Proposition 1

Tag 0DV4, permanent. Cite it as [Bourbaki, Tag 0DV4].

Algebra, chapter X, § 4. Produits de torsion

If the complexes $C$ and $C'$ are zero on the right, $C \otimes_A C'$ is zero on the right and the canonical $k$ -linear mapping

\gamma_{0,0}(C, C') : H_0(C) \otimes_A H_0(C') \to H_0(C \otimes_A C')

is bijective.

Let $u : (C, d) \to (C_1, d_1)$ be a morphism of complexes of right $A$ -modules and $u' : (C', d') \to (C'_1, d'_1)$ a morphism of complexes of left $A$ -modules; then $u \otimes u' : C \otimes_A C' \to C_1 \otimes_A C'_1$ is a morphism of complexes of $k$ -modules; indeed, it is graded of degree 0, and if we denote by $D$ and $D_1$ the differentials of $C \otimes C'$ and $C_1 \otimes C'_1$ , we have for $p, q \in \mathbf{Z}, x \in C_p, x' \in C'_q$ ,

(u \otimes u') (D(x \otimes x')) = u(dx) \otimes u'(x') + (-1)^p u(x) \otimes u'(d'x') = = d_1 u(x) \otimes u'(x') + (-1)^p u(x) \otimes d'_1 u'(x') = D_1(u(x) \otimes u'(x')) .

Moreover the following diagram is commutative :

\begin{array}{ccc} H(C) \otimes_A H(C') & \xrightarrow{\gamma(C, C')} & H(C \otimes_A C') \\ H(u) \otimes H(u') \downarrow & & \downarrow H(u \otimes u') \\ H(C_1) \otimes_A H(C'_1) & \xrightarrow{\gamma(C_1, C'_1)} & H(C_1 \otimes_A C'_1) \end{array}

Let $A^\circ$ be the $k$ -algebra opposite to $A$ , $C^\circ$ (resp. ${C'}^\circ$ ) the complex $C$ (resp. $C'$ ) considered as a complex of left $A^\circ$ -modules (resp. right). Let us denote by

\sigma(C, C') : C \otimes_A C' \to {C'}^\circ \otimes_{A^\circ} C^\circ

the unique graded

k

-linear mapping of degree 0 such that, for

x \in C_p, x' \in C'_q, p, q \in \mathbf{Z}

, we have

\sigma(C, C') (x \otimes x') = (-1)^{pq} x' \otimes x .

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

Cited by

Nothing.

Cites

Nothing.