Concatenation of Contours is Contour

Theorem
Let $C$ and $D$ be contours.

That is, $C$ is a finite sequence of directed smooth curves $C_1, \ldots, C_n$.

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

Similarly, $D$ is a finite sequence of directed smooth curves $D_1, \ldots, D_m$.

Let $D_i$ be parameterized by the smooth path $\sigma_i: \left[{c_i\,.\,.\,d_i}\right] \to \C$ for all $i \in \left\{ {1, \ldots, m}\right\}$.

Suppose $\gamma_n \left({b_n}\right) = \sigma_1 \left({c_1}\right)$.

Then the finite sequence:


 * $C_1, \ldots, C_n, D_1, \ldots, D_m$

defines a contour.

Proof
By definition of contour, each $C_i$ and $D_j$ is a directed smooth curve for all $i \in \left\{ {1, \ldots, n}\right\}, j \in \left\{ {1, \ldots, m}\right\}$.

By definition of contour, $\gamma_i \left({b_i}\right) = \gamma_{i+1} \left({a_{i+1} }\right)$ and $\sigma_j \left({d_j}\right) = \sigma_{j+1} \left({c_{j+1} }\right)$ for all $i \in \left\{ {1, \ldots, n-1}\right\}, j \in \left\{ {1, \ldots, m-1}\right\}$.

By assumption, $\gamma_n \left({b_n}\right) = \sigma_1 \left({c_1}\right)$.

Hence, $C_1, \ldots, C_n, D_1, \ldots, D_m$ defines a contour.