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.

$T$ generates $F(S)$.

$T$ 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 $S$.