Proof of Associativity of Operation * on Path-homotopy Classes

(Continued from https://mathtuition88.com/2015/06/25/the-groupoid-properties-of-operation-on-path-homotopy-classes-proof/)

Earlier we have proved the properties (2) Right and left identities, (3) Inverse, leaving us with (1) Associativity to prove.

For this proof, it will be convenient to describe the product f*g in the language of positive linear maps.

First we will need to define what is a positive linear map. We will elaborate more on this since Munkres’ books only discusses it briefly.

Definition: If [a,b] and [c,d] are two intervals in $\mathbb{R}$ , there is a unique map $p:[a,b]\to [c.d]$ of the form p(x)=mx+k that maps a to c and b to d. This is called the positive linear map of [a,b] to [c,d] because its graph is a straight line with positive slope.

Why is it a positive slope? (Not mentioned in the book) It turns out to be because we have:

p(a) = ma+k=c

p(b) = mb+k=d

Hence, d-c = mb-ma = m(b-a)

Thus, m=(d-c)/(b-a), which is positive since d-c and b-a are all positive quantities.

Note that the inverse of a positive linear map is also a positive linear map, and the composite of two such maps is also a positive linear map.

Now, we can show that the product f*g can be described as follows: On [0,1/2], it is the positive linear map of [0,1/2] to [0,1], followed by f; and on [1/2,1] it equals the positive linear map of [1/2,1] to [0,1], followed by g.

Let’s see why this is true. The positive linear map of [0,1/2] to [0,1] is p(x)=2x. fp(x) = f(2x).

The positive linear map of [1/2,1] to [0,1] is p(x)=2x-1. gp(x)=g(2x-1).

If we look back at the earlier definition of f*g, that is precisely it!

Now, given paths, f, g, and h in X, the products f*(g*h) and (f*g)*h are defined if and only if f(1)=g(0) and g(1)=h(0), i.e. the end point of f = start point of g, and the end point of g = start point of h. If we assume that these two conditions hold, we can also define a triple product of the paths f, g, and h as follows:

Choose points a and b of I so that 0<a<b<1. Define a path $k_{a,b}$ in X as follows: On [0,a] it equals the positive linear map of [0,a] to I=[0,1] followed by f; on [a,b] it equals the positive linear map of [a,b] to I followed by g; on [b,1] it equals the positive linear map of [b,1] to I followed by h. This path $k_{a,b}$ depends on the choice of the values of a and b, but its path-homotopy class turns out to be independent of a and b.

We can show that if c and d are another pair of points of I with 0<c<d<1, then $k_{c,d}$ is path homotopic to $k_{a,b}$ .

Let $p:I\to I$ be the map whose graph is pictured in Figure 51.9 (taken from Munkre’s Book)

On the intervals [0,a], [a,b], [b,1], it equals the positive linear maps of these intervals onto [0,c],[c,d],[d,1] respectively. It follows that $k_{c,d} \circ p = k_{a,b}$ . Let’s see why this is so.

On [0,a] $k_{c,d}\circ p$ is the positive linear map of [0,a] to [0,c], followed by the positive linear map of [0,c] to I, followed by f. This equals the positive linear map of [0,a] to I, followed by f, which is precisely $k_{a,b}$ . Similar logic holds for the intervals [a,b] and [b,1].

$p$ is a path in I from 0 to 1, and so is the identity map $i: I\to I$ . Since I is convex, there is a path homotopy P in I between p and i. Then, $k_{c,d}\circ P$ is a path homotopy in X between $k_{a,b}$ and $k_{c.d}$ .

Now the question many will be asking is: What has this got to do with associativity. According to the author Munkres, “a great deal”! We check that the product $f*(g*h)$ is exactly the triple product $k_{a,b}$ in the case where $a=1/2$ and $b=3/4$ .

By definition,

$(g*h)(s)=\begin{cases} g(2s)\ &\text{for }s\in [0,\frac{1}{2}]\\ h(2s-1)\ &\text{for }s\in [\frac{1}{2},1] \end{cases}$

Thus, $f*(g*h)(s)=\begin{cases} f(2s)\ &\text{for }s\in [0,\frac{1}{2}]\\ (g*h)(2s-1)\ &\text{for }s\in [\frac{1}{2},1] \end{cases} =\begin{cases} f(2s)\ &\text{for }s\in [0,\frac{1}{2}]\\ g(4s-2)\ &\text{for }s\in [\frac{1}{2},\frac{3}{4}]\\ h(4s-3) &\text{for }s\in [\frac{3}{4},1] \end{cases}$

We can also check in a very similar way that $(f*g)*h)=k_{c,d}$ when c=1/4 and d=1/2. Thus, the these two products are path homotopic, and we have finally proven the associativity of *.