In fact, put $g = f' - k \circ \alpha'$; we have

This implies $\mathrm{Im}\ (g) \subset \mathrm{Ker}\ (\beta) = \mathrm{Im}\ (\beta')$. Since $M'$ is projective, there therefore exists an A-homomorphism $k': M' \to N'$ such that $\beta' \circ k' = g$, whence the lemma.

It is enough to put $k'' = u'', k = -u, f = 0, f' = 0$ and $u' = k'$ in Lemma 1.

Lemma 1 bis. — Consider a diagram of A-modules and homomorphisms

such that $f \circ \alpha' = \beta \circ f'$, $\mathrm{Ker}\, \alpha = \mathrm{Im}\, \alpha'$, $\beta \circ \beta' = 0$, and $f' = k \circ \alpha' + \beta' \circ k'$ and where $N''$ is injective. There exists an $A$-homomorphism $k'' : M'' \to N''$ such that $f = k'' \circ \alpha + \beta \circ k$.

In fact, put $g = f - \beta \circ k$, we have

This implies $\mathrm{Ker}\, g \supset \mathrm{Im}\, \alpha' = \mathrm{Ker}\, \alpha$. Since $N''$ is injective, there therefore exists (X, p. 16, remark) an $A$-homomorphism $k' : M'' \to N''$ such that $g = k'' \circ \alpha$, whence the lemma.

Lemma 2 bis. — If, in the commutative diagram of $A$-modules and homomorphisms

one has $\mathrm{Ker}\, \alpha = \mathrm{Im}\, \alpha', \beta \circ \beta' = 0$ and if $N''$ is injective, there exists an $A$-homomorphism $u'' : M'' \to N''$ such that $u'' \circ \alpha = \beta \circ u$. It is enough to put $u' = k', u = -k, f = 0, f' = 0$ and $k'' = u''$ in Lemma 1 bis.

a) Let $(u_i : P_i \to E_i)_{i \leq r}$ be a family of homomorphisms such that $d_E \circ u_i = u_{i-1} \circ d_P$ for $i \leq r$. Suppose that $P_i$ is projective for $i > r$ and that $H_i(E) = 0$ for $i \geq r$. Then the family of the $u_i$ extends to a morphism of complexes from $P$ into $E$; two such extensions are homotopic.

b) Let $(u^i : P^i \to E^i)_{i \leq r}$ be a family of homomorphisms such that $u^i \circ d_P = d_E \circ u^{i-1}$ for $i \leq r$. Suppose that $E^i$ is injective for $i > r$ and that $H^i(P) = 0$ for $i \geq r$. Then the family of the $u^i$ extends to a morphism of complexes from $P$ into $E$; two such extensions are homotopic.

Let us prove a). The existence of an extension $v$ of the family $(u_i)_{i \leq r}$ follows at once from Lemma 2 by induction. Let $v'$ be another extension; put $f = v' - v$, and construct by induction on the integer $n$ a homomorphism $k_n : P_n \to E_{n+1}$ such that $f_n = d_E \circ k_n + k_{n-1} \circ d_P$. For $i \leq r$, take $k_i = 0$. Let $n \geq r$ and suppose that the $k_i$ have been constructed for $i \leq n$. Consider then the diagram

The hypotheses of Lemma 1 are satisfied; there therefore exists an $A$-homomorphism $k_{n+1} : P_{n+1} \to E_{n+2}$ such that $f_{n+1} = d_E \circ k_{n+1} + k_n \circ d_P$, whence a).

The proof of b) is analogous, via Lemmas 1 bis and 2 bis.

(1)

be exact.

In fact, to say that $p:P\to M$ is a homomorphism means that $H_i(P)=0$ for $i>0$ and that $p_0$ induces an isomorphism of $\operatorname{Coker}(d_P:P_1\to P_0)$ onto $M$.

Similarly:

Proposition 2bis. — Let $E$ be a complex null on the left and $e:M\to E$ a morphism. In order that $(e,E)$ be a right resolution of $M$, it is necessary and sufficient that the sequence

(1 bis)

be exact.

By abuse of language, one often says that sequence (1) (resp. (1 bis)) is a left (resp. right) resolution of $M$.

Let us consider the complex $\overline{P}$ defined as follows: $\overline{P}_n = P_n$ for $n \neq -1$, $\overline{P}_{-1} = M$, $d_{\overline{P},n} = d_{P,n}$ for $n \neq 0, -1$, $d_{\overline{P},0} = p_0$, $d_{\overline{P},-1} = 0$, and the complex $\overline{P}'$ defined analogously. Applying to the complexes $\overline{P}$ and $\overline{P}'$ Prop. 1 a) with $r = 0$, $u_i = 0$ for $i < -1$ and $u_{-1} = f$, we obtain Prop. 3.

In fact, there exists a morphism $\alpha : P' \to P$ (resp. $\beta : P \to P'$) such that $p \circ \alpha = p'$ (resp. $p' \circ \beta = p$). Since $p \circ \alpha \circ \beta = p$ (resp. $p' \circ \beta \circ \alpha = p'$), $\alpha \circ \beta$ is homotopic to $1_P$ (resp. $\beta \circ \alpha$ is homotopic to $1_{P'}$).

PROPOSITION 3 bis. — Let $g : N \to N'$ be a homomorphism of a-modules, $e' : N' \to E'$ a morphism from $N'$ into a left zero and injective complex $E'$, and $e : N \to E$ a right resolution of N. There exists a morphism of complexes $\tilde{g} : E \to E'$, and only one up to homotopy, such that $\tilde{g} \circ e = e' \circ g$.

This is proved like Prop. 3 with the aid of Prop. 1 b).

It is a matter of proving, for every integer $n > 0$, the formula

One has first

When $n > 1$, one has successively

One has at once

(7)

On the other hand, if $g : N \to P$ is an A-homomorphism of a-modules, one has

(8)

In fact, for $m \in M$, one has

hence $L_0(g \circ f) = L(g) \circ L(f)$; consequently $Z_0(g \circ f) = Z_0(g) \circ Z_0(f)$; whence at once $L_n(g \circ f) = L_n(g) \circ L_n(f)$, for $n \geqslant 0$, by induction on $n$, whence (8).

One has (14)

and for every homomorphism $g : N \to P$ of A-modules (15)

b) Suppose $A$ left noetherian and the $A$-modules $H_n(C)$, $n\in \mathbf{Z}$, finitely generated. There exists a free $A$-complex $L$ such that $L_n=0$ for $n<a$ and that $L_n$ is a finitely generated $A$-module for every $n$, and a homomorphism $f:L\to C$.

Let $C'$ be the subcomplex of $C$ such that $C'_n=C_n$ for $n>a$, $C'_a=Z_a(C)$, $C'_n=0$ for $n<a$; then the canonical injection of $C'$ into $C$ is a homomorphism. Replacing $C$ by $C'$, one may therefore suppose that $C_n=0$ for $n<a$. The statement then results from repeated application of the following lemma, for $r=a,a+1,\ldots$ :

a) Let first $h:M\to C_r$ be a homomorphism of $A$-modules; let $d=(d_n)$ denote the differential of $C$. Let $N$ be the submodule of $M\times C_{r+1}$ formed by the pairs $(m,x)$ such that $h(m)=d_{r+1}(x)$; define a complex $(C',d')$ by $C'_n=C_n$ for $n\ne r,r+1$, $C'_r=M$, $C'_{r+1}=N$, $d'_n=d_n$ for $n\ne r,r+1$, $d'_r=h$, $d'_{r+1}(m,x)=m$ for $(m,x)\in N$ and $d'_{r+2}(y)=(0,d_{r+2}(y))$ for $y\in C_{r+2}$. Also consider the morphism of complexes $f:C'\to C$ such that $f_n=1_{C_n}$ for $n\ne r,r+1$, $f_r=h$, $f_{r+1}(m,x)=x$.

b) The complex $\operatorname{Ker}f$ is zero in degree $\ne r,r+1$ and the differential $d'_{r+1}$ induces an isomorphism of $\operatorname{Ker}f_{r+1}$ onto $\operatorname{Ker}f_r$, hence $H(\operatorname{Ker}f)=0$.

c) When the composite mapping $M\xrightarrow{h}C_r\to C_r/B_r(C)$ is surjective, one sees analogously that $H(\operatorname{Coker}f)=0$, and $f$ is then a homomorphism (X, p. 31, cor. 2).

d) When $A$ is assumed noetherian and the $A$-modules $H_r(C)$ and $C_{r-1}$ are of finite type, then $C_r/B_r(C)$ is of finite type, by virtue of the exact sequence (X, p. 25)

there then exists a free $A$-module of finite type $M$ and a homomorphism $h:M\to C_r$ such that the condition of c) is satisfied; in the general case, there exists a free module $M$ and a surjective homomorphism $h:M\to C_r$. This completes the proof.