Jordan curve; is the inside mapped to the inside by a homeomorphism?

Question

Suppose that

• $\gamma:[0,1]\to \mathbb{R}^2$ is a closed simple curve;
• $U$ is an open set of $\mathbb{R}^2$ that contains both the image $|\gamma|$ of $\gamma$ and the inside $D$ of $\gamma$;
• $f:U\to V$ is a homeomorphism, where $V$ is an open set of $\mathbb{R}^2$.

Is it true that the inside of the closed simple curve $f\circ\gamma $ is $f(D)$?

Intuitively, it seems true. But I can neither prove nor disprove it. Can anyone answer this question? Any help is appreciated.

Answer 1

$U\setminus |\gamma|$ may have several components $C$ (it's not given that $U$ is connected), but only one with the property that $C\cup|\gamma|$ is compact, namely $C=D$. The image of $D$ under $f$ must have the same property with respect to $V$, so $f(D)$ is the interior of $f(\gamma)$.