Homotopy is an Equivalence Relation

Theorem

Homotopy is an equivalence relation.

Proof

We examine each condition for equivalency.

Reflexivity

For any function $f:X \to Y$, define $F:X\times [0,1] \to Y$ as $F(x,t)=f(x)$.

This yields a smooth homotopy between $f$ and itself.

Symmetry

Given a homotopy:

$F: X \times [0,1] \to Y$ from $f(x)=F(x,0)$ to $g(x)=F(x,1)$

the function:

$G(x,t)=F(x,1-t)$

is a homotopy from $g$ to $f$.

Thus $G$ is smooth whenever $F$ is.

Transitivity

The continuous case admits of a simpler solution than the smooth case, but the smooth case implies the continuous case, so we examine only the smooth case.

Define the function:

$\beta (x) = \begin{cases} e^{-1/(1-x^2)} & : |x| < 1 \\ 0 & : \mbox{ otherwise} \end{cases}$

This function is known to be smooth.

Define:

$\displaystyle \phi (t) = \dfrac{\displaystyle \int_0^t \beta \left({\dfrac{x+2} 4}\right) \ \mathrm d x} {\displaystyle \int_0^1 \beta \left({\dfrac{x+2} 4}\right) \ \mathrm d x}$

By construction, $\phi$ is a smooth function which is $0$ for all $t \leq 1/4$, $1$ for all $t \geq 3/4$, and rises smoothly from $0$ to $1$ in $\left({\dfrac 1 4, \dfrac 3 4}\right)$.

Let $f,g,h$ be smooth functions such that $f$ is homotopic to $g$, which is in turn homotopic to $h$.

Then we can define the smooth homotopies:

$A(x,t) = \phi(t)g(x)+(1-\phi(t))f(x)$, which satisfies $A(x,0)=f(x)$ and $A(x,1)=g(x)$

and:

$B(x,t) = \phi(t)h(x)+(1-\phi(t))g(x)$, which satisfies $B(x,0)=g(x)$ and $B(x,1)=h(x)$

We then define a smooth function:

$C(x,t) = \begin{cases} A \left({x, \dfrac t 2}\right) & : t \le \dfrac 1 2 \\ B \left({x, \dfrac{t+1} 2}\right) & : t > \dfrac 1 2 \end{cases}$

$C$ is a smooth function satisfying $C(x,0)=f(x)$ and $C(x,1)=h(x)$, so it is a homotopy from $f$ to $h$.

$\blacksquare$