We examine each condition for equivalence.   === Reflexivity === 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. {{explain|Prove the above.}} So $\sim$ has been shown to be reflexive. {{qed|lemma}} === Symmetry === 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 $f$. {{explain|a) The above needs to be proved, and b) there is nothing in the above that states that smoothness has been preserved.}} Thus $G$ is smooth whenever $F$ is. So $\sim$ has been shown to be symmetric. {{qed|lemma}} === 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.   {{explain|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.}} 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. {{Handwaving|"is known" -- not here it's not. Link to a page proving it.}} Define: : $\displaystyle \phi \left({t}\right) = \dfrac {\displaystyle \int_0^t \beta \left({\dfrac {x + 2} 4}\right) \, \mathrm d x} {\displaystyle \int_0^1 \beta \left({\dfrac{x + 2} 4}\right) \, \mathrm d x}$ 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 $A \left({x, 1}\right) = g \left({x}\right)$ and: :$B \left({x, t}\right) = \phi \left({t}\right) h \left({x}\right) + \left({1 - \phi \left({t}\right)}\right) g \left({x}\right)$, 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. {{qed|lemma}} $\sim$ has been shown to be reflexive, symmetric and transitive. Hence by definition it is an equivalence relation. {{qed}} Category:Homotopy Theory e7wfmr9v3s72m3cvcb8dqdve3ep9l8g