Defining a map on a subgroup of a free group

Question

Given a set $S$, we write $G(S)$ for the free abelian group on the basis $S$. Given a subset $T\subseteq S$, let $H$ be the subgroup of $G(S)$ generated by $T$.

I wonder if the following is true: Can we define a group morphism of abelian groups $H \to K$ by specifying the image of this map on $T$?

I.e., does every map $T \to K$ where $K$ is an abelian group extend uniquely to a group homomorphism $H \to K.$

Observation: If $H = G(T)$, we are done, but this is the crux.

Thanks in advance.

Answer 1

Since $T \subseteq S$ and $S$ is a basis, the subgroup $H$ generated by $T$ is also a free commutative group. Thus, yes, every map from $T$ to a commutative group $K$ extends uniquely to a group homomorphism from $H$ to $K$.