Lemma 2

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

Algebra, chapter X, § 4. Produits de torsion

If $u:C\to C'$ is a morphism of complexes of right A-modules and $E$ a complex of left A-modules, then the complexes $\operatorname{Con}(u)\otimes_A E$ and $\operatorname{Con}(u\otimes 1_E)$ are isomorphic.

By definition, $\operatorname{Con}(u)\otimes_A E$ is the graded module $(C'(-1)\oplus C)\otimes_A E$ endowed with the differential $D$ such that, for $x\in C_p$ , $y'\in C'(-1)_p=C'_{p-1}$ , $z\in E_q$ , we have

(4)

D(y',x)\otimes z=(-dy',dx-u(y'))\otimes z+(-1)^p(y',x)\otimes dz,

while $\operatorname{Con}(u\otimes 1_E)$ is the graded module $(C'\otimes_A E)(-1)\oplus(C\otimes_A E)$ endowed with the differential $D_1$ such that, for $x\in C_p$ , $y'\in C'_{p-1}$ , $z\in E_q$ , we have

D_1(y'\otimes z,x\otimes z)=(-dy'\otimes z-(-1)^{p-1}y'\otimes dz,dx\otimes z+(-1)^p x\otimes dz-u(y')\otimes z)

=(-dy'\otimes z,(dx-u(y'))\otimes z)+(-1)^p(y'\otimes dz,x\otimes dz),

whence the assertion.

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

Cited by

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Cites

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