Simple Loop Image Equals Set Homeomorphic to Circle

Theorem
Let $\struct { X, \tau_X }$ be a Hausdorff space.

Let $C \subseteq X$ be a subset of $X$.

Let $\tau_C$ be the subspace topology on $C$ induced by $\tau_X$.

Let $\mathbb S^1$ denote the unit circle whose center is at the origin of the Euclidean space $\R^2$.

Let $\tau_{\mathbb S^1}$ be the subspace topology on $\mathbb S^1$ induced by the Euclidean topology on $\R^2$.

Then $\struct{ C, \tau_C }$ is homeomorphic to $\struct { \mathbb S^1, \tau_{\mathbb S^1} }$, there exists a simple loop $\gamma : \closedint 0 1 \to X$ with image equal to $C$.

Sufficient condition
Let $\sim$ be the equivalence relation on the closed real interval $\closedint 0 1$ defined by:

Let $q : \closedint 0 1 \to \closedint 0 1 / \sim$ be the canonical surjection induced by $\sim$.

Let $\tau_q$ be the quotient topology on the quotient space $\closedint 0 1 / \sim$ induced by $q$.

By definition of quotient topology, $\tau_q$ is the identification topology on $\closedint 0 1 / \sim$ by the identification mapping $q$.

Identification Mapping is Continuous shows that $q$ is continuous.

Parameterization of Unit Circle is Simple Loop shows that there exists a simple loop in $\R^2$ with image equal to $\mathbb S^1$.

Euclidean Space is Complete Metric Space and Metric Space is Hausdorff shows that $\R^2$ is a Hausdorff space.

Simple Loop in Hausdorff Space is Homeomorphic to Quotient Space of Interval shows that there exists a homeomorphism $f: \closedint 0 1 / \sim \to \mathbb S^1$.

By assumption, there exists a homeomorphism $h: \mathbb S^1 \to C$.

Composite of Continuous Mappings is Continuous shows that the mapping:


 * $h \circ f \circ q : \closedint 0 1 \to C$

is continuous.

By definition of homeomorphism, $h$ and $f$ are bijections.

Quotient Mapping is Surjection shows that $q$ is a surjective mapping.

Composite of Surjections is Surjection shows that $h \circ f \circ q$ is a surjective mapping, so $\Img {h \circ f \circ q} = C$.

By definition of $\sim$, we see that:


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

It follows that $h \circ f \circ q$ is a simple loop, as:


 * $\map {h \circ f \circ q }{t_1} \ne \map {h \circ f \circ q}{t_2}$ for all $t_1 ,t_2 \in \hointr 0 1$ with $t_1 \ne t_2$
 * $\map {h \circ f \circ q} 0 = \map {h \circ f \circ q} 1$

Define $\gamma : \closedint 0 1 \to X$ by $\map \gamma t = \map {h \circ f \circ q} t$.

Continuity of Composite with Inclusion/Inclusion on Mapping shows that $\gamma$ is continuous.

It follows that $\gamma$ is a simple loop.

Necessary Condition
By assumption, there exists a simple loop $\gamma : \closedint 0 1 \to X$ with $\Img \gamma = C$.

Simple Loop in Hausdorff Space is Homeomorphic to Quotient Space of Interval shows that there exists a homeomorphism $g : \closedint 0 1 / \sim \to C$.

As shown above, there exists a homeomorphism $f: \closedint 0 1 / \sim \to \mathbb S^1$.

Composite of Homeomorphisms is Homeomorphism shows that $f \circ g^{-1} : C \to \mathbb S^1$ is a homeomorphism.