Corollary 1

Suppose that the diagram (4) is commutative and has exact rows. Then:

(i) If $u', f$ and $h$ are injective, $g$ is injective. (ii) If $v, f$ and $h$ are surjective, $g$ is surjective.

Assertion (i) is a consequence of assertion (i) of prop. 2: indeed we have $\mathrm{Ker}\,(f) = 0$ and $\mathrm{Ker}\,(h) = 0$, hence $\mathrm{Ker}\,(g) = 0$.

Assertion (ii) is a consequence of assertion (ii) of prop. 2: indeed, we have $\mathrm{Coker}\,(f) = 0$ and $\mathrm{Coker}\,(h) = 0$, hence $\mathrm{Coker}\,(g) = 0$.