Proposition 2

Suppose that in the commutative diagram (4), the sequences $(u, v)$ and $(u', v')$ are exact. Then: (i) We have $v_1 \circ u_1 = 0$; if $u'$ is injective, the sequence $(u_1, v_1)$ is exact. (ii) We have $v_2 \circ u_2 = 0$; if $v$ is surjective, the sequence $(u_2, v_2)$ is exact. (iii) Suppose that $u'$ is injective and $v$ is surjective. There exists then one and only one homomorphism $d : \text{Ker } (h) \to \text{Coker } (f)$ having the following property: if $z \in \text{Ker } (h), y \in N$ and $x' \in M'$ satisfy the relations $v(y) = k(z)$ and $u'(x') = g(y)$, we have $d(z) = p(x')$. Moreover the sequence (*) $\text{Ker } (f) \xrightarrow{u_1} \text{Ker } (g) \xrightarrow{v_1} \text{Ker } (h) \xrightarrow{d} \text{Coker } (f) \xrightarrow{u_2} \text{Coker } (g) \xrightarrow{v_2} \text{Coker } (h)$ is exact.

Let us prove (i). Since $u_1$ and $v_1$ have the same graphs as the restrictions of $u$ and $v$ to $\operatorname{Ker}(f)$ and $\operatorname{Ker}(g)$ respectively, we have $v_1\circ u_1=0$. We have

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

Let us prove (ii). Since $u_2$ and $v_2$ are obtained from $u$ and $v$ by passing to quotients, it is clear that $v_2\circ u_2=0$. Suppose that $v$ is surjective; since $q$ and $p$ are surjective, we have, by virtue of the hypotheses and prop. 1 (ii)

Let us finally prove (iii). For $z\in\operatorname{Ker}(h)$, there exists $y\in N$ such that $v(y)=k(z)$ since $v$ is surjective; moreover, we have $v'(g(y))=h(k(z))=0$, and consequently there exists a unique $x'\in M'$ such that $u'(x')=g(y)$ since $u'$ is injective. Let us show that the element $p(x')\in\operatorname{Coker}(f)$ is independent of the element $y\in N$ such that $v(y)=k(z)$. Indeed, if $y_1\in N$ is a second element such that $v(y_1)=k(z)$, we have $y_1=y+u(x)$ where $x\in M$; let us show that if $x'_1\in M'$ is such that $u'(x'_1)=g(y_1)$, we have $x'_1=x'+f(x)$; indeed, we have $u'(x'+f(x))=u'(x')+u'(f(x))=g(y)+g(u(x))=g(y+u(x))=g(y_1)$. Finally, we conclude that $p(x'_1)=p(x')+p(f(x))=p(x')$. We can therefore set $d(z)=p(x')$ and we have thus defined a mapping $d:\operatorname{Ker}(h)\to\operatorname{Coker}(f)$.

If now $z_1,z_2$ are elements of $\operatorname{Ker}(h)$, if $\lambda_1,\lambda_2\in A$ and $z=\lambda_1z_1+\lambda_2z_2$, we take elements $y_1$ and $y_2$ of $N$ such that $v(y_1)=k(z_1)$ and $v(y_2)=k(z_2)$ and we choose for $y\in N$ the element $\lambda_1y_1+\lambda_2y_2$; it is then immediate that

so $d$ is a homomorphism.

Suppose that $z=v_1(t)$ for some $t\in\operatorname{Ker}(g)$; we then take for $y\in N$ the element $j(t)$. Since $g(j(t))=0$, we conclude that $d(z)=0$, hence $d\circ v_1=0$. Conversely, suppose that $d(z)=0$. With the preceding notations, we thus have $x'=f(x)$, where x ∈ M. In this case, we have $g(y) = u'(f(x)) = g(u(x))$, or equivalently $g(y - u(x)) = 0$. The element $y - u(x)$ is therefore of the form $j(n)$ for $n \in \mathrm{Ker}\,(g)$, and we have

as $k$ is injective, $z = v_1(n)$, which proves that the sequence (*) is exact at $\mathrm{Ker}\,(h)$.

Finally, we have (with the same notations)

Conversely, suppose that an element $w = p(x')$ of $\mathrm{Coker}\,(f)$ is such that

We thus have $q(u'(x')) = 0$, and consequently $u'(x') = g(y)$ for a $y \in \mathbf{N}$; since $v'(u'(x')) = 0$, we have $v'(g(y)) = 0$, hence $h(v(y)) = 0$, in other words $v(y) = k(z)$ for a $z \in \mathrm{Ker}\,(h)$, and by definition $w = d(z)$, which shows that the sequence (*) is exact at $\mathrm{Coker}\,(f)$. We have seen in (i) that it is exact at $\mathrm{Ker}\,(g)$ and in (ii) that it is exact at $\mathrm{Coker}\,(g)$, which completes the proving of (iii).