Reversed Contour is Contour

Theorem
Let $\R^n$ be a real cartesian space of $n$ dimensions.

Let $C$ be a contour in $\R^n$ that is defined as a concatenation of a finite sequence $C_1, \ldots, C_n$ of directed smooth curves in $\R^n$.

Then the finite sequence of reversed directed smooth curves:


 * $-C_n, -C_{n - 1}, \ldots, -C_1$

defines a contour that is independent of the parameterization of $C_1, \ldots, C_n$.

Proof
Let $C_i$ be parameterized by the smooth path $\gamma_i: \left[{a_i \,.\,.\, b_i}\right] \to \C$ for all $i \in \left\{ {1, \ldots, n}\right\}$.

From Reversed Directed Smooth Curve is Directed Smooth Curve, it follows that $-C_i$ is independent of the paraterization $\gamma_i$ of $C_i$.

We now prove that the end point of $-C_i$ is equal to the start point of $-C_{i-1}$ for all $i \in \left\{ {2, \ldots, n}\right\}$.

By definition of reversed directed smooth curve, $-C_i$ is parameterized by $\rho_i: \left[{a_i \,.\,.\, b_i}\right] \to \C$.

Here, $\rho_i \left({t}\right) = \gamma_i \left({a_i + b_i - t}\right)$.

From Reparameterization of Directed Smooth Curve Maps Endpoints To Endpoints, it follows that the endpoints $\rho_i \left({a_i}\right)$ and $\rho_i \left({b_i}\right)$ are independent of the parameterization $\rho_i$.

Then:

By definition, it follows that $-C_n, \ldots, -C_1$ can be concatenated to form a contour.