Let $f:D=\overline{D(a,r)}\subset \mathbb{C} \to \mathbb{C}$ be a continuous function whose image of $D'=\{z \ | \ |z-a|=r\}$ is homeomorphic to circle $S^1$.

Then I guess the image of $f$ should be in the interior of $f(D')$. But I couldn't prove it.

I tried to find a counterexample. A almost counterexample is $f:\overline{D(0,1)}-\{0\}\to \mathbb{C}$ defined by $z \to \frac{1}{z}$. It sends a punctured disc $D(0,1)-\{0\}$ to unbounded region. But the domain is not disc.

My guess is true? If it is wrong, please suggest any counter example.

Thank you very much!

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    $\begingroup$ Let $f$ map the point with polar coordinates $(r,\phi)$ to $(2-r,\phi)$ if $r\geq1/2$ and to $(3r,\phi)$ if $r\leq1/2$. $\endgroup$ – Andreas Blass Aug 1 '19 at 21:52
  • $\begingroup$ I'm confused... are you asking if a continuous function can send a bounded domain to an unbounded domain, or are you concerned about the image of the boundary of a closed disk? What, precisely, is your question? $\endgroup$ – Xander Henderson Aug 1 '19 at 21:53
  • $\begingroup$ Dear Blass, thank you. Your example is what I am expecting to. $\endgroup$ – user29422 Aug 2 '19 at 4:57
  • $\begingroup$ Just play around with the example $\tfrac{1}{z}$ by shifting around the $z$ variable and scaling appropriately so it blows up along the boundary. $\endgroup$ – Brevan Ellefsen Aug 3 '19 at 4:17

Firstly, $\overline{D(a,r)}$ is compact, so the image of it by $f$ is compact, and consequently bounded.

Secondly, $f(\overline{D(a,r)})$ needs not to be bounded by $f(D')$ (nevertheless it is possible to show that $f(\overline{D(a,r)})$ must contain the bounded component of $C \backslash f(D')$). Intuitively, the image of $f$ might "go out and go in".

For instance it is possible to construct $f : \overline{D(0, 2)} \rightarrow C$ such that $f$ sends $\overline{D(0,1)}$ to $\overline{D(0,2})$ by multiplication by $2$, $\overline{D(0,2)} \backslash D(0,1)$ to $\overline{D(0,2)}\backslash D(0,1)$ by the natural inversion (i.e $z\rightarrow 2\frac{z}{|z|^2}$). Thus $f(D')$ is the circle of radius $1$, and $f(\overline{D(0, 2)})$ goes a little out of this circle (it goes to the circle of radius 2).

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  • $\begingroup$ I think $f(D’)$ is the circle of radius of 2 in your instance. $\endgroup$ – user29422 Aug 2 '19 at 4:48

Consider $$ h(x, t) =\begin{cases} x & x < 0\\ \min(xt, 1) & x \ge 0 \end{cases} $$ for $t = 1$, this is the identity. For $t = 2$, it sends all of $[1/2, 1]$ to $1$. It sends the interval $[-1, 1]$ to $[-1, 1]$, no matter what value of $t\ge 1$ you pick.

Now define $$ f(x, y) = (h(x, 2 - |y| )\sqrt{1 - y^2}, y) $$ That sends the unit disk to the unit disk. But the entire segment $[0.5, 1] \times \{0\}$ gets sent to $(1, 0)$, so the interior of the disk does not map to the interior of the disk.

Yeah, this is needlessly complicated, but it was the first thing I thought up.

Here's the second thing I thought up:

In polar coordinates, send $(r, \theta)$ to $(\min(2r, 1), \theta)$.

That maps the unit disk to itself continuously, but the annulus between $r = 0.5$ and $r = 1$ is all sent to the unit circle in the codomain, hence NOT in the interior of the image.

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