For every integer p≥0, there exists a unique homomorphism fp of the K-module g⊗KPp into the K-module P satisfying the following conditions:
(Ap)(Bp)(Cp)fp(xλ⊗zM)=zλzMfor λ≤M,zM∈Pp;fp(xλ⊗zM)−zλzM∈Pqfor zM∈Pq,q≤p;fp(xλ⊗fp(xμ⊗zN))=fp(xμ⊗fp(xλ⊗zN))+fp([xλ,xμ]⊗zN) for zN∈Pp−1. (The terms appearing in (Cp) are meaningful by (B_p).)*
Moreover, the restriction of fp to g⊗Pp−1 coincides with fp−1.
The last assertion follows from the others since the restriction of fp to g⊗KPp−1 satisfies conditions (A_{p-1}), (B_{p-1}) and (C_{p-1}). We shall prove the existence and uniqueness of fp by induction on p. For p=0, condition (A_0) gives f0(xλ⊗1)=zλ and conditions (B_0) and (C_0) are then obviously satisfied. Suppose now that the existence and uniqueness of fp−1 are proved. We show that fp−1 admits a unique extension fp to g⊗KPp satisfying conditions (A_p), (B_p) and (C_p).
We must define fp(xλ⊗zM) for an increasing sequence M of p elements.
If λ≤M, the value is given by condition (A_p). Otherwise, M can be written uniquely in the form (μ,N), where μ<λ,μ≤N. Then
zM=zμzN=fp−1(xμ⊗zN) by (A_{p-1}), so that the left hand side of (C_p) is
fp(xλ⊗zM). Now the right hand side of (C_p) is already defined: for (B_{p-1}) allows us to write:
fp(xλ⊗zN)=fp−1(xλ⊗zN)=zλzN+w where
w∈Pp−1; hence the right hand side of (C_p) becomes:
zλzμzN+fp−1(xμ⊗w)+fp−1([xλ,xμ]⊗zN).
Thus fp is defined uniquely and obviously satisfies conditions (A_p) and (B_p). Condition (C_p) is satisfied if μ<λ,μ≤N. As [xμ,xλ]=−[xλ,xμ], condition (C_p) is also satisfied for λ<μ,λ≤N. As (C_p) is trivially satisfied for λ=μ, (C_p) is therefore satisfied if λ≤N or μ≤N. If none of these inequalities holds, N=(ν,Q), where ν≤Q,ν<λ,ν<μ. Writing henceforth to abbreviate fp(x⊗z)=xz for x∈g and z∈Pp, we have by the induction hypothesis:
xμzN=xμ(xνzQ)=xν(xμzQ)+[xμ,xν]zQ.
Now zμzQ is of the form zμzQ+w, where w∈Pp−2. (C_p) can be applied to xλ(xν(zμzQ)), since ν≤Q and ν<μ, and to xλ(xνw) by the induction hypothesis, and hence to xλ(xν(xμzQ)). Hence:
xλ(xμzN)=xν(xλ(xμzQ))+[xλ,xν](xμzQ)+[xμ,xν](xλzQ)+[xλ,[xμ,xν]]zQ.
Exchanging λ and μ and subtracting:
xλ(xμzN)−xμ(xλzN)=xv(xλ(xμzQ)−xμ(xλzQ))+[xλ,[xμ,xv]]zQ−[xμ,[xλ,xv]]zQ=xv([xλ,xμ]zQ)+[xλ,[xμ,xv]]zQ+[xμ,[xv,xλ]]zQ=[xλ,xμ](xvzQ)+([xv,[xλ,xμ]]+[xλ,[xμ,xv]]+[xμ,[xv,xλ]])zQ
and hence by the Jacobi identity
xλ(xμzN)−xμ(xλzN)=[xλ,xμ]zN
which completes the proof of Lemma 1.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.