Bourbaki

Definition 1

Tag 0667, permanent. Cite it as [Bourbaki, Tag 0667].

Algebra, chapter III, § 10. Derivations

Given the above and a commutation factor ε\varepsilon on Δ×Δ\Delta \times \Delta, an ε\varepsilon-derivation (or (K,ε)(K, \varepsilon)-derivation) of degree δ∈Δ\delta \in \Delta of (A,A′,A′′)(A, A', A'') into (B,B′,B′′)(B, B', B'') is a triple of graded KK-linear mappings of degree δ\delta:

d:A→B,d′:A′→B′,d′′:A′′→B′′ d : A \to B, \qquad d' : A' \to B', \qquad d'' : A'' \to B''
such that, for every homogeneous element a∈Aa \in A and every element a′∈A′a' \in A' (4)
d′′(a.a′)=(da).a′+εδ,deg⁡(a)a.(d′a′). d''(a.a') = (da).a' + \varepsilon_{\delta, \deg(a)}a.(d'a').

It obviously suffices by linearity to verify relation (4) when aa and a′a' run through respective generating systems of AA and A′A'.

It is often convenient to denote the three mappings d,d′,d′′d, d', d'' by the same letter dd (which can be justified by denoting equally by dd the graded KK-linear mapping of degree δ\delta)

(a,a′,a′′)↦(da,d′a′,d′′a′′) (a, a', a'') \mapsto (da, d'a', d''a'')

of A⊕A′⊕A′′A \oplus A' \oplus A'' into B⊕B′⊕B′′B \oplus B' \oplus B''). Relation (4) can then be written more simply

d(a.a′)=(da).a′+εδ,deg⁡(a)a.(da′). d(a.a') = (da).a' + \varepsilon_{\delta,\deg(a)} a.(da').
The ε\varepsilon-derivations of (A,A′,A′′)(A, A', A'') into (B,B′,B′′)(B, B', B'') of given degree form a sub-K-module of the K-module of graded linear mappings
Hom⁡K(A⊕A′⊕A′′,B⊕B′⊕B′′). \operatorname{Hom}_{\mathbf{K}}(A \oplus A' \oplus A'', B \oplus B' \oplus B'').
When ε(α,β)=1\varepsilon(\alpha, \beta) = 1 for all α,β\alpha, \beta in Δ\Delta, we say simply derivation (or K-derivation) instead of ε\varepsilon-derivation. Derivations form a sub-K-module of
Hom⁡K(A⊕A′⊕A′′,B⊕B′⊕B′′). \operatorname{Hom}_{\mathbf{K}}(A \oplus A' \oplus A'', B \oplus B' \oplus B'').
When Δ=Z\Delta = \mathbf{Z} and ε(p.q)=(−1)pq\varepsilon(p.q) = (-1)^{pq}, every ε\varepsilon-derivation of even degree is a derivation; every ε\varepsilon-derivation of odd degree is often called an antiderivation (or K-antiderivation); an antiderivation dd therefore satisfies
d(a.a′)=(da).a′+(−1)deg⁡(a)a.(da′) d(a.a') = (da).a' + (-1)^{\deg(a)} a.(da')
for a homogeneous element a∈Aa \in A.

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

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Transcribed from the printed edition, 1998, Springer. Not translated.