$\operatorname{Aut}(G)$ construction is not functorial

Question

I'm trying to prove what the title says: There is no functor $F$ such that $G \mapsto \operatorname{Aut}(G)$. I know that there is an equivalent question: Taking the automorphism group of a group is not functorial. but I really don't understand the proof, and the author's answer was last seen several years ago. Do you have another proof, or could you give me a little explanation of that?

Answer 1

If there is such a functor $F$, then for every sequence of group homomorphisms $$G_1\ \stackrel{f}{\longrightarrow}\ G_2\ \stackrel{g}{\longrightarrow}\ G_3,$$ we get another sequence of group homomorphisms $$\operatorname{Aut}G_1\ \stackrel{F(f)}{\longrightarrow}\ \operatorname{Aut}G_2\ \stackrel{F(g)}{\longrightarrow}\ \operatorname{Aut}G_3,$$ where $F(g)\circ F(f)=F(g\circ f)$ because $F$ is a functor. Also $F(\operatorname{id}_G)=\operatorname{id}_{\operatorname{Aut}G}$ for every group $G$ because $F$ is a functor.

In particular, it follows that if $G_3=G_1$ and $g\circ f=\operatorname{id}_{G_1}$, then also $F(g\circ f)=\operatorname{id}_{\operatorname{Aut}G_1}$. So we could prove that no such functor exists if we can find explicit groups $G_1$ and $G_2$ with explicit group homomorphisms $$G_1\ \stackrel{f}{\longrightarrow}\ G_2\ \stackrel{g}{\longrightarrow}\ G_1,$$ such that $g\circ f=\operatorname{id}_{G_1}$, but such that there exist no group homomorphisms $$\operatorname{Aut}G_1\ \stackrel{F(f)}{\longrightarrow}\ \operatorname{Aut}G_2\ \stackrel{F(g)}{\longrightarrow}\ \operatorname{Aut}G_1,$$ such that $F(g\circ f)=\operatorname{id}_{\operatorname{Aut}G_1}$. The answer to the question you link gives precisely such groups and homomorphisms.