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

Proof: define $H(s, t)=u((1-s)t+sphi(t))$, 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 $phi:Ito I$ by rescaling the speed according to the time of travel through each curve. More explicitly, we have $[alphacdot(betacdotgamma)] (t) = [(alphacdotbeta) cdotgamma] (phi(t)) $ if we define: $$phi(t) =begin{cases}frac 12 t, & tin[0,frac 12]\ t-frac14, & tin [frac12, frac34] \ 2t-1, & tin [frac34, 1]end{cases} $$

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 $(alphacdotbeta) cdotgamma$ is piecewisely defined. To make them match we need to "shear" $phi(t)$ to the left.

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

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

The homotopy you provided in the post is then $H(s, t) =[(alphacdot beta)cdot gamma](tildephi_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.)