Theorem 1

Let A be a ring, E', E, E'' three A-modules and u : E' \to E, v : E \to E'' two homomorphisms. For the sequence

to be exact, it is necessary and sufficient that, for every A-module F, the sequence

(where $\bar{u} = \mathrm{Hom}(u, 1_F)$, $\bar{v} = \mathrm{Hom}(v, 1_F)$) be exact.

Suppose that sequence (1) is exact. If $w \in \mathrm{Hom}(E'', F)$ and $\bar{v}(w) = w \circ v = 0$, then $w = 0$ since $v$ is surjective. Sequence (2) is therefore exact at $\mathrm{Hom}(E'', F)$. We show that it is exact at $\mathrm{Hom}(E, F)$. $\bar{u} \circ \bar{v} = \mathrm{Hom}(v \circ u, 1_F)$ (§ 1, no. 2, formula (10)) and $v \circ u = 0$ since sequence (1) is exact at E. Therefore $\bar{u} \circ \bar{v} = 0$, that is $\mathrm{Im}(\bar{v}) \subset \mathrm{Ker}(\bar{u})$. On the other hand, if $w \in \mathrm{Ker}(\bar{u})$, then $w \circ u = 0$ and hence $\mathrm{Ker}(w) \supset \mathrm{Im}(u)$. But as sequence (1) is exact at E, $\mathrm{Im}(u) = \mathrm{Ker}(v)$ and hence $\mathrm{Ker}(w) \supset \mathrm{Ker}(v)$; as $v$ is surjective, it follows from § 1, no. 3, Remark that there exists a $w' \in \mathrm{Hom}(E'', F)$ such that w = w' \circ v = \bar{v}(w'). Therefore Ker(\bar{u}) \subset \operatorname{Im}(\bar{v}), which completes the proof that sequence (2) is exact.

Conversely, suppose that (2) is exact for every A-module F. As $\bar{u} \circ \bar{v} = \operatorname{Hom}(v \circ u, 1_F) = 0, w \circ v \circ u = 0$ for every homomorphism $w : E'' \to F$. Taking $F = E''$ and $w = 1_{E''}$, it is seen first that $v \circ u = 0$ and hence $u(E') \subset \operatorname{Ker}(v)$. Now take $F = \operatorname{Coker}(u)$ and let $\phi : E \to F = E/u(E')$ be the canonical mapping. Then $\bar{u}(\phi) = \phi \circ u = 0$ by definition and hence there exists a $\psi \in \operatorname{Hom}(E'', F)$ such that $\phi = \bar{v}(\psi) = \psi \circ v$; this obviously implies $u(E') = \operatorname{Ker}(\phi) \supset \operatorname{Ker}(v)$, which proves that sequence (1) is exact at E. Finally, let $\theta$ be the canonical homomorphism of $E''$ onto $F = E''/v(E)$; then $\bar{v}(\theta) = \theta \circ v = 0$, hence $\theta = 0$; therefore, $F = \{0\}$ and $v$ is surjective. Sequence (1) is therefore exact at $E''$.