Smooth Homotopy is an Equivalence Relation

== 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 ===
Suppose there exists an arbitrary smooth mapping $f: X \to Y$.
Let $m := \dim X$ and $n := \dim Y$. 
Let $H$ be a mapping such that for every chart:
:$\struct {U, \kappa}$ on $X$
and every chart:
:$\struct {V, \xi}$ on $Y$
we have:
:$V \cap \map H U \ne \O$
where:
:$\kappa: U \to X$ is a homeomorphism of $U$ onto an open subset $X$ of Euclidean space $\R^m$
:$\xi: V \to Y$ is a homeomorphism of $V$ onto an open subset $Y$ of Euclidean space $\R^n$.
Let:
:$\ds \xi \circ H \circ \kappa^{-1}: \map \kappa U \subseteq \R^m \to \map \xi {V \cap \map H U} \subseteq \R^n$
be a real function of differentiability class $C^\infty$.
and:
: $H: X \times \left[{0 \,.\,.\, 1}\right] \to Y$
: $H \left({x, 0}\right) = f \left({x}\right)$
: $H \left({x, 1}\right) = f \left({x}\right)$
Then it follows that $H$ is a smooth mapping which yields a smooth homotopy between $f$ and itself.
Furthermore, $H$ is a free homotopy between $f$ and itself such that:
:$(1): \quad \forall x \in K: \map f x = \map f x$
:$(2): \quad \forall x \in K, t \in \closedint 0 1: \map H {x, t} = \map f x$
So $f$ and itself are homotopic relative to $K$.
So $\sim$ has been shown to be reflexive.
=== 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 $f$.
Thus $G$ is smooth whenever $F$ is.
So $\sim$ has been shown to be symmetric.
=== 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:
:$\map \beta x = \begin{cases}
e^{-1 / \paren {1 - x^2} } & : \size x < 1 \\
0 & : \text { otherwise} 
\end{cases}$
This function is known to be smooth.
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.
$\sim$ has been shown to be reflexive, symmetric and transitive.
Hence by definition it is an equivalence relation.
Category:Homotopy Theory
Category:Examples of Equivalence Relations