Let $F(S)$ be the free group on a (possibly infinite) set $S$. Let $T$ be a subset of $F(S)$ with the following two properties.


1. $T$ generates $F(S)$.
2. $T$ injectively projects to a basis for the free abelian group $H\_1(F(S);\mathbb{Z})$.


Question : Must $T$ be a free basis for $F(S)$?


If $S$ is finite, then this follows from the standard fact that any generating set for $F(S)$ of cardinality $|S|$ is a free basis for $F(S)$ (for example, this is Proposition 2.7 in Lyndon and Schupp's book "Combinatorial Group Theory"). However, I don't see how to adapt this to the case where $S$ is infinite.


I really only care about the case where $S$ is countable, but I can't imagine that this is true for countable $S$ but false for uncountable $S$.



Bases for infinitely generated free groups
If I understand the question correctly, you are concerned that there might be hidden relations between the elements of $T$. Moreover, since the images of $T$ are linearly independent in the abelianization, these would have to be commutator relations.


This kind of thing can't happen: if there were a nontrivial relation involving elements of $T$, then said relation would involve only finitely many elements of $T$, so it would be a relation on the free subgroup generated by that finite set of elements. Essentially, condition (2) guarantees that the elements of any finite subset of $T$ are algebraically independent, and therefore all of the elements of $T$ are algebraically independent.


