I'm struggling with the following question:
Let $\gamma$ be a (not necessarily smooth) closed Jordan curve and let $\operatorname{int} \gamma, \operatorname{ext}\gamma$ denote the interior and the exterior respectivelly.
Is it true that $\gamma$ is homotopic to some circle in $\operatorname{ext} \gamma$? More precise:
Is there some continuous mapping
$$
H: \left[ 0,1 \right] \times \left[ 0,1 \right] \longrightarrow \operatorname{ext} \gamma
$$
such that $H(0, t) = \gamma(t)$ and $H(1,t) = Re^{it}$ for all $t \in \left[ 0,1 \right]$ and some large $R >0$?
If found this related question and it seems to me that the answer to my question is given in the second comment. However I'm not able to elaborate the details.
What I know so far:
By a theorem of Caratheodory the conformal map
$$
\phi : \lbrace \vert z \vert > 1\rbrace \longrightarrow \operatorname{ext} \gamma
$$
extends to a homeomorphism from $\lbrace \vert z \vert \geq 1\rbrace$ to $\overline{\operatorname{ext} \gamma}$.
Putting for example $H(s,t) = \phi((1 + s) e^{it})$ should transform $\gamma$ to some curve in $\operatorname{ext} \gamma$, but I don't know how the proceed.