Given the above and a commutation factor ε on Δ×Δ, an ε-derivation (or (K,ε)-derivation) of degree δ∈Δ of (A,A′,A′′) into (B,B′,B′′) is a triple of graded K-linear mappings of degree δ:
d:A→B,d′:A′→B′,d′′:A′′→B′′ such that, for every homogeneous element a∈A and every element a′∈A′ (4)
d′′(a.a′)=(da).a′+εδ,deg(a)a.(d′a′).
It obviously suffices by linearity to verify relation (4) when a and a′ run through respective generating systems of A and A′.
It is often convenient to denote the three mappings d,d′,d′′ by the same letter d (which can be justified by denoting equally by d the graded K-linear mapping of degree δ)
(a,a′,a′′)↦(da,d′a′,d′′a′′)
of A⊕A′⊕A′′ into B⊕B′⊕B′′). Relation (4) can then be written more simply
d(a.a′)=(da).a′+εδ,deg(a)a.(da′). The
ε-derivations of
(A,A′,A′′) into
(B,B′,B′′) of
given degree form a sub-K-module of the K-module of graded linear mappings
HomK(A⊕A′⊕A′′,B⊕B′⊕B′′). When
ε(α,β)=1 for all
α,β in
Δ, we say simply
derivation (or
K-derivation) instead of
ε-derivation. Derivations form a sub-K-module of
HomK(A⊕A′⊕A′′,B⊕B′⊕B′′). When
Δ=Z and
ε(p.q)=(−1)pq, every
ε-derivation of
even degree is a derivation; every
ε-derivation of
odd degree is often called an
antiderivation (or
K-antiderivation); an antiderivation
d therefore satisfies
d(a.a′)=(da).a′+(−1)deg(a)a.(da′) for a
homogeneous element
a∈A.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.