Let $F$ be a free $R$-module with a basis $B$. We know that $B$ satisfies the following property:
For any $R$-module $M$ and any $g:B\rightarrow M$, there exists a unique $R$-map $\varphi:F\rightarrow M$ that extends $g$.
Now suppose that $F$ is any $R$-module and $B$ is any subset of $F$. If $B$ satisfies the above property, is $B$ a basis of $F$?
I think this is true for vector spaces. If $B$ doesn't span $F$ then there is no unique extension of $g$, and if $B$ is linearly dependent then the extension might not exist at all. But I'm having trouble extending my reasoning to modules, due to the lack of division and the presence of torsion elements.
Best Answer
Juan S's comment is correct. For any set $B$ let me denote by $F(B)$ the free module on that set. In this case there is a canonical map $f : F(B) \to F$, and your hypotheses say that the induced maps
$$f^{\ast} : \text{Hom}_{R\text{-Mod}}(F, M) \to \text{Hom}_{R\text{-Mod}}(F(B), M) \cong \text{Hom}_{\text{Set}}(B, M)$$
are bijections for every $M$. By the Yoneda lemma, $f$ must be an isomorphism. Explicitly, take $M = F(B)$ in the above; then $(f^{\ast})^{-1}(\text{id}_{F(B)})$ is an inverse to $f$.