Quotient group maps to outer automorphism group of normal subgroup

Statement

Suppose $G$ is a group and $N$ is a normal subgroup of $G$. There is a natural choice of homomorphism of groups from the quotient group $G/N$ to the outer automorphism group $\operatorname {Out} (N)$:

$G/N\to \operatorname {Out} (N)$

defined as follows: for any element of $G/N$, pick an element $g\in G$ in that coset of $N$. Conjugation by $g$ induces an automorphism of $N$, i.e., an element of the automorphism group $\operatorname {Aut} (N)$. Although the automorphism depends on the choice of $g$ in the coset of $N$, the coset of $\operatorname {Inn} (N)$ in $\operatorname {Aut} (N)$ for that element is independent of $g$.

More explicitly, there is a composite map:

$G\to \operatorname {Inn} (G)\to \operatorname {Aut} (N)$

Under this map, the image of the subgroup $N$ of $G$ lies inside $\operatorname {Inn} (N)$. Thus, we get an induced map from the quotient group $G/N$ to the quotient group $\operatorname {Aut} (N)/\operatorname {Inn} (N)=\operatorname {Out} (N)$.