Directed Smooth Curve Relation is Equivalence

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

Let $\gamma: \closedint a b \to \C$ and $\sigma : \closedint c d \to \C$ be smooth paths.

Define $\sim$ as follows:


 * $\gamma \sim \sigma$ there exists a bijective differentiable strictly increasing real function $\phi: \closedint c d \to \closedint a b$ 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 $\gamma: \closedint a b \to \C$ be a smooth path.

Define $\phi: \closedint a b \to \closedint a b$ as the identity function, so: $\map \phi t = t$ for all $t \in \closedint a b$

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

From Derivative of Identity Function, it follows that:
 * $\map {\phi'} t = 1 > 0$

for all $t \in \closedint a b$.

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

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

Symmetric
Let $\gamma: \closedint a b \to \C$ and $\sigma: \closedint c d \to \C$ be smooth paths such that $\gamma \sim \sigma$.

That is, $\sigma = \gamma \circ \phi$ where $\phi: \closedint c d \to \closedint a b$ is bijective, differentiable and strictly increasing.

From Bijection iff Inverse is Bijection, it follows that $\phi$ has an inverse function:
 * $\phi^{-1}: \closedint a b \to \closedint c d$

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

From Derivative of Inverse Function:
 * $D \map {\phi^{-1} } t = \dfrac 1 {\map {\phi'} {\map {\phi^{-1} } t} }$

for all $t \in \closedint a b$.

From Derivative of Monotone Function:
 * $\map {\phi'} {\map {\phi^{-1} } t} > 0$

Then:
 * $D \map {\phi^{-1} } t > 0$

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

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

Transitive
Let $\gamma: \closedint a b \to \C$, $\sigma: \closedint c d \to \C$ and $\rho: \closedint g h \to \C$ be smooth paths such that $\gamma \sim \sigma$ and $\sigma \sim \rho$.

That is:
 * $\sigma = \gamma \circ \phi$ and $\rho = \sigma \circ \psi$

where $\phi: \closedint c d \to \closedint a b$ and $\psi: \closedint g h \to \closedint c d$ are bijective, differentiable and strictly increasing.

Then:
 * $\rho = \gamma \circ \paren {\phi \circ \psi}$

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

From Derivative of Composite Function, it follows that:
 * $D \map {\paren {\phi \circ \psi} } t = \map {\phi'} {\map \psi t} \map {\psi'} t$

for all $t \in \closedint g h$.

As $\map {\phi'} {\map \psi t} > 0$ and $\map {\psi'} t > 0$, this implies that:
 * $D \map {\paren {\phi \circ \psi} } t > 0$

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

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

Therefore, $\sim$ is an equivalence relation.