User:Anghel/Sandbox

Theorem
Let $D \subseteq \R^2$ be an open path-connected subset of the Euclidean plane.

Then $D$ is simply connected, the following condition holds:


 * For all Jordan curves $f : \closedint 0 1 \to \R^2$ with $\Img f \subseteq D$, we have $\Int f \subseteq D$.

Here $\Img f$ denotes the image of $f$, and $\Int f$ denotes the interior of $f$.

Sufficient condition
Suppose $D$ is simply connected.

Let $f : \closedint 0 1 \to D$ be a Jordan curve.

By definition of loop, $f$ is a loop in $D$.

By definition of simple connectedness by null-homotopy, there exists a path homotopy $H_0 : \closedint 0 1 \times \closedint 0 1 \to D$ between $f$ and a constant loop $c: \closedint 0 1 \to \set { \map f 0 }$.

From Simple Loop Image Equals Set Homeomorphic to Circle, it follows that $\Img f$ is homeomorphic to $\Bbb S^1$, which denotes the unit circle in $\R^2$.

Let the homeomorphism between $\Bbb S^1$ and $\Img f$ be $h: \Bbb S^1 \to \Img f$.

Be definition of homeomorphism, it follows that $h$ is a continuous injective mapping.

Define $H : \Bbb S^1 \times \closedint 0 1 \to D$ by $\map H {\mathbf x, t}$

qed}}