Let K,M,N be left A-modules, P,Q,R right A-modules, α∈ExtAn(P,Q), β∈ExtAp(Q,R), λ∈ExtAr(K,M), μ∈ExtAs(M,N), γ∈TormA(P,K). Then (13)
(β∘α)∘γ=β∘(α∘γ)inTorm−p−nA(R,K), (14)
(μ∘λ)∘γ=μ∘(λ∘γ)inTorn−r−sA(P,N), (15)
α∘(λ∘γ)=(−1)nrλ∘(α∘γ)inTorm−p−rA(Q,M).
Formulas (13) and (14) follow immediately from the definitions. Let us prove (15). Let z=∑zij, zij∈Li(P)⊗Lj(K) be a cycle representing γ, f:L(P)→L(Q)(−n) and g:L(K)→L(M)(−r) morphisms representing α and λ. Then λ∘(α∘γ) is the class of ∑(−1)(i−n)r(f⊗g)(zij) and α∘(λ∘γ) is the class of
∑(−1)ir(f⊗g)(zij),whence (15).
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.
Nothing.