User:Anghel/Sandbox

Theorem
Let $C$ be a simple closed contour in the complex plane.

Let $f : \closedint 0 1 \to \R^2$ be a Jordan curve.

Let $\phi : \R^2 \to \C$, defined by:


 * $\map{ \phi }{ x, y } = x + iy$

be the standard homeomorphism between $\R^2$ and $\C$.

Let $\Img C = \map{ \phi }{ \Img f }$, where $\Img C$ denotes the image of $C$, and $\Img f$ denotes the image of $f$.

Then the interior of $C$:


 * $\Int C = \map{ \phi }{ \Int f }$

is well-defined.

Here, $\Int f$ denotes the interior of $f$.

Proof
Let $C$ be defined as a concatenation of a (finite) sequence of directed smooth curves $\sequence {C_1, \ldots, C_n}$.

We show existence of a Jordan curve $f : \closedint 0 1 \to \R^2$ that fulfills the criteria $\map{ \phi }{ \Img f } = \Img C$.

Reparameterization of Directed Smooth Curve with Given Domain shows that $C_i$ can be reparameterized by a smooth path:


 * $\gamma_i: \closedint { \dfrac{ i-1 }{ n } } { \dfrac i n } \to \C$ for all $i \in \set {1, \ldots, n}$

Define $\gamma: \closedint 0 1 \to \C$ by:


 * $\map \gamma t = \map { \gamma_i } t$ for all $t \in \closedint { \dfrac{ i-1 }{ n } } { \dfrac i n } $

Reparameterization of Directed Smooth Curve Preserves Image shows that $\Img \gamma = \Img C$.

Pasting Lemma for Pair of Continuous Mappings on Closed Sets shows that $\gamma$ is continuous.

By definition of simple contour, it follows that:


 * $\map \gamma { t_1 } \ne \map \gamma { t_2 }$ for all $t_1 \in \hointr 0 1, t_2 \in \hointr 0 1$ with $t_1 \ne t_2$

By definition of closed contour, it follows that:


 * $\map \gamma 0 = \map \gamma 1$

Composite of Continuous Mappings between Metric Spaces is Continuous shows that $\phi^{-1} \circ \gamma : \closedint 0 1 \to \R^2$ is continuous.

As $\phi$ is a homeomorphism, it follows by that $\phi^{-1} \circ \gamma$ is a Jordan curve.

Set $f = \phi^{-1} \circ \gamma$.

We now have $\Img C = \map{ \phi }{ \Img f }$.

We show uniqueness of the definition of $\Int C$.

Let $g: \closedint 0 1$ be a Jordan curve such that $\Img g = \Img f$.

Jordan Curve Theorem shows that the only depends on the image of the Jordan curve.

Then:


 * $\Int g = \Int f$

RemCategory:Contour Integration