Let G1′ be a group, and let w:G→G1′ and τ1:G1′→ Aut(F′) be group homomorphisms such that τ′=τ1∘w. Let E1′=(Γ1′,ι1′,π1′) be a τ1-extension of G1′ by F′, and let φ1:Γ→Γ1′ be a group homomorphism such that the following diagram commutes:
F ι // Γ π // G
vφ1w
F′ι1′ // Γ1′π1′ // G1′.
Then there exists a unique group homomorphism ψ:Γ′→Γ1′ such that the diagram
F′ι′ // Γ′π′ // G
ψw
F′ι1′ // Γ1′π1′ // G1′ commutes and φ1=ψ∘φ.
For any (f,γ)∈F′×Γ, denote the class of (f,γ) in Γ′ by (f,γ). If the group homomorphism ψ:Γ′→Γ1′ has the desired properties, then it satisfies the relations
ψ(f′,γ)=ψ(ι′(f′)φ(γ))=ι1′(f′)φ1(γ)
for every f′∈F′ and γ∈Γ. Conversely, the mapping ψ from F′×τ′∘πΓ to Γ1′ given by (f,γ)↦ι1′(f)φ1(γ) is a group homomorphism. Indeed, we have the relations
for f,f′∈F′ and γ,γ′∈Γ. The kernel of ψ contains the image of j because ι1′(v(f))=φ1(ι(f)) for f∈F, and the morphism ψ deduced from ψ by passing to the quotient has the desired properties.
Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.