# How to see that an induced Outer automorphism does not depend on the choice of the Out-Inverse function?t

A marked graph is a pair $(G,τ)$ where $G$ is a graph and $τ: R_n \to G$ is a homotopy equivalence . $R_n$ is the Rose with n petals which is isomorphic to $F_n$ ,the free group with $n$ generators.

A homotopy equivalence $σ: G_2 \to G_1$ is called Out-inverse to a homotopy equivalence $τ:G_1 \to G_2$ if for any vertices $u_1 \in V(G_1)$ and $u_2 \in V(G_2)$ the maps $σ \circ τ$ and $τ \circ σ$ induce the identity outer Automorphism of the groups $π_1(G_1,u_1)$ and $π_1(G_2,u_2)$ respectively.

Let $(G,τ)$ be a marked graph and let $σ : G \to R_n$ a homotopy equivalence Out-inverse to $τ$. Then every homotopy equivalence $f: G \to G$ determines the outer automorphism $(σ \circ f \circ t)_\circledast$ of the group $π_1(R_n,\ast)= F_n$. This outer automorphism does not depend on the choice of $σ$.

My main question is: why this outer automorphism does not depend on choice of $σ$ ?

This statement is the same to say that that the homotopy equivalnce $σ \circ f \circ t: R_n \to R_n$ determines the outer automorphism $(σ \circ f \circ t)_\circledast$ of the group $π_1(R_n,\ast)$ ?

The concept of $\sigma : G \to R_n$ being "a homotopy equivalence Out-inverse to $\tau$" is exactly the same as $\sigma$ being "a homotopy inverse of $\tau$". Any two homotopy inverses $\sigma_1,\sigma_2 : G \to R_n$ of $\tau$ are homotopic to each other. Therefore $\sigma_1 \circ f \circ t$ and $\sigma_2 \circ f \circ t$ are homotopic to each other. Therefore they induce the same outer automorphism of $\pi_1(R_n,*)$.