Directed Smooth Curve Relation is Equivalence

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

Let $\sim$ denote a relation of smooth paths.

Define $\sim$ as follows:


 * $\gamma \sim \sigma$ iff 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 of smooth paths.

Proof
Checking in turn each of the criteria for 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 follow that $\phi$ is bijective.

From Derivative of Identity Function, it follows that $\phi' \left({t}\right) = 1 > 0$ for all $t \in \left[{ a \,.\,.\, b }\right]$.

From Derivative of Monotone Function, it follows that $\phi$ is strictly increasing.

Hence, $\gamma \sim \gamma$.

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

From Derivative of Inverse Function, it follows that $D \phi^{-1} \left({t}\right) = \dfrac 1 {\phi \left({t}\right)} $ for all $t \in \left[{a \,.\,.\, b}\right]$.

From Derivative of Monotone Function, it follows that $\phi' \left({t}\right) > 0$.

It follows that $D \phi^{-1} \left({t}\right) > 0$, so $\phi^{-1}$ is strictly increasing.

Hence, $\sigma \sim \gamma$.

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 $\psi' \left({t}\right) > 0$, it follows that $D \left({\phi \circ \psi}\right) \left({t}\right) > 0$, so $\phi \circ \psi$ is strictly increasing.

Hence, $\gamma \sim \rho$.