Smooth Homotopy is an Equivalence Relation

This article needs to be linked to other articles. In particular: Links needed throughout to definitions and results. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{MissingLinks}} from the code.

Theorem

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

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

Let $\map {\CC^\infty} {X, Y}$ be the set of all smooth mappings from $X$ to $Y$.

Define a relation $\sim$ on $\map \CC {X, Y}$ by $f \sim g$ if $f$ and $g$ are smoothly homotopic relative to $K$.

Then $\sim$ is an equivalence relation.

Proof

We examine each condition for equivalence.

Reflexivity

For any function $f: X \to Y$, define $F: X \times \closedint 0 1 \to Y$ as:

$\forall \tuple {x, t} \in X \times \closedint 0 1: \map F {x, t} = \map f x$

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

This article, or a section of it, needs explaining. In particular: Prove the above. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

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

$\Box$

Symmetry

Given a homotopy:

$F: X \times \closedint 0 1 \to Y$ from $\map f x = \map F {x, 0}$ to $\map g x = \map F {x, 1}$

the mapping:

$\map G {x, t} = \map F {x, 1 - t}$

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

This article, or a section of it, needs explaining. In particular: a) The above needs to be proved, and b) there is nothing in the above that states that smoothness has been preserved. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

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

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

$\Box$

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.

This article, or a section of it, needs explaining. In particular: The smooth case implies the continuous case: Needs to be a link to a page proving this - even if it follows trivially from the definitions this still needs to be shown - so that people who don't know this fact learn from it. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

Define the function:

$\map \beta x = \begin{cases}

e^{-1 / \paren {1 - x^2} } & : \size x < 1 \\ 0 & : \text { otherwise} \end{cases}$

This function is known to be smooth.

This article contains statements that are justified by handwavery. In particular: "is known" -- not here it's not. We need a link to a page proving it. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding precise reasons why such statements hold. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Handwaving}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Define:

$\ds \map \phi t = \dfrac {\ds \int_0^t \map \beta {\dfrac {x + 2} 4} \rd x} {\ds \int_0^1 \map \beta {\dfrac {x + 2} 4} \rd x}$

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 $\closedint {\dfrac 1 4} {\dfrac 3 4}$.

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:

$\map A {x, t} = \map \phi t \map g x + \paren {1 - \map \phi t} \map f x$, which satisfies $\map A {x, 0} = \map f x$ and $\map A {x, 1} = \map g x$

and:

$\map B {x, t} = \map \phi t \map h x + \paren {1 - \map \phi t} \map g x$, which satisfies $\map B {x, 0} = \map g x$ and $\map B {x, 1} = \map h x$

We then define a smooth function:

$\map C {x, t} = \begin {cases} \map A {x, \dfrac t 2} & : t \le \dfrac 1 2 \\

\map B {x, \dfrac {t + 1} 2} & : t > \dfrac 1 2 \end {cases}$

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

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

$\Box$

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

Hence by definition it is an equivalence relation.

$\blacksquare$