Theorem 1

Let $n$ be an integer $\geqslant 1$, $M$ and $N$ two $A$-modules.

a) Every element of $\mathrm{Ext}^n_A(M, N)$ is the class of an exact sequence (X, p. 117, def. 1).

b) Let $0 \to N \xrightarrow{f_{n+1}} R_n \xrightarrow{f_n} \ldots \to R_1 \xrightarrow{f_1} M \to 0$ and $0 \to N \xrightarrow{f_{n+1}} R'_n \xrightarrow{f_n} \ldots \to R'_1 \xrightarrow{f_1} M \to 0$

be exact sequences, $\theta$ and $\theta'$ the associated classes. The following conditions are equivalent:

(i) $\theta = \theta'$;

(ii) there exists a commutative diagram with exact rows:

(iii) there exists a commutative diagram with exact rows:

Let us prove $a$. Let $\alpha \in \mathrm{Ext}_A^n(M, N)$, and let $P$ be a projective resolution of $M$. Let $a : P(n) \to N$ be a morphism of complexes representing $\alpha$; its unique non-zero component is an $A$-homomorphism $u : P_n \to N$ which satisfies $u \circ d_{n+1} = 0$, hence factors as $u = \overline{u} \circ \delta_n$, where $\delta_n : P_n \to Z_{n-1}$ is the mapping induced by $d_n$ (we put $Z_{n-1} = \mathrm{Im}\, d_n$) and $\overline{u}$ is an $A$-homomorphism from $Z_{n-1}$ into $N$. By Remark 1, p. 117, the class $\theta \in \mathrm{Ext}_A^n(M, Z_{n-1})$ of the exact sequence

is equal to the homotopy class of the morphism $(-1)^{n(n+1)/2} \delta_n$. We therefore have

which makes it possible, by Cor. 3, p. 120, to represent $\alpha$ as the class of an exact sequence.

Let us prove $b$. It follows from Cor. 1 of X, p. 120 that (ii) $\Rightarrow$ (i) and that (iii) $\Rightarrow$ (i). Suppose that (i) is satisfied, and let $P$ be a projective resolution of $M$. There exists a commutative diagram

The morphisms from $P(n)$ into $N$ defined by $u_n$ and $u'_n$ are homotopic, since they both belong to the class $(-1)^{n(n+1)/2} \theta$, hence $u'_n - u_n$ is of the form $w \circ d_n$, where $w : P_{n-1} \to N$ is an $A$-homomorphism. Replacing $u'_{n-1}$ by $u'_{n-1} - f'_{n+1} \circ w$ and $u'_n$ by $u_n$, we reduce to the case where $u_n = u'_n$. This makes it possible to construct a new commutative diagram with exact rows:

where $N'$ is the quotient of $P_{n-1} \oplus N$ by the submodule formed by the couples $(d_n(x), -u_n(x))$ for $x \in P_n$, and where $v$ (resp. $v'$) is defined by passing to the quotient from the mapping $u_{n-1} \oplus f_{n+1}$ (resp. $u'_{n-1} \oplus f'_{n+1}$). Condition (ii) is therefore satisfied.

Suppose again that condition (i) is satisfied, and let $E$ be an injective resolution of $N$. There exists a commutative diagram

and we show as above that we can assume $v'_n = v_n$. We then have a commutative diagram with exact rows

with $M' = M \times_{\mathbf{R}_1} E_{n-1}$ (cf. X, p. 120, cor. 3, b)). Condition (iii) is therefore satisfied, which completes the proof of the theorem.