Let G and G′ be groups, and let F and F′ be abelian groups. Let τ:G→ Aut(F),τ′:G→ Aut(F′),u:G′→G, and v:F→F′ be group homomorphisms such that
τ′(g)⋅v(f)=v(τ(g)⋅f)
for g∈G and f∈F. We write τ′′=τ′∘u. Let E be a τ-extension of G by F. We denote by φu(resp. φv,φu′,φv′) the canonical homomorphism corresponding to the τ∘u-extension u∗(E)(resp. to the τ′-extension v∗(E), to the τ′′-extensions u∗(v∗(E)) and v∗(u∗(E))). Then there exists a unique morphism ψ of τ′′-extensions from v∗(u∗(E)) to u∗(v∗(E)) such that φv∘φu= φu′∘ψ∘φv′.
We denote the τ∘u-extension u∗(E) (resp. the τ′′-extension u∗(v∗(E))) by (Γu,ιu,πu) (resp. (Γu′,ιu′,πu′)). Applying Lemma 2 of VIII, p. 287 to φv∘φu, we find that there exists a group homomorphism ψ1:Γu→Γu′ such that the diagram
F ιu // Γuπu // G′
vψ1
F′ιu′ // Γu′πu′ // G′
commutes and φu′∘ψ1=φv∘φu. The existence of ψ follows from Proposition 2 applied to ψ1. Conversely, if ψ′ also has the desired properties, then we have ψ′∘φv′=ψ1 by Lemma 2, so ψ′=ψ (Proposition 2).
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.