Proposition 2

Suppose that in the commutative diagram (10) the sequences $(u, v)$ and $(u', u')$ are exact. Then:

(i) $v_1 \circ u_1 = 0$; if $u'$ is injective, the sequence $(u_1, v_1)$ is exact. (ii) $v_2 \circ u_2 = 0$; if $v$ is surjective, the sequence $(u_2, v_2)$ is exact. (iii) Suppose that $u'$ is injective and $v$ is surjective. Then there exists one and only one homomorphism $d : \mathrm{Ker}(c) \to \mathrm{Coker}(a)$ with the following property: if $x \in \mathrm{Ker}(c)$, $y \in B$ and $t' \in A'$ satisfy the relations $v(y) = k(x)$ and $u'(t') = b(y)$, then $d(x) = p(t')$. Moreover the sequence

(*)

is exact.

Let us prove (i). As $u_1$ and $v_1$ have the same graphs as the restrictions of $u$ and $v$ to $\mathrm{Ker}(a)$ and $\mathrm{Ker}(b)$ respectively, we have $v_1 \circ u_1 = 0$. Then

But, by Proposition 1, (i), $\mathrm{Ker}(v_1) = \mathrm{Im}(j \circ u_1) = \mathrm{Im}(u_1)$ if $u'$ is injective.

Let us prove (ii). As $u_2$ and $v_2$ are obtained from $u$ and $v$ by taking quotients, it is clear that $v_2 \circ u_2 = 0$. Suppose that $v$ is surjective; as $q$ and $p$ are surjective, it follows from the hypotheses and Proposition 1, (ii) that

Finally let us prove (iii). For $x \in \mathrm{Ker}(c)$ there exists $y \in B$ such that $v(y) = k(x)$ since $v$ is surjective; moreover $v'(b(y)) = c(k(x)) = 0$ and consequently there exists a unique $t' \in A'$ such that $u'(t') = b(y)$, since $u'$ is injective. We now show that the element $p(t') \in \mathrm{Coker}(a)$ is independent of the element $y \in B$ such that $v(y) = k(x)$. For if $y' \in B$ is another element such that $v(y') = k(x)$, then $y' = y + u(z)$, where $z \in A$; we show that if $t'' \in A'$ is such that $u'(t'') = b(y')$ then $t'' = t' + a(z)$; for

Finally, it follows that $p(t'') = p(t') + p(a(z)) = p(t')$. Then it is possible to set $d(x) = p(t')$ and the mapping $d : \mathrm{Ker}(c) \to \mathrm{Coker}(a)$ has thus been defined.

If now $x_1, x_2$ are elements of $\mathrm{Ker}(c)$ and $x = x_1 + x_2$, we take elements $y_1, y_2$ of $B$ such that $v(y_1) = k(x_1)$ and $v(y_2) = k(x_2)$ and choose for $y \in B$ the element $y_1 + y_2$; it is then immediate that $d(x) = d(x_1) + d(x_2)$ and hence $d$ is a homomorphism.

Suppose that $x = v_1(x')$ for some $x' \in \mathrm{Ker}(b)$; then take $y \in B$ to be the element $j(x')$. As $b(j(x')) = 0$, it follows that $d(x) = 0$, hence $d \circ v_1 = 0$. Conversely, suppose that $d(x) = 0$. In the above notation we then have $t' = a(s)$, where $s \in A$. In this case we have $b(y) = u'(t') = u'(a(s)) = b(u(s))$, or $b(y - u(s)) = 0$. The element $y - u(s)$ is therefore of the form $j(n)$, where $n \in \mathrm{Ker}(b)$, and we have $k(x) = v(y) = v(u(s) + j(n)) = v(j(n)) = k(v_1(n))$; as $k$ is injective, $x = v_1(n)$, which proves that the sequence (*) is exact at $\mathrm{Ker}(c)$.

Finally, we have (always in the same notation)

and hence $u_2 \circ d = 0$. Conversely, suppose that an element $w = p(t')$ of $\mathrm{Coker}(a)$ is such that $u_2(w) = u_2(p(t')) = 0$ (where $t' \in A$). Then $q(u'(t')) = 0$ and consequently $u'(t') = b(y)$ for some $y \in B$; as $v'(u'(t')) = 0$, we have $v'(b(y)) = 0$, hence $c(v(y)) = 0$, in other words $v(y) = k(x)$ for some $x \in \mathrm{Ker}(c)$, and by definition $w = d(x)$, which shows that the sequence (*) is exact at

Coker(a). It has been seen in (i) that it is exact at Ker(b) and in (ii) it is exact at Coker(b), which completes the proof of (iii).