Concatenation of Contours is Contour

Theorem
Let $\left[{ a \,.\,.\, b }\right]$ and $\left[{ c \,.\,.\, d }\right]$ be closed real intervals.

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

Let $\gamma+\sigma: \left[{ a \,.\,.\, b-c+d }\right] \to \C$ denote the concatenation of the contours $\gamma$ and $\sigma$.

Then $\gamma+\sigma$ is a contour.

Proof
From the definition of contour, it follows that there exists a subdivision $a_0, a_1, \ldots , a_n$ of $\left[{ a \,.\,.\, b }\right]$ such that $\gamma \restriction_{ I_i }$ is a smooth path, where $I_i = \left[{ a_{ i - 1 } \,.\,.\, a_i }\right]$.

Also, there exists a subdivision $b_0, b_1, \ldots , b_m$ of $\left[{ c \,.\,.\, d }\right]$ such that $\sigma \restriction_{ J_i }$ is a smooth path, where $J_i = \left[{ b_{ i - 1 } \,.\,.\, b_i }\right]$.

Define a subdivision of $\left[{ a \,.\,.\, b-c+d }\right]$ by:


 * $a = a_0 < a_1 < a_2 < \cdots < a_n = b < b + b_1 < b + b_2 < \cdots < b + b_m = b + d$

Put $\rho = \gamma + \sigma$.

Put $K_i = \left[{ b + b_{ i - 1 } \,.\,.\, b + b_i }\right]$.

Then for all $i \in \left\{ { 1, \ldots, n }\right\}$ and $t \in I_i$, we have:


 * $\rho' \restriction_{I_i} \left({t}\right) = \gamma' \restriction_{I_i} \left({t}\right)$

For all $i \in \left\{ { 1, \ldots, m }\right\}$ and $t \in K_i$, we have:


 * $\rho' \restriction_{K_i} \left({t}\right) = \sigma' \restriction_{J_i} \left({t - a + c }\right)$

which follows from Derivative of Complex Composite Function, as $\rho \left({t}\right) = \sigma \left({t - a + c}\right)$.

Hence $\rho = \gamma + \sigma$ is a contour.