Exercise A0.III.6.09

Proposition

Let $R$ be a ring, $F$ a non-zero free $R$-module and let $\varphi : M \to N$ be a homomorphism. Then $\varphi$ is surjective if and only if for all homomorphisms $\alpha : F \to N$ there exists a homomorphism $\beta : F \to M$ such that $\alpha = \varphi \circ \beta.$

Proof

First, we will prove the forward direction. Assume that $\varphi$ is surjective and let $\alpha : F \to N$ be arbitrary.

Let $A$ be a non-empty subset of $F$ such that $F$ is a free $R$-module on $A$ (such $A$ must exist, since $F$ is non-zero). We can form a set function $\tau : A \to \mathscr{P}(M)$ as $a \mapsto \varphi^{-1}(\alpha(a))$ where $\varphi^{-1} : N \to \mathscr{P}(M)$ is the preimage function. Since $\varphi$ is surjective, this preimage is always non-empty. The axiom of choice allows us to take $\tau': A \to M$ which is just $a \mapsto \operatorname{Any}(\varphi^{-1}(\alpha(a))),$ taking an arbitrary element from each preimage.

The universal property of $F$ as a free object gives us the unique homomorphism $\overline{\tau'} : F \to M$ extending $\tau'.$ Take any $a \in A$ and note that

$$\varphi(\overline{\tau'}(a)) = \varphi(\operatorname{Any}(\varphi^{-1}(\alpha(a)))) = \alpha(a).$$

Since $A$ is the generating set of $F,$ any element of $F$ can be written down as an $R$-linear combination of elements of $A.$ Because of this, any homomorphism $F \to X$ is uniquely determined by the values it takes on the elements of $A.$ In particular, since they agree on all elements of $A$, $\varphi \circ \overline{\tau'} = \alpha.$ This proves the forward direction.

Conversely, assume that for all $\alpha : F \to N$ there exists a $\beta : F \to M$ such that $\alpha = \varphi \circ \beta.$

Let $v \in N$ be arbitrary. Then we can define $\alpha : F \to N$ by $a \mapsto v$, sending every generator to $v$. Now, let $a_0 \in A$ and note that $\alpha(a_0) = \varphi(\beta(a_0))$, which implies that $v \in \im(\varphi).$ Thus, $\varphi$ must be surjective. This proves the backward direction.

$\blacksquare$