The stabilizer of $5$ is a subgroup $H$ of index $5$ (which, e.g. by covering space arguments, is a free group $F_6$, but we don't need this). We'll try to construct a counterexample by picking $2$ elements of $H$ such that the induced map $\phi : F_2 \to F_2$ is an isomorphism on abelianizations; since $H$ is a proper subgroup this map cannot be onto. (The alternating group was chosen here because it's the smallest perfect group; I'm not actually sure if this is necessary but it seemed worth a try.)
By construction $H$ contains $b$ which we'll choose to be one of the elements; this makes our job choosing the other element easy, we only have to ensure that it's a word in $a$ and $b$ stabilizing $5$ such that the total exponent of $a$ is $1$ or $-1$. Some fiddling around in WolframAlpha gives $a^4 b a^{-3} b = (2 3 4)$ which therefore also lies in $H$. So the map sending $a, b$ to $a^4 b a^{-3} b, b$ (from the free group on $a$ and $b$ to itself) is our desired $\phi$. The induced map on the abelianization sends $a, b$ to $a + 2b, b$ so is clearly invertible.
