On the generators of free $R$-modules.

Question

Let $R$ be a ring and $M$ be an $R$-module. Let $(F(M),\eta)$ be the free $R$-module over $M$. By that I mean that, for every $R$-module $N$ and every function $f\colon M\to N$ there exists exactly one $R$-linear mapping $\hat{f}\colon F(M)\to N$ such that $f=\hat{f}\circ \eta$. Now let $X$ be a set of generators for $M$. Is it possible to prove that $\eta(X)$ is a set of generators for $F(M)$, using just the universal property of $F(M)$? I don't want to use any realisation of the free $R$-module F(M), such as the direct sum of $|M|$ copies of $R$, I would like to use only universality. Is that possible?

Answer 1

The generators of $M$ have nothing to do with generators of $F(M)$. In fact, if $X \subsetneq M$ is a proper subset of $M$, then $\eta(X)$ does not generate $F(M)$. The point here is that the free module structure $F(M)$ does not in any way depend on module structure on $M$, it only sees $M$ as a set.