# Directed Smooth Curve Relation is Equivalence

## Theorem

Let $\sim$ denote a relation on the set of all smooth paths: $\left\{ {\gamma: I \to \C : \text{$I$is a closed real interval,$\gamma$is a smooth path} }\right\}$.

Let $\gamma : \left[{ a \,.\,.\, b }\right] \to \C$ and $\sigma : \left[{c \,.\,.\, d}\right] \to \C$ be smooth paths.

Define $\sim$ as follows:

$\gamma \sim \sigma$ if and only if there exists a bijective differentiable strictly increasing real function $\phi: \left[{c \,.\,.\, d}\right] \to \left[{a \,.\,.\, b}\right]$ such that $\sigma = \gamma \circ \phi$.

Then $\sim$ is an equivalence relation on the set of all smooth paths.

## Proof

Checking in turn each of the criteria for an equivalence:

### Reflexive

Let $\gamma: \left[{ a \,.\,.\, b }\right] \to \C$ be a smooth path.

Define $\phi: \left[{ a \,.\,.\, b }\right] \to \left[{ a \,.\,.\, b }\right]$ as the identity function, so : $\phi \left({t}\right) = t$ for all $t \in \left[{ a \,.\,.\, b }\right]$

From Identity Mapping is Bijection, it follows that $\phi$ is bijective.

From Derivative of Identity Function, it follows that: $\phi' \left({t}\right) = 1 > 0$

for all $t \in \left[{ a \,.\,.\, b }\right]$.

Then Derivative of Monotone Function shows that $\phi$ is strictly increasing.

Hence, $\gamma \sim \gamma$, so $\sim$ is reflexive.

$\Box$

### Symmetric

Let $\gamma : \left[{ a \,.\,.\, b }\right] \to \C$ and $\sigma : \left[{ c \,.\,.\, d }\right] \to \C$ be smooth paths such that $\gamma \sim \sigma$.

That is, $\sigma = \gamma \circ \phi$ where $\phi: \left[{c \,.\,.\, d}\right] \to \left[{a \,.\,.\, b}\right]$ is bijective, differentiable and strictly increasing.

From Bijection iff Inverse is Bijection, it follows that $\phi$ has an inverse function $\phi^{-1}: \left[{a \,.\,.\, b}\right] \to \left[{c \,.\,.\, d}\right]$.

Then $\gamma = \sigma \circ \phi^{-1}$.

$D \phi^{-1} \left({t}\right) = \dfrac 1 {\phi' \left({\phi^{-1} \left({t}\right) }\right)}$

for all $t \in \left[{a \,.\,.\, b}\right]$.

$\phi' \left({\phi^{-1} \left({t}\right) }\right) > 0$

Then:

$D \phi^{-1} \left({t}\right) > 0$

so $\phi^{-1}$ is strictly increasing.

Hence, $\sigma \sim \gamma$, so $\sim$ is symmetric.

$\Box$

### Transitive

Let $\gamma : \left[{ a \,.\,.\, b }\right] \to \C$, $\sigma : \left[{ c \,.\,.\, d }\right] \to \C$ and $\rho : \left[{ g \,.\,.\, h }\right] \to \C$ be smooth paths such that $\gamma \sim \sigma$ and $\sigma \sim \rho$.

That is:

$\sigma = \gamma \circ \phi$ and $\rho = \sigma \circ \psi$

where $\phi: \left[{c \,.\,.\, d}\right] \to \left[{a \,.\,.\, b}\right]$ and $\psi: \left[{g \,.\,.\, h}\right] \to \left[{c \,.\,.\, d}\right]$ are bijective, differentiable and strictly increasing.

Then:

$\rho = \gamma \circ \left({\phi \circ \psi}\right)$

From Composite of Bijections is Bijection, it follows that $\phi \circ \psi$ is bijective.

From Derivative of Composite Function, it follows that:

$D \left({\phi \circ \psi}\right) \left({t}\right) = \phi' \left({\psi \left({t}\right) }\right) \psi' \left({t}\right)$

for all $t \in \left[{ g \,.\,.\, h }\right]$.

As $\phi'\left({\psi \left({t}\right) }\right) > 0$ and $\psi' \left({t}\right) > 0$, this implies that:

$D \left({\phi \circ \psi}\right) \left({t}\right) > 0$

so $\phi \circ \psi$ is strictly increasing.

Hence, $\gamma \sim \rho$, so $\sim$ is transitive.

$\Box$

Therefore, $\sim$ is an equivalence relation.

$\blacksquare$