Concatenation of paths and loops (homotopy)

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Flesh out, link to other parts, make sure loop (topology) and path (topology) both link back here

Definition

Let $p,q:[0,1]\rightarrow X$ be paths (possibly loops) in a topological space $(X,\mathcal{ J })$ such that $p(1)=q(0)$ - the terminal point of $p$ is the initial point of $q$ - then we define their concatenation (AKA: composition, product) as follows:

• $f*g:[0,1]\rightarrow X$ given by $f*g:t\mapsto\left\{\begin{array}{lr}f(2t)& \text{if }t\in[0,\frac{1}{2}]\\g(2t-1)& \text{if }t\in[\frac{1}{2},1]\end{array}\right.$ - we claim this is a path.
  • in words this is the path that goes first around $f$ (at double the speed of $f$) and then around $g$ (again at double the speed of $g$)
  • Note that $t=\frac{1}{2}$ is in both parts of the piecewise function, this is to emphasise that $(f*g)(\frac{1}{2})$ is the same in either case.

Note: that if $f$ and $g$ are loops based at $x_0$ then so is $f*g$, and also that if $f(0)=g(1)$ (in addition to the $f(1)=g(0)$ required for concatenation) then $f*g$ is a loop.

Caution

Don't be over-eager and think "I see the group structure!" the constant loop is the identity and for a path $p$ it done backwards is the inverse!

Not quite. Mainly because if you do $f*\text{backwards}(f)$ you do not end up with the constant loop based at $f(0)$, you end up with a loop that goes around $f$ then back again!

See the "see next" section below.

Proof of claims

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Elementary from pasting lemma

See next

• First homotopy group
  • Homotopic paths
  • Homotopy (overview)