Corollary 2

We keep the notation of the proposition. Let $F''$ be an abelian group, and let $\tau '': G\rightarrow$ Aut(F$''$) and $v': F'\rightarrow F''$ be group homomorphisms such that

for $g\in G$ and $f\in F'$. Let $\mathscr{E}$ be a $\tau$-extension of G by F, and denote by $\varphi ($resp. $\varphi ',\varphi '')$ the canonical homomorphism associated with $v_*(\mathscr{E}) ($resp. $v'_*(v_*(\mathscr{E}))$, $(v'\circ v)_*(\mathscr{E}))$. Then there exists a unique morphism $\psi$ from the $\tau ''$-extension $v'_*(v_*(\mathscr{E}))$ of G by $F''$ to the $\tau ''$-extension $(v'\circ v)_*(\mathscr{E})$ such that $\varphi ''=\psi \circ \varphi '\circ \varphi$.