Bourbaki

Lemma 3

Tag 0DW4, permanent. Cite it as [Bourbaki, Tag 0DW4].

Algebra, chapter X, § 4. Produits de torsion

The connecting homomorphism H(B⊗AC′)→H(Z⊗AC′)\mathrm{H}(B \otimes_A C') \to \mathrm{H}(Z \otimes_A C') associated with the exact sequence (12) is equal to H(i⊗1)\mathrm{H}(i \otimes 1).

Indeed, let a∈H(B⊗AC′)a \in \mathrm{H}(B \otimes_A C'); since BB is flat, aa belongs to the image of B⊗AZ(C′)B \otimes_A Z(C'), hence can be written as ∑λdaλ⊗bλ\sum_\lambda da_\lambda \otimes b_\lambda, where aλ∈Ca_\lambda \in C, bλ∈C′b_\lambda \in C', and dbλ=0db_\lambda = 0. The image of the class of aa under the connecting homomorphism is, by definition, the class of D(∑aλ⊗bλ)=∑daλ⊗bλ=(i⊗1)(a)D(\sum a_\lambda \otimes b_\lambda) = \sum da_\lambda \otimes b_\lambda = (i \otimes 1)(a), which proves the lemma.

The associated homology exact sequence to (12) is therefore

H(B⊗AC′)→H(i⊗1)H(Z⊗AC′)→H(j⊗1)H(C⊗AC′) \mathrm{H}(B \otimes_A C') \xrightarrow{\mathrm{H}(i \otimes 1)} \mathrm{H}(Z \otimes_A C') \xrightarrow{\mathrm{H}(j \otimes 1)} \mathrm{H}(C \otimes_A C')
→H(δ⊗1)H(B⊗AC′)(−1)→H(i⊗1)H(Z⊗AC′)(−1). \xrightarrow{\mathrm{H}(\delta \otimes 1)} \mathrm{H}(B \otimes_A C')(-1) \xrightarrow{\mathrm{H}(i \otimes 1)} \mathrm{H}(Z \otimes_A C')(-1) .

On the other hand, since ZZ is flat, from (II) we obtain a sequence of graded kk-modules

0⟶Tor⁡1A(H,H′)→∂(II,H′)B⊗AH′→i⊗1Z⊗AH′→p⊗1H⊗AH′⟶0; 0 \longrightarrow \operatorname{Tor}_1^A(H, H') \xrightarrow{\partial(\mathrm{II}, H')} B \otimes_A H' \xrightarrow{i \otimes 1} Z \otimes_A H' \xrightarrow{p \otimes 1} H \otimes_A H' \longrightarrow 0 ;

finally, we have the canonical homomorphisms of No. 1

γB=γ(B,C′):B⊗AH′→H(B⊗AC′) \gamma_B = \gamma(B, C') : B \otimes_A H' \to \mathrm{H}(B \otimes_A C')
γZ=γ(Z,C′):Z⊗AH′→H(Z⊗AC′) \gamma_Z = \gamma(Z, C') : Z \otimes_A H' \to \mathrm{H}(Z \otimes_A C')
γC=γ(C,C′):H⊗AH′→H(C⊗AC′), \gamma_C = \gamma(C, C') : H \otimes_A H' \to \mathrm{H}(C \otimes_A C') ,

whence a diagram of graded kk-modules, with exact rows:

B⊗H′→i⊗1Z⊗H′→p⊗1H⊗H′⟶0↓γB↓γZ↓γCH(B⊗C′)→H(i⊗1)H(Z⊗C′)→H(j⊗1)H(C⊗AC′)→H(δ⊗1)H(B⊗C′)(−1)→H(i⊗1)H(Z⊗C′)(−1) \begin{array}{ccccccccc} B \otimes H' & \xrightarrow{i \otimes 1} & Z \otimes H' & \xrightarrow{p \otimes 1} & H \otimes H' & \longrightarrow & 0 \\ \downarrow \gamma_B & & \downarrow \gamma_Z & & \downarrow \gamma_C & & \\ H(B \otimes C') & \xrightarrow{H(i \otimes 1)} & H(Z \otimes C') & \xrightarrow{H(j \otimes 1)} & H(C \otimes_A C') & \xrightarrow{H(\delta \otimes 1)} & H(B \otimes C')(-1) & \xrightarrow{H(i \otimes 1)} & H(Z \otimes C')(-1) \end{array}
0⟶Tor⁡1A(H,H′)(−1)→∂(II,H′)(B⊗H′)(−1)→i⊗1(Z⊗H′)(−1), 0 \longrightarrow \operatorname{Tor}_1^A(H, H')(-1) \xrightarrow{\partial(\mathrm{II}, H')} (B \otimes H')(-1) \xrightarrow{i \otimes 1} (Z \otimes H')(-1),

which is commutative by definition of the homomorphisms γ\gamma. But since the complexes BB and ZZ are split and flat, γB\gamma_B and γZ\gamma_Z are bijective (X, p. 65, Corollary 1). From this, on the one hand, we deduce that γC\gamma_C is injective and has image equal to Ker⁡H(δ⊗1)\operatorname{Ker} H(\delta \otimes 1), and on the other hand, that γB∘∂(II,H′)\gamma_B \circ \partial(\mathrm{II}, H') is injective with image equal to Im⁡H(δ⊗1)\operatorname{Im} H(\delta \otimes 1). The theorem follows immediately from this.

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

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