A group extension induces a well-defined outer action on its kernel

For $q\in Q$, choose $x\in E$ with $\pi (x)=q$. Since $i(N)=\ker \pi$ is normal, conjugation by $x$ restricts to an automorphism of $i(N)$, hence of $N$, by [L2]. Let ${\omega }_E(q)$ be its class in $\mathrm{Out}(N)$ from [L1].

If $x^{\mathrm{\prime }}=xi(n)$ is another lift of $q$, then for $m\in N$, $x^{\mathrm{\prime }}i(m)x^{\mathrm{\prime }-1}=xi(nm{n}^{-1})x^{-1}\mathrm{.}$ Thus the automorphism from $x^{\mathrm{\prime }}$ differs from the automorphism from $x$ by the inner automorphism of $N$ defined by $n$. So step 1.1 is independent of the chosen lift as an element of $\mathrm{Out}(N)$.

If $q_1,q_2\in Q$ are lifted by $x_1,x_2\in E$, then $x_1{x}_2$ lifts $q_1{q}_2$, and conjugation by $x_1{x}_2$ is the composite of conjugation by $x_1$ and by $x_2$. Passing to classes modulo inner automorphisms with [L1], ${\omega }_E(q_1{q}_2)={\omega }_E(q_1){\omega }_E(q_2)$. So ${\omega }_E$ is a homomorphism.

If $\phi :E\to E^{\mathrm{\prime }}$ is an equivalence of extensions as in [L3], then $\phi$ identifies the chosen copy of $N$ with itself and carries lifts of $q$ in $E$ to lifts of $q$ in $E^{\mathrm{\prime }}$. Therefore the conjugation automorphisms correspond and define the same class in $\mathrm{Out}(N)$. Hence ${\omega }_E$ depends only on the extension class.