Corollary 3

Tag 0E00, permanent. Cite it as [Bourbaki, Tag 0E00].

Algebra, chapter X, § 7. Produit de composition

If $\theta\in\operatorname{Ext}_A^n(M,N)$ is the class of the exact sequence $(\mathscr{S})$ , we have

\delta^m(P,\mathscr{S})(\alpha)=\theta\circ\alpha,\qquad \delta^m(\mathscr{S},P)(\beta)=(-1)^{mn+n(n+1)/2}\beta\circ\theta.

If $\theta_i\in\operatorname{Ext}_A^1(K_{i-1},K_i)$ is the class associated with the exact sequence (9), we have by Proposition 5

\delta^m(P,\mathscr{S})(\alpha)=\theta_n\circ\cdots\circ\theta_2\circ\theta_1\circ\alpha

\delta^m(\mathscr{S},P)(\beta)=(-1)^{(m+1)+\cdots+(m+n)}\beta\circ\theta_n\circ\cdots\circ\theta_1.

Moreover, by Proposition 3 (X, p. 118), we have $\theta=\theta_n\circ\cdots\circ\theta_1$ . The corollary follows immediately from this, and from the relation (E, III, p. 44)

(m+1)+\cdots+(m+n)=mn+n(n+1)/2.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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