Jordan Curve and Jordan Arc form Two Jordan Curves/Corollary

Corollary to Jordan Curve and Jordan Arc Form Two Jordan Curves
Let $\gamma: \left[{a \,.\,.\, b}\right] \to \R^2$ be a Jordan curve, where $\left[{a \,.\,.\, b}\right]$ is a closed real interval.

Denote the interior of $\gamma$ as $\operatorname{Int} \left({\gamma}\right)$, and denote the image of $\gamma$ as $\operatorname{Im} \left({\gamma}\right)$.

Let $\sigma: \left[{c \,.\,.\, d}\right] \to \R^2$ be a Jordan arc such that $\sigma \left({c}\right) \ne \sigma \left({d}\right)$, $\sigma \left({c}\right), \sigma \left({d}\right) \in \operatorname{Im} \left({\gamma}\right)$, and $\sigma \left({ t }\right) \in \operatorname{Int} \left({\gamma}\right)$ for all $t \in \left({c \,.\,.\, d}\right)$.

Put $t_1 = \gamma^{-1} \left({ \sigma \left({c}\right) }\right)$, and $t_2 = \gamma^{-1} \left({ \sigma \left({d}\right) }\right)$.

Suppose that $t_1 > t_2$. Define $-\sigma: \left[{c \,.\,.\, d}\right] \to \operatorname{Im} \left({\sigma}\right)$ by $-\sigma \left({t}\right) = \sigma \left({c + d - t}\right)$.

Let $*$ denote concatenation of paths, and let $\gamma \restriction_{ \left[{a \,.\,.\, t_1}\right] }$ denote the restriction of $\gamma$ to $\left[{a \,.\,.\, t_1}\right]$. Now, define $\gamma_1 = \gamma {\restriction_{ \left[{a \,.\,.\, t_2}\right] } } * \left({ -\sigma }\right) * \gamma{ \restriction_{ \left[{t_1 \,.\,.\, b}\right] } }$, and define $\gamma_2 = \gamma{ \restriction_{ \left[{t_2 \,.\,.\, t_1}\right] } } * \sigma$.

Then $\gamma_1$ and $\gamma_2$ are Jordan curves with $\operatorname{Int} \left({\gamma_1}\right) \subseteq \operatorname{Int} \left({\gamma}\right)$, and $\operatorname{Int} \left({\gamma_2}\right) \subseteq \operatorname{Int} \left({\gamma}\right)$.

Proof
The corollary can be proven in the same way as the main theorem.