Smooth Homotopy is an Equivalence Relation

From ProofWiki
Jump to navigation Jump to search


Let $X$ and $Y$ be smooth manifolds.

Let $K \subseteq X$ be a (possibly empty) subset of $X$.

Let $\mathcal C^\infty \left({X, Y}\right)$ be the set of all smooth mappings from $X$ to $Y$.

Define a relation $\sim$ on $\mathcal C \left({X, Y}\right)$ by $f \sim g$ if $f$ and $g$ are smoothly homotopic relative to $K$.

Then $\sim$ is an equivalence relation.


We examine each condition for equivalence.


For any function $f: X \to Y$, define $F: X \times \left[{0 \,.\,.\, 1}\right] \to Y$ as:

$\forall \left({x, t}\right) \in X \times \left[{0 \,.\,.\, 1}\right]: F \left({x, t}\right) = f \left({x}\right)$

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

So $\sim$ has been shown to be reflexive.



Given a homotopy:

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

the mapping:

$G \left({x, t}\right) = F \left({x, 1 - t}\right)$

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

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

So $\sim$ has been shown to be symmetric.



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:

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

This function is known to be smooth.


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

By construction, $\phi$ is a smooth function which is $0$ for all $t \le 1/4$, $1$ for all $t \ge 3/4$, 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:

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


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

We then define a smooth function:

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

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

So $\sim$ has been shown to be transitive.


$\sim$ has been shown to be reflexive, symmetric and transitive.

Hence by definition it is an equivalence relation.