Homotopy between constant loop and composite of loop with inverse

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Existential version

Suppose x_0 is a point in a topological space X. Suppose f is a loop based at x_0, i.e., f is a continuous map from the closed unit interval [0,1] to X such that f(0) = f(1) = x_0. Denote by f^{-1} the loop defined as f^{-1}(t) = f(1 - t). Denote by e the constant loop at the point x_0.

Denote by * the composition of loops by concatenation. Then, the loops f * f^{-1} and f^{-1} * f are both homotopic to e.

Note that since (f^{-1})^{-1} = f, it suffices to show that f * f^{-1} is homotopic to e.

This is the inverses part of the proof that the Fundamental group (?) of a based topological space is indeed a group.

Constructive/explicit version

We note that f^{-1}(t) = f(1 - t), so we have:

\! (f * f^{-1})(t) = \lbrace\begin{array}{rl} f(2t), & 0 \le t < 1/2 \\ f(2 - 2t), & 1/2 \le t \le 1 \\\end{array}

The map e to which we want to homotope this is:

\! e(t) = x_0, 0 \le t \le 1

The homotopy is given by:

\! F(t,s) = \lbrace\begin{array}{rl} f(2t), & 0 \le t \le s/2 \\ f(s), & s/2 < t < (2 - s)/2 \\ f(2 - 2t), & (2 - s)/2 \le t \le 1 \\\end{array}

Graphical version

Here is a pictorial description of the homotopy:


Uniform version

Suppose (X,x_0) is a based topological space. Let L = \Omega(X,x_0) be the loop space. Consider the following two maps from L to itself:

A: L \to L, \qquad A(f) = f * f^{-1}


E:L \to L, \qquad E(f) = e

Then, the maps A and E are homotopic maps.