1
$\begingroup$

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 $π_1(R_n,\ast)$ ?

$\endgroup$
1
$\begingroup$

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 $\pi_1(R_n,*)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.