Homotopy between composites associated in different ways

Statement

Existential version

Suppose $f_{1},f_{2},f_{3}$ are loops based at a point $x_{0}$ in a topological space $X$. We can consider two differently associated products of these three loops:

$\!a=(f_{1}*f_{2})*f_{3},b=f_{1}*(f_{2}*f_{3})$

$a$ and $b$ are homotopic loops, i.e., they are in the same homotopy class of loops based at $x_{0}$.

This version is essentially the associativity part of showing that the Fundamental group (?) of a based topological space is indeed a group.

Constructive/explicit version

We first note the explicit piecewise definitions of $a$ and $b$:

$a=\lbrace {\begin{array}{rl}f_{1}(4t),&0\leq t\leq 1/4\\f_{2}(4t-1),&1/4<t\leq 1/2\\f_{3}(2t-1),&1/2<t\leq 1\\\end{array}}$

and:

$b=\lbrace {\begin{array}{rl}f_{1}(2t),&0\leq t\leq 1/2\\f_{2}(4t-2),&1/2<t\leq 3/4\\f_{3}(4t-3),&3/4<t\leq 1\\\end{array}}$

If we denote the homotopy by $H$, we want $H(t,0)=a(t),H(t,1)=b(t)$ and $H(0,s)=H(1,s)=x_{0}$. This homotopy is explicitly given by:

$H(t,s)=\lbrace {\begin{array}{rl}f_{1}\left({\frac {4t}{1+s}}\right),&0\leq t\leq (1+s)/4\\f_{2}(4t-1-s),&(1+s)/4<t\leq (2+s)/4\\f_{3}\left({\frac {4t-2-s}{2-s}}\right),&(2+s)/4<t\leq 1\\\end{array}}$

Graphical version

Uniform version

This version is a little stronger than the other versions. Let $L=\Omega (X,x_{0})$ be the loop space of $(X,x_{0})$, i.e., the space of all loops in $X$ based at $x_{0}$ under the compact-open topology. Then, consider the following two maps:

$\!A:L\times L\times L\to L,\qquad A(f_{1},f_{2},f_{3})=(f_{1}*f_{2})*f_{3}$

and:

$\!B:L\times L\times L\to L,\qquad B(f_{1},f_{2},f_{3})=f_{1}*(f_{2}*f_{3})$

Then, the maps $A$ and $B$ are homotopic maps. This is part of the proof that $\Omega (X,x_{0})$ is a H-space, which is a homotopy variant of topological monoid.