Proposition 1

Let $\mathscr{E}: F \xrightarrow{i} E \xrightarrow{p} G$ and $\mathscr{E}': F \xrightarrow{i'} E' \xrightarrow{p'} G$ be extensions of $G$ by $F$. If $u: E \to E'$ is a morphism of $\mathscr{E}$ into $\mathscr{E}'$, $u$ is an isomorphism of $E$ onto $E'$ and $u^{-1}$ is a morphism of $\mathscr{E}'$ into $\mathscr{E}$.

Let $x \in E$ be such that $u(x) = e$. Then $p(x) = p'(u(x)) = e$, whence $x \in i(F)$. Let $y \in F$ be such that $x = i(y)$; then $i'(y) = u(i(y)) = e$. As $i'$ is injective, $y = e$ and $x = e$. Therefore $u$ is injective. By virtue of § 4, no. 6, Corollary 1 to Proposition 7, $u$ is surjective since $u(i(F)) = i'(F)$. The last assertion is immediate.

In other words, the extensions $\mathscr{E}$ and $\mathscr{E}'$ are isomorphic if and only if there exists a morphism of $\mathscr{E}$ into $\mathscr{E}'$.

Let $F$ and $G$ be two groups and write $E_0 = F \times G$; let $i: F \to E_0$ be the canonical injection and $p: E_0 \to G$ be the canonical projection. Every extension of $G$ by $F$ isomorphic to the extension $\mathscr{E}_0: F \xrightarrow{i} E_0 \xrightarrow{p} G$ is called a trivial extension.