Suppose that the two lines of (4) are exact. Then:

(i) If $h$ is injective, one has

(ii) If $f$ is surjective, one has

Let us prove (i). It is clear that one has

Inverse support, let $y'\in\operatorname{Im}(g)\cap\operatorname{Im}(u')$. There exists $y\in N$ such that $y'=g(y)$. Since $v'\circ u'=0$, we have $0=v'(y')=v'(g(y))=h(v(y))$, whence $v(y)=0$ since $h$ is injective. Since $(u,v)$ is an exact sequence, there exists $x\in M$ such that $y=u(x)$, whence $y'=g(u(x))$.

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

Conversely, let $y\in\operatorname{Ker}(v'\circ g)$. Then $g(y)\in\operatorname{Ker}(v')$, and there exists $x'\in M'$ such that $u'(x')=g(y)$ since the sequence $(u',v')$ is exact. Since $f$ is surjective, there exists $x\in M$ such that $f(x)=x'$, whence $g(y)=u'(f(x))=g(u(x))$; we conclude that $y-u(x)\in\operatorname{Ker}(g)$, which completes the proof.

Then there exists one and only one homomorphism $u_1:\operatorname{Ker}(f)\to\operatorname{Ker}(g)$, and one and only one homomorphism $u_2:\operatorname{Coker}(f)\to\operatorname{Coker}(g)$, such that the diagrams

and

are commutative, $i$ and $j$ denoting the canonical injections, $p$ and $q$ the canonical surjections.

Indeed, if $x \in \text{Ker } (f)$, we have $f(\bar{x}) = 0$ and $g(u(x)) = u'(f(x)) = 0$, hence $u(x) \in \text{Ker } (g)$, and the existence and uniqueness of $u_1$ are then immediate. Similarly, we have

hence $u'$ gives by passing to the quotients a homomorphism which is the only homomorphism for which (9) is commutative.

Let us now start from a commutative diagram (4) of A-modules; by virtue of Lemma 1, there corresponds to it a commutative diagram

where $i, j, k$ are the canonical injections, $p, q, r$ the canonical surjections, $u_1, u_2$ (resp. $v_1, v_2$) the homomorphisms deduced from $u, u'$ (resp. $v, v'$) by Lemma 1.

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).

(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 $\mathrm{Coker}\,(g) = 0$.

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

To prove (i), consider the diagram

where $f'$ is the mapping having the same graph as the restriction of $g$ to $u(\mathbf{M})$, $w$ and $w'$ being the canonical injections; it is clear that this diagram is commutative and has exact rows. Moreover $w'$ is injective, and by assumption $v$ is surjective; we therefore have, by prop. 2 (iii), an exact sequence

since $g$ is injective and $f'$ is surjective, we therefore have $\mathrm{Ker}\,(h) = 0$.

To prove (ii), consider the diagram

where this time $h'$ is the mapping having the same graph as the restriction of $h$ to $\nu(N)$, and $w$ and $w'$ have respectively the same graphs as $\nu$ and $\nu'$; this diagram is commutative and its rows are exact. Moreover $w$ is surjective and by assumption $u'$ is injective; we therefore have, by prop. 2 (iii), an exact sequence

since $g$ is surjective and $h'$ is injective, we therefore have $\operatorname{Coker}(f)=0$.

Corollary 3 (Five Lemma). — Consider a commutative diagram of $A$-modules

where the rows are exact.

(i) If $f_2$ and $f_4$ are injective and $f_1$ surjective, $f_3$ is injective.

(ii) If $f_2$ and $f_4$ are surjective and $f_5$ injective, $f_3$ is surjective.

In particular, if $f_1$, $f_2$, $f_4$ and $f_5$ are isomorphisms, the same is true of $f_3$.

To prove (i), put $\widetilde M_2=\operatorname{Coker}(u_1)$, $\widetilde M'_2=\operatorname{Coker}(u'_1)$ and denote by $\widetilde f_2:\widetilde M_2\to\widetilde M'_2$ the mapping deduced from $f_2$. It follows from cor. 2 (i) that $\widetilde f_2$ is injective. Applying cor. 1 (i) to the diagram

where $\widetilde u_2$ and $\widetilde u'_2$ are deduced from $u_2$ and $u'_2$, we see that $f_3$ is injective.

To prove (ii), set $\widetilde M_4=\operatorname{Ker}(u_4)$, $\widetilde M'_4=\operatorname{Ker}(u'_4)$ and denote by $\widetilde f_4:\widetilde M_4\to\widetilde M'_4$ the mapping induced by $f_4$. It follows from cor. 2 (ii) that $\widetilde f_4$ is surjective. Applying cor. 1 (ii) to the diagram

where $\widetilde u_3$ and $\widetilde u'_3$ have the same graph as $u_3$ and $u'_3$, one sees that $f_3$ is surjective.

If E is flat and $u : M' \to M$ is injective, the sequence $0 \to M' \xrightarrow{u} M$ is exact, hence also the sequence $0 \longrightarrow M' \otimes_A E \xrightarrow{u \otimes 1} M \otimes_A E$, and $u \otimes 1$ is injective. Conversely, consider the exact sequence (11); put $M''_1 = v(M)$, and let $i : M''_1 \to M''$ be the canonical injection and $p : M \to M''_1$ the mapping $m \mapsto v(m)$. The sequence $M' \xrightarrow{u} M \xrightarrow{p} M''_1 \longrightarrow 0$ is exact; according to II, p. 58, prop. 5, the sequence $M' \otimes_A E \xrightarrow{u \otimes 1} M \otimes_A E \xrightarrow{p \otimes 1} M''_1 \otimes_A E$ is therefore exact. Moreover, one has $v = i \circ p$, hence $(v \otimes 1) = (i \otimes 1) \circ (p \otimes 1)$; if E satisfies the condition of the statement, then $i \otimes 1$ is injective, hence

and the sequence (12) is exact.

(ii) Let I be a right filtered preordered set, $(E_\alpha, f_{\beta \alpha})$ an inductive system of a-modules relative to I, $E = \varinjlim E_\alpha$ its inductive limit. If each of the a-modules $E_\alpha$ is flat, then E is flat.

Let $M' \to M \to M''$ be an exact sequence of right a-modules.

(i) In order that the sequence $\bigoplus_{i \in I} (M' \otimes_A E_i) \to \bigoplus_{i \in I} (M \otimes_A E_i) \to \bigoplus_{i \in I} (M'' \otimes_A E_i)$ be exact, it is necessary and sufficient that each of the sequences $M' \otimes_A E_i \to M \otimes_A E_i \to M'' \otimes_A E_i$ be so (II, p. 13, prop. 7), which proves (i) since $\bigoplus (M \otimes_A E_i)$ identifies canonically with $M \otimes_A E$ (II, p. 61, prop. 7).

(ii) By assumption, each of the sequences $M' \otimes_A E_i \to M \otimes_A E_i \to M'' \otimes_A E_i$ is exact, and hence so is the sequence $M' \otimes_A E \to M \otimes_A E \to M'' \otimes_A E$, since passage to the inductive limit commutes with the tensor product (II, p. 93, prop. 7) and preserves exactness (II, p. 91, prop. 3).

Assertion (i) follows trivially from the definitions. Suppose A is left noetherian and E is finitely generated. There then exists a surjective homomorphism $u : L_0 \to E$, where $L_0$ is a free a-module having a finite basis; the kernel R of $u$ is finitely generated, hence there is a surjective homomorphism $v : L_1 \to R$ where $L_1$ is free with finite basis, and the exact sequence $L_1 \xrightarrow{v} L_0 \xrightarrow{u} E \to 0$ is a finite presentation of E; whence (ii).

Finally, suppose that E is a finitely generated projective module; it is then a direct factor of a finitely generated free module L_0 (II, p. 40, cor. 1); the kernel R of the surjective homomorphism L_0 → E is then isomorphic to a quotient of L_0, hence is finitely generated, and we conclude as above.