Bourbaki

Proposition 6

Tag 0E05, permanent. Cite it as [Bourbaki, Tag 0E05].

Algebra, chapter X, § 7. Produit de composition

Let K,M,NK, M, N be left AA-modules, P,Q,RP, Q, R right AA-modules, α∈ExtAn(P,Q)\alpha \in \mathrm{Ext}_A^n(P, Q), β∈ExtAp(Q,R)\beta \in \mathrm{Ext}_A^p(Q, R), λ∈ExtAr(K,M)\lambda \in \mathrm{Ext}_A^r(K, M), μ∈ExtAs(M,N)\mu \in \mathrm{Ext}_A^s(M, N), γ∈TormA(P,K)\gamma \in \mathrm{Tor}_m^A(P, K). Then (13)

(β∘α)∘γ=β∘(α∘γ)inTorm−p−nA(R,K), (\beta \circ \alpha) \circ \gamma = \beta \circ (\alpha \circ \gamma) \quad \text{in} \quad \mathrm{Tor}_{m-p-n}^A(R, K),
(14)
(μ∘λ)∘γ=μ∘(λ∘γ)inTorn−r−sA(P,N), (\mu \circ \lambda) \circ \gamma = \mu \circ (\lambda \circ \gamma) \quad \text{in} \quad \mathrm{Tor}_{n-r-s}^A(P, N),
(15)
α∘(λ∘γ)=(−1)nrλ∘(α∘γ)inTorm−p−rA(Q,M). \alpha \circ (\lambda \circ \gamma) = (-1)^{nr} \lambda \circ (\alpha \circ \gamma) \quad \text{in} \quad \mathrm{Tor}_{m-p-r}^A(Q, M).

Formulas (13) and (14) follow immediately from the definitions. Let us prove (15). Let z=∑zijz = \sum z_{ij}, zij∈Li(P)⊗Lj(K)z_{ij} \in L_i(P) \otimes L_j(K) be a cycle representing γ\gamma, f:L(P)→L(Q)(−n)f : L(P) \to L(Q) (-n) and g:L(K)→L(M)(−r)g : L(K) \to L(M) (-r) morphisms representing α\alpha and λ\lambda. Then λ∘(α∘γ)\lambda \circ (\alpha \circ \gamma) is the class of ∑(−1)(i−n)r(f⊗g)(zij)\sum (-1)^{(i-n)r} (f \otimes g)(z_{ij}) and α∘(λ∘γ)\alpha \circ (\lambda \circ \gamma) is the class of

∑(−1)ir(f⊗g)(zij),whence (15). \sum (-1)^{ir} (f \otimes g)(z_{ij}), \quad \text{whence (15)}.

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

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Cites

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