# The Groupoid Properties of * on Path-homotopy Classes

This is one of the first instances where algebra starts to appear in Topology. We will continue our discussion of material found in Topology (2nd Economy Edition) by James R. Munkres.

First, we need to define the binary operation *, that will later make * satisfy properties that are very similar to axioms for a group.

Definition: If f is a path in X from $x_0$ to $x_1$, and if g is a path in X from $x_1$ to $x_2$, we define the product f*g of f and g to be he path h given by the equations

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

Well-defined: The function h is well-defined, at s=1/2, f(1)=x_1, g(0)=x_1.

Continuity: h is also continuous by the pasting lemma.

h is a path in X from x_0 to x_2. We think of h as the path whose first half is the path f and whose second half is the path g.

We will verify that the product operation on paths induces a well-defined operation on path-homotopy classes, defined by the equation $[f]*[g]=[f*g]$

Let F be a path homotopy between f and f’, and let G be a path homotopy between g and g’.

i.e. we have F(s,0)=f(s), F(s,1)=f'(s)
F(0,t)=x_0, F(1,t)=x_1
G(s,0)=g(s), G(s,1)=g'(s)
G(0,t)=x_1, G(1,t)=x_2

We can define:

$H(s,t)=\begin{cases}F(2s,t) &\text{for }s\in[0,\frac{1}{2}],\\ G(2s-1,t)&\text{for }s\in[\frac{1}{2},1]\end{cases}$.

We can check that F(1,t)=x_1=G(0,t) for all t, hence the map H is well-defined. H is continuous by the pasting lemma.

Let’s check that H is the required path homotopy between f*g and f’*g’.

For s in [0,1/2],

H(s,0) = F(2s,0) =f(2s)=h(s)

H(s,1) = F(2s,1) =f'(2s)=h'(s)
h’ := f’ * g’

H(0,t) = F(0,t) = x_0

s in [1/2,1] works fine too:

H(s,0) = G(2s-1,0) = g(2s-1)=h(s)

H(s,1) = G(2s-1,1)= g'(2s-1) = h'(s)

H(1,t) = G(1,t)= x_2

Thus, H is indeed the required path homotopy between f*g and f’*g’. * is almost like a binary operation for a group. The only difference is that [f]*[g] is not defined for every pair of classes, but only for those pairs [f], [g] for which f(1) = g(0), i.e. the end point of f is the starting point of g.

## Author: mathtuition88

https://mathtuition88.com/

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