Universal property of free module, "converse"

Solution 1:

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 $f$.

Solution 2:

I'm pretty sure that it also works without Yoneda lemma:

Just consider $F/N$, where $N = (B)$, the R-module generated by B and a map $\psi: B \rightarrow F/N $. $\psi$ sends all $ x \in B$ to $0$, so it's the zero map.

We now extend $\psi$ in two ways:

• set $\phi_{1}: F \rightarrow F/N$ by mapping everything to $0$. This extends $\psi$
• let $\phi_{2}: F \rightarrow F/N$ be the canonical epimorphism. This also extends $\psi$

By the above property, the extension has to be unique, so $\phi_{1} = \phi_{2}$. Since the canonical epimorphism is then the zero map, it holds: $F = N$. For R-linear independence let $$\sum_{\text{finite}} r_{i}x_{i} = 0\;,\qquad r_{i}\in R, \:\, x_{i} \in B$$ Now let $r_{j}$ be one of the coefficients and define the map

$$\gamma_{j}:B\rightarrow R:x\mapsto\left\{\begin{array}{ll} 1, & x = x_{j} \\ 0, & x\neq x_{j}\end{array}\right.$$

Then again by above property this extends to: $F_{j}:F\rightarrow R$. And finally it holds $$0 = F_{j}(0) = F_{j}\left(\sum_{\text{finite}} r_{i}x_{i}\right) = \sum_{\text{finite}}r_{i}F_{j}(x_{i}) = r_{j}$$

So B is a basis of F.