# Construct homotopy from $(\alpha \cdot \beta) \cdot \gamma$ to $\alpha \cdot (\beta \cdot \gamma)$ explicitly

I understand the general idea of a homotopy, but I'm a little lost on how to create them myself. For example if I wanted to show

$$\text{Let} \: \alpha, \beta, \text{and} \: \gamma \: \text{be paths} \: I \to X, \: \text{from} \: x_{0} \: \text{to} \: y_{0}, y_{0} \: \text{to} \: z_{0}, \: \text{and} \: z_{0} \: \text{to} \: u_{0}. \: \text{Then} \: \\ (\alpha \cdot \beta) \cdot \gamma \sim \alpha \cdot (\beta \cdot \gamma)$$

A possible homotopy is $$F: I \times I \to X$$, given by $$\\ F(t,s) = \begin{cases} \alpha(\frac{4t}{1+s}) & 0 \leq t \leq \frac{s+1}{4} \\ \beta(4t-1-s) & \frac{s+1}{4} \leq t \leq \frac{s+2}{4} \\ \gamma(\frac{4t - 2 - s}{2-s}) & \frac{s+2}{4} \leq t \leq 1 \\ \end{cases}$$

What I don't understand is where this comes from. What is the intuition here and how can I form explicit homotopies like this?

See the picture below.

The idea is that, for any choice of $$s$$, the loop $$t \mapsto F(t, s)$$ consists of walking along $$\alpha$$, $$\beta$$ and $$\gamma$$, in that order. The difference between the various choices of $$s$$ is in the "time schedule": each choice of $$s$$ allocates different amounts of time to $$\alpha$$, $$\beta$$ and $$\gamma$$.

• If $$s = 0$$, you walk path $$\alpha$$ in time $$[0, \tfrac 1 4]$$, then walk $$\beta$$ in time $$[\tfrac 1 4, \tfrac 1 2 ]$$, then walk $$\gamma$$ in time $$[\tfrac 1 2 , 1]$$.

• If $$s = 1$$, you walk path $$\alpha$$ in time $$[0, \tfrac 1 2]$$, then walk $$\beta$$ in time $$[\tfrac 1 2, \tfrac 3 4]$$, then walk $$\gamma$$ in time $$[\tfrac 3 4 , 1]$$.

For intermediate choices of $$s$$, the schedule is given by interpolation.

So for example, if $$s = \tfrac 1 2$$, you walk $$\alpha$$ in time $$[0, \tfrac 3 8]$$, then walk $$\beta$$ in time $$[\tfrac 3 8, \tfrac 5 8]$$, then walk $$\gamma$$ in time $$[\tfrac 5 8, 1]$$. And so on.

This is basically a reparametrization of the curve, but the parametrization is changing continuously. We may come up with the homotopy in two steps.

Let $$u:I\to X$$ be a curve, let $$\phi:I\to I$$ be a continuous map with $$\phi(0)=0$$ and $$\phi(1)=1$$, then $$u$$ is homotopic to $$u\circ\phi$$.

Proof: define $$H(s, t)=u((1-s)t+s\phi(t))$$, then $$H(0,t)=u(t)$$, $$H(1,t)=u(\phi(t))$$ and $$H(s, 0)=u(0)$$, $$H(s, 1)=u(1)$$. Essentially this is just mapping a homotopy between $$\operatorname{Id}$$ and $$\phi$$ in $$I$$ to the homotopy in $$X$$ by $$u$$.

We could construct $$\phi:I\to I$$ by rescaling the speed according to the time of travel through each curve. More explicitly, we have $$[\alpha\cdot(\beta\cdot\gamma)] (t) = [(\alpha\cdot\beta) \cdot\gamma] (\phi(t))$$ if we define: $$\phi(t) =\begin{cases}\frac 12 t, & t\in[0,\frac 12]\\ t-\frac14, & t\in [\frac12, \frac34] \\ 2t-1, & t\in [\frac34, 1]\end{cases}$$

However if you apply the formula above, while we still get a homotopy, but the formula is not as clean (as in we need to divide into 5 cases). To obtain the formula you stated, let's look at the graphs of the intermediate reparametrizations, for each fixed $$s_0$$, the graph of $$\phi_{s_0}(t) =1-s_0)t + s_0 \phi(t)$$ would look like a "sheared up" version of $$\phi(t)$$. So image of the points at which $$\phi_{s_0}$$ are piecewisely defined do not match with the points at which $$(\alpha\cdot\beta) \cdot\gamma$$ is piecewisely defined. To make them match we need to "shear" $$\phi(t)$$ to the left.

So we should take $$\tilde\phi_s(t)$$ the inverse function of $$(1-s)t+s\phi^{-1}(t)$$, i.e. $$\tilde\phi_s(t)$$ is the inverse of $$\begin{cases}(1+s)t, & t\in[0,\frac14] \\ t+\frac s4, & t\in[\frac 14,\frac 12]\\ \frac 12(2-s)t +\frac s2, & t\in[\frac 12,1]\end{cases}$$

Which is given by: $$\tilde\phi_s(t) =\begin{cases} \frac t{1+s}, & t\in[0,\frac {s+1}4]\\ t-\frac s4, & t\in[\frac{s+1}4,\frac{s+2}4]\\ \frac 2{2-s}t - \frac s{2-s}, & t\in[\frac {s+2}4,1]\end{cases}$$

The homotopy you provided in the post is then $$H(s, t) =[(\alpha\cdot \beta)\cdot \gamma](\tilde\phi_s(t))$$.

I have checked the definition and all axioms of the first homotopy group in this note and in this I also explain the exact homotopy you describe. I hope it helps you...

There is an easier way to do this by using the notion of a "Moore path" as a pair $$(f,r)$$ where $$r \in \mathbb R, r \geqslant 0$$ and $$f:[0, \infty) \to X$$ is constant $$[r, \infty)$$. This definition is used in the quite old book on Knot Theory by Crowell and Fox. Alternatively, as in Topology and Groupoids, one considers a path of length $$r \geqslant 0$$ to be a map $$f:[0, r] \to X$$. The composite of a path of length $$r$$ with a path of length $$s$$ is then, when defined, of length $$r+s$$. This corresponds intuitively to the idea of path as a "journey". Composition is then associative. We also have paths of length $$0$$ and in both definitions the composition of paths in $$X$$ gives a category $$PX$$.

We also write $$s$$ for a constant path of length $$s$$ at $$y \in X$$ and say two paths $$f,g$$ from $$x$$ to $$y$$ in $$X$$ are equivalent if there are real numbers $$s,t \geqslant 0$$ such that $$f+s, g+t$$ are homotopic rel end points. This gives the fundamental groupoid $$\pi_1(X)$$.

One easily proves that any path is equivalent to a path of length $$1$$. This is called normalisation, which is not a process one uses for journeys.

The formulae needed for all this work out much easier than the usual ones, which seems to me a good thing. (I've said all this somewhere else on this site.)