Proposition 1

Consider a commutative diagram of commutative groups:

Suppose that the two rows of (10) are exact. Then:

(i) If c is injective, we have

(11) $\operatorname{Im}(b) \cap \operatorname{Im}(u') = \operatorname{Im}(u' \circ a) = \operatorname{Im}(b \circ u).$

(ii) If a is surjective, we have

(12) $\operatorname{Ker}(b) + \operatorname{Im}(u) = \operatorname{Ker}(v' \circ b) = \operatorname{Ker}(c \circ v).$

Let us prove (i). Clearly

Conversely, let $x \in \operatorname{Im}(b) \cap \operatorname{Im}(u')$. There exists $y \in B$ such that $x = b(y)$. As $v' \circ u' = 0$, we have $0 = v'(x) = v'(b(y)) = c(v(y))$, whence $v(y) = 0$ since c is injective. As $(u, v)$ is an exact sequence, there exists $z \in A$ such that $y = u(z)$, whence $x = b(u(z))$.

Let us prove (ii). As $v \circ u = 0$ and $v' \circ u' = 0$, it is clear that

Conversely, let $x \in \operatorname{Ker}(v' \circ b)$. Then $b(x) \in \operatorname{Ker}(v')$ and there exists $y' \in A'$ such that $u'(y') = b(x)$, since the sequence $(u', v')$ is exact. As a is surjective, there exists $y \in A$ such that $a(y) = y'$, whence $b(x) = u'(a(y)) = b(u(y))$; it follows that $x - u(y) \in \operatorname{Ker}(b)$, which completes the proof.