Homotopy between constant loop and composite of loop with inverse

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 fact about::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:

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

and:

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

Then, the maps $$A$$ and $$E$$ are homotopic maps.