Lemma 3

Tag 0E2T, permanent. Cite it as [Bourbaki, Tag 0E2T].

Algebra, chapter X, § 9. Complexes de Koszul

Let $K$ be a complex such that $K_i = 0$ for $i \neq 0, 1$ , and let $C$ be a complex and $p$ an integer.

a) If $K$ is flat, there is for every $p \in \mathbf{N}$ an exact sequence

0 \to H_0(K \otimes_A H_p(C)) \to H_p(K \otimes_A C) \to H_1(K \otimes_A H_{p-1}(C)) \to 0.

b) If $K$ is projective, we have for every $p \in \mathbf{N}$ an exact sequence

0 \to H^1(\mathrm{Homgr}_A(K, H^{p-1}(C))) \to H^p(\mathrm{Homgr}_A(K, C))

\to H^0(\mathrm{Homgr}_A(K, H^p(C))) \to 0.

Let us prove a), the proof of b) being analogous. For every $i$ , denote by $K_{(i)}$ the complex $K_i(-i)$ . We have an exact sequence of complexes, split as a sequence of $A$ -modules

0 \to K_{(0)} \xrightarrow{\alpha} K \xrightarrow{\beta} K_{(1)} \to 0;

the sequence

\begin{array}{ccccccccc} 0 & \longrightarrow & K_{(0)} \otimes_A C & \xrightarrow{\alpha \otimes 1} & K \otimes_A C & \xrightarrow{\beta \otimes 1} & K_{(1)} \otimes_A C & \longrightarrow & 0 \end{array}

is exact, and $K$ being flat, the homomorphisms

\gamma_{0,p}(K_{(0)}, C) : K_{(0)} \otimes_A H_p(C) \to H_p(K_{(0)} \otimes_A C)

\gamma_{1,p-1}(K_{(1)}, C) : K_{(1)} \otimes_A H_{p-1}(C) \to H_p(K_{(1)} \otimes_A C)

are bijective (X, p. 66, cor. 2). Let us calculate the connecting homomorphism $\partial(\alpha \otimes 1,\beta \otimes 1)$ ; by definition, it maps the class of the cycle $\sum a_i \otimes b_i$ onto that of $\sum d a_i \otimes b_i$ , which means that

\partial(\alpha \otimes 1,\beta \otimes 1)\circ\gamma(K_{(1)},\mathbf C)=\gamma(K_{(0)},\mathbf C)\circ(d_K\otimes 1).

The exact homology sequence associated with (14) therefore has the form

K_1\otimes H_p(\mathbf C)\xrightarrow{d_k\otimes 1}K_0\otimes H_p(\mathbf C)\longrightarrow H_p(K\otimes\mathbf C)

\hspace{7cm}\longrightarrow K_1\otimes H_{p-1}(\mathbf C)\xrightarrow{d_k\otimes 1}K_0\otimes H_{p-1}(\mathbf C),

whence a).

Applying Lemma 3, a) to the complex $K(u)$ , and using the commutation isomorphisms, we obtain:

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

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