Reversed Directed Smooth Curve is Directed Smooth Curve

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Theorem

Let $C$ be a directed smooth curve in $\C$.

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


Define $\psi: \left[{a \,.\,.\, b}\right] \to \left[{a \,.\,.\, b}\right]$ by $\psi \left({t}\right) = a + b - t$.

Define $\rho: \left[{a \,.\,.\, b}\right] \to \C$ by $\rho = \gamma \circ \psi$.


Then $\rho$ is a smooth path which parameterizes a directed smooth curve $-C$.

The directed smooth curve $-C$ is independent of the parameterization $\gamma$.


Proof

First, we prove that $\rho$ is a smooth path:

\(\ds \rho' \left({t}\right)\) \(=\) \(\ds \gamma' \left({\psi\left({t}\right) }\right) \psi' \left({t}\right)\) Derivative of Complex Composite Function
\(\ds \) \(=\) \(\ds -\gamma' \left({\psi\left({t}\right) }\right)\) Derivatives of Function of $a x + b$: $\psi' \left({t}\right) = -1$
\(\ds \) \(\ne\) \(\ds 0\) Definition of Smooth Path: $\gamma' \left({\psi\left({t}\right)}\right) \ne 0$

As $\gamma'$ is continuously differentiable, and $\psi$ is continuous, it follows that $\rho'$ is continuous.

Then $\rho$ is a parameterization of a directed smooth curve $-C$.

$\Box$


Next, we prove that the definition of $-C$ is independent of the parameterization $\gamma$.

Suppose $\sigma$ is another parameterization of $C$:

$\sigma = \gamma \circ \phi$

Here $\phi: \left[{c \,.\,.\, d}\right] \to \left[{a \,.\,.\, b}\right]$ is a bijective differentiable strictly increasing function.

Define $\tilde \psi: \left[{c \,.\,.\, d}\right] \to \left[{c \,.\,.\, d}\right]$ by $\tilde \psi \left({t}\right) = c + d - t$.

Define $\tilde \rho: \left[{c \,.\,.\, d}\right] \to \C$ by $\tilde \rho = \sigma \circ \tilde{\psi}$.

We now prove that $\rho$ and $\tilde \rho$ both are parameterizations of the same directed smooth curve $-C$.

Both $\psi$ and $\tilde \psi$ are bijective with $\psi^{-1} = \psi$ and $\tilde \psi^{-1} = \tilde \psi$.

Define $\tilde \phi: \left[{c \,.\,.\, d}\right] \to \left[{a \,.\,.\, b}\right]$ by $\tilde \phi = \psi^{-1} \circ \phi \circ \tilde \psi$.

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

From Derivative of Composite Function, it follows that $\tilde \phi$ is differentiable with

\(\ds \tilde \phi' \left({t}\right)\) \(=\) \(\ds \dfrac 1 {\psi' \left({\psi^{-1} \left({\phi \left({\tilde \psi \left({t}\right) }\right) }\right) }\right) } \phi' \left({\tilde \psi \left({t}\right) }\right) \tilde \psi' \left({t}\right)\) Derivative of Inverse Function
\(\ds \) \(=\) \(\ds \phi' \left({\tilde \psi \left({t}\right)}\right)\) Derivatives of Function of $a x + b$
\(\ds \) \(>\) \(\ds 0\) Derivative of Monotone Function

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

As:

$\rho \circ \tilde \phi = \gamma \circ \psi \circ \psi^{-1} \circ \phi \circ \tilde \psi = \gamma \circ \phi \circ \tilde \psi = \sigma \circ \tilde \phi = \tilde \rho$

it follows that $\rho$ and $\tilde{\rho}$ are parameterizations of $-C$.

$\blacksquare$


Sources

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