$ \widehat{\alpha} : {\pi}_1 (X, x_0) \longrightarrow {\pi}_1 (X, x_1) $ is well defined?

Question

Definition: Let $X$ be a topological space and $ \alpha : I \longrightarrow X $ a path from $x_0$ to $x_1$. We define a map

$ \widehat{\alpha} : {\pi}_1 (X, x_0) \longrightarrow {\pi}_1 (X, x_1) $

by the eqution:

$ \widehat{\alpha}([f]):= [\bar{\alpha}]*[f]*[\alpha] $.

I want to show that this map is well defined. by definition I have to show that

If $ [f], [f^´] \in {\pi}_1 (X, x_0) $ such that $ [f] = [f^´] $, then $ \widehat{\alpha}([f]) = \widehat{\alpha}([f^´]) $

to do this, I know that it is enough to prove that

$ f \; {\simeq}_p \; f^´ $ imply $ \widehat{\alpha}([f]) \; {\simeq}_p \;\widehat{\alpha}([f^´]) $.

but I do not know how to proceed from that point!

Answer 1

You need only to check that $\ast$ is well-defined on the homotopy classes. Munkres does it like this:

First, if $f$ is a path from $x$ to $y$ and $g$ is a path from $y$ to $z$, define $\ast$ by

$(f\ast g)(s):=h(s)=\begin{cases} f(2s) & 0\le s\le 1/2 \\ g(2s-1)& 1/2\leq s\leq 1 \\ \end{cases}$

Now, define an operation, also denoted $\ast$, on the homotopy classes, $[f]\ast[g]= [f\ast g].$

To show that $\ast$ is well-defined, let $F$ be a path homotopy between $f$ and $f'$, and $G$ a path homotopy between $g$ and $g'$.

Then, check that there is a path homotopy between $f\ast g$ and $f'\ast g'$ given by

$H(s,t)=\begin{cases} F(2s,t) & 0\le s\le 1/2 \\ G(2s-1,t)& 1/2\leq s\leq 1 \\ \end{cases}.$

Answer 2

It is useful to have the algebraic, abstract, formulation of this "change of base point", which is actually a result on groupoids (e.g. 6.3.1 and 6.3.2 of Topology and Groupoids) applied to the fundamental groupoid $\pi_1(X)$ of a topological space $X$.

There is a not too well known formulation which links this result with another idea, that of crossed module.

Let $G$ be a groupoid with object set $X$, The object group of $G$ at an object $x$ is written $G(x)$. Then there is a groupoid $Aut(G)$ with object set $X$ and where $$Aut(G)(x,y)= Iso(G(x),G(y)), $$ the set of isomorphisms $G(x) \to G(y)$.

There is a morphism of groupoids $$\chi:G \to Aut(G) $$ which sends $a:x \to y$ to the "change of base" conjugation $G(x) \to G(y)$. (Some care is needed with conventions for conjugation and composition of morphisms to get $\chi$ to be a morphism rather than an antimorphism.)

The rules that $\chi$ satisfies are just like the "crossed module" rules given in this book, Nonabelian Algebraic Topology, and indeed if $G$ is a group then $\chi$ is a standard example of a crossed module. But I have seen few uses of the generalisation, and that is why I am advertising it here.