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


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

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

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


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

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

See the "see next" section below.

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Elementary from pasting lemma

See next


  1. Or, if they're both loops, we could just say "both loops have the same basepoint"


  1. Introduction to Topological Manifolds - John M. Lee