A morphism between two extension of a group $G$ by a group $F$ is an isomorphism. (Not sure if the terminology is standard.)

Question

Let $F,G$ be two groups. An extension of $G$ by $F$ is a triple $\mathscr{E}=(E,i,p)$, where $E$ is a group, $i:F\rightarrow E$ is an injective homomorphism, and $p:E\rightarrow G$ is a surjective homomorphism such that $Im(i)=Ker(p)$.

Let $\mathscr{E}:F\xrightarrow{i} E\xrightarrow{p} G$ denote the extension $\mathscr{E}=(E,i,p)$ of $G$ by $F$.

Let $\mathscr{E}:F\xrightarrow{i} E\xrightarrow{p} G$ and $\mathscr{E'}:F\xrightarrow{i'} E'\xrightarrow{p'} G$ be two extensions of $G$ by $F$. A morphism of $\mathscr{E}$ into $\mathscr{E'}$ is a homomorphism $u:E\rightarrow E'$ such that $u\circ i=i'$ and $p'\circ u=p$.

Claim. Let $u:E\rightarrow E'$ be a morphism of $\mathscr{E}$ into $\mathscr{E'}$. $u$ is an isomorphism.

Question: why is $u$ surjective? I imagine that one would have to use one of the isomorphism theorems; but I am not sure.

Any hints will be appreciated.

Answer 1

Let $x'\in E'$, and $x\in E$ such that $p(x)=p'(x')$, we have $p'(x'(u(x))^{-1})=1$, implies that there exists $y\in F$ such that $x'=i'(y)u(x)=u(i(y))u(x)=u(i(y)x).$