Interior of Jordan Curve is Subset of Image of Null-Homotopy

Theorem

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

Let $H : \closedint 0 1 \times \closedint 0 1 \to \R^2$ be a path homotopy between $f$ and a constant loop.

Then:

$\Int f \subseteq \Img H$

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

Proof

Let $\map \Omega {\R^2, \map f 0}$ denote the set of all loops based at $\map f 0$, where $\R^2$ is the real number plane with euclidean topology.

Let $c_{\map f 0}: \closedint 0 1 \to \set { \map f 0 }$ be the constant loop.

That is, $c_{\map f 0}$ is the loop $c_{\map f 0} \in \map \Omega {\R^2, \map f 0}$ such that:

$\forall t \in \closedint 0 1 : \map {c_{ \map f 0 }} t = \map f 0 $

Let $c_{\map f 0}$ be path-homotopic to $f$.

We are given that $H : \closedint 0 1 \times \closedint 0 1 \to \R^2$ is a path homotopy between $f$ and the constant loop $c_{\map f 0}$.

That is:

$\forall s \in \closedint 0 1 : \map H {s, 0} = \map f {s} $

$\forall s \in \closedint 0 1 : \map H {s, 1} = \map {c_{\map f 0}} {s} $

and:

$\forall t \in \closedint 0 1 : \map H {0, t} = \map f 0 = \map f {0} $

$\forall t \in \closedint 0 1 : \map H {1, t} = \map f 1 = \map {c_{\map f 0}} {1} $

Since $c_{\map f 0}$ is a constant loop, from Constant Loop is Loop, $c_{\map f 0}$ is a constant mapping.

We also have the image of $c_{\map f 0}$ is equal to $\set {\map f 0}$.

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By definition of null-homotopy, $H$ is a null-homotopy between $f$ and $c_{\map f 0}$.

Let $\sim$ be the equivalence relation on $\closedint 0 1$ defined by:

\(\ds \forall s_1 \in \openint 0 1 , s_2 \in \closedint 0 1: \, \)

\(\ds s_1 \sim s_2\)

\(\iff\)

\(\ds s_2 = s_1\)

\(\ds \forall s_1 \in \set {0, 1} , s_2 \in \closedint 0 1: \, \)

\(\ds s_1 \sim s_2\)

\(\iff\)

\(\ds s_2 \in \set {0,1}\)

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From Simple Loop in Hausdorff Space is Homeomorphic to Quotient Space of Interval, it follows that $\Img f$ is homeomorphic to $\closedint 0 1 / \sim$.

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Let the homeomorphism between $\closedint 0 1 / \sim$ and $\Img f$ be $h: \closedint 0 1 / \sim \to \Img f$.

Let $h$ be map, defined by:

$\map h {\eqclass s \sim } = \map f s$

where $\eqclass s \sim$ denotes the equivalence class defined by $\sim$.

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It follows that $h$ is a continuous injective mapping.

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Define $H_0 : \paren{ \closedint 0 1 / \sim } \times \closedint 0 1 \to D$ for all $s \in \closedint 0 1, t \in \closedint 0 1$ by:

$\map {H_0} { \eqclass s \sim,t } = \map H {s,t}$

The map $H$ is well-defined, as:

$\map {H}{0,t} = \map {H}{1,t}$

for all $t \in \closedint 0 1$

It follows that:

$\Img {H_0} = \Img H$

From Composite of Continuous Mappings is Continuous, it follows that $H_0$ is continuous.

As $H$ is a null-homotopy between $f$ and $c_{\map f 0}$, it follows that $H_0$ is a null-homotopy between $h$ and the constant mapping:

$c_0: \closedint 0 1 / \sim \to \set { \map f 0 }$

From Closed Real Interval is Compact, it follows that $\closedint 0 1$ is compact.

From Continuous Image of Compact Space is Compact, it follows that $\Img f$, as well as $\closedint 0 1 / \sim$, are compact.

From Topological Product of Compact Spaces, it follows that:

$\paren{ \closedint 0 1 / \sim } \times \closedint 0 1$ is compact.

From Continuous Image of Compact Space is Compact, it follows that $\Img {H_0}$ is compact.

From Euclidean Space is Complete Metric Space, it follows that $\closedint 0 1$ is a metric space

From Compact Subspace of Metric Space is Bounded, it follows that $\Img {H_0}$ is bounded.

From the Jordan Curve Theorem, it follows that $\R^2 \setminus \Img f$ is a union of two disjoint connected components.

These components are the interior $\Int f$, which is bounded, and the exterior $\Ext f$, which is unbounded.

By definitions of bounded and unbounded, it follows that there exists some $\mathbf a \in \Ext f \setminus \Img {H_0}$.

Let $\mathbf b \in \R^2 \setminus \Img {H_0}$.

From Borsuk Null-Homotopy Lemma:Corollary, it follows that $\mathbf a$ and $\mathbf b$ lie in the same component of $\R^2 \setminus \Img f$.

That is, $\mathbf b \in \Ext f$.

It follows that if $\mathbf x \in \Int f$, then $\mathbf x \in \Img {H_0}$.

Hence, $\Int f \subseteq \Img H$.

$\blacksquare$