Lemma 3

Let $C$ be a complex and let $r\in \mathbf{Z}$. There exists a complex $C'$ and a homomorphism $f:C'\to C$ such that $f_n:C'_n\to C_n$ is an isomorphism for $n<r$ and such that $C'_r$ is a free $A$-module. If $A$ is noetherian and the $A$-modules $H_r(C)$ and $C_{r-1}$ are of finite type, one can require that $C'_r$ be of finite type.

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.