Corollary 3

Let $0 \to N \xrightarrow{f_{n-1}} R_n \xrightarrow{f_n} \cdots \to R_1 \xrightarrow{f_1} M \to 0$ be an exact sequence, $\theta$ its class in $\mathrm{Ext}_A^n(M, N)$, $u \colon M' \to M$ and $v \colon N \to N'$ homomorphisms of $A$-modules.

a) The element $v \circ \theta$ of $\mathrm{Ext}_A^n(M, N')$ is equal to the class of the exact sequence

where $R'_n$ is the $A$-quotient module of $R_n \oplus N'$ by the submodule formed by the pairs $(f_{n+1}(x), -v(x))$ for $x \in N$, and where $f_{n'+1}'$ (resp. $f_{n'}'$) is deduced from the canonical injection (resp. from $(f_n, 0)$) by passing to quotients.

b) The element $\theta \circ u$ of $\mathrm{Ext}_A^n(M', N)$ is the class of the exact sequence

where $R'_1$ is the fibre product $R_1 \times_M M'$, that is to say (I, p. 44) the submodule of $R_1 \times M'$ formed by the pairs $(x, y)$ such that $f_1(x) = u(y)$, and where $f_2''$ (resp. $f_1''$) is deduced from $(f_2, 0)$ (resp. from the second projection).

Let us prove for example $a$. Let $z$ be an element of $R'_n$ such that $f'_n(z) = 0$; if $z$ is the class of a pair $(x, y)$, with $x \in R_n$, $y \in N'$, one has $f_n(x) = 0$, so that there exists an element $t \in N$ such that $x = f_{n+1}(t)$. Then one has $z = f'_{n+1}(y + v(t))$, which proves that $\mathrm{Ker}\ f'_n = \mathrm{Im}\ f'_{n+1}$. The injectivity of $f'_{n+1}$ follows from that of $f_{n+1}$.

Let $j : R_n \to R'_n$ be the homomorphism deduced from the canonical injection; one has a commutative diagram of exact sequences:

assertion $a$ then follows from the proposition.

The proof of $b$ is analogous.