If all the modules R i R_i R i i = 1 , . . . , n i = 1, ..., n i = 1 , ... , n α ↦ θ ∘ α \alpha \mapsto \theta \circ \alpha α ↦ θ ∘ α T o r m + n A ( P , M ) \mathrm{Tor}_{m+n}^A(P, M) Tor m + n A ( P , M ) T o r m A ( P , N ) \mathrm{Tor}_m^A(P, N) Tor m A ( P , N ) A A A P P P m > 0 m > 0 m > 0
This follows from Cor. 2 and the exact sequences
T o r m + n − i + 1 A ( P , R i ) → T o r m + n − i + 1 A ( P , K i − 1 ) → ∂ ^ T o r m + n − i A ( P , K i ) → T o r m + n − i A ( P , R i )
\mathrm{Tor}_{m+n-i+1}^A(P, R_i) \to \mathrm{Tor}_{m+n-i+1}^A(P, K_{i-1}) \xrightarrow{\hat{\partial}} \mathrm{Tor}_{m+n-i}^A(P, K_i) \to \mathrm{Tor}_{m+n-i}^A(P, R_i)
Tor m + n − i + 1 A ( P , R i ) → Tor m + n − i + 1 A ( P , K i − 1 ) ∂ ^ Tor m + n − i A ( P , K i ) → Tor m + n − i A ( P , R i ) where the extreme terms are zero by hypothesis.
Analogously, if
0 → Q → S n → S n − 1 → … → S 1 → P → 0 0 \to Q \to S_n \to S_{n-1} \to \ldots \to S_1 \to P \to 0 0 → Q → S n → S n − 1 → … → S 1 → P → 0 is an exact sequence of right A A A M M M A A A iterated connecting homomorphisms
∂ m ( S 1 , M ) : T o r m A ( P , M ) → T o r m − n A ( Q , M ) \partial^m(\mathscr{S}_1, M) : \mathrm{Tor}_m^A(P, M) \to \mathrm{Tor}_{m-n}^A(Q, M) ∂ m ( S 1 , M ) : Tor m A ( P , M ) → Tor m − n A ( Q , M ) and we have:
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
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