1
$\begingroup$

Is there a well known homeomorphism between the open unit half-disk and the open unit disk (all in $\mathbb{R}^2$)?

Intuitively all we need to do is "double" the open unit half-disk, but this is just me guessing with inspiration from the fact that $\mathbb{N}$ is homeomorphic to $\mathbb{Z}$.

$\endgroup$
2
$\begingroup$

Let $(x, y) \in D^2$ be a point in the open unit disc. Then $$ \left(x, \frac{y + \sqrt{1-x^2}}{2}\right) $$ is in the upper half of the open unit disc. To go the other way, if $(x, y)$ is in the upper half of the unit disc, then $$ (x, 2y-\sqrt{1-x^2}) $$ is in the unit disc.

Added intuition: What i've basically done is divide the circle into lines parallel to the $y$-axis. Then I've raised every line so that the lower edge of the unit circle lies on the $x$-axis (i.e. I've added, to the $y$ coordinate, the distance between the $x$-axis and the lower edge). This moves every point on the upper edge of the circle to twice its height, so I've divided the height by $2$ again to take it back down.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

We can use complex analysis!

Suppose $G = \{ z \in \mathbb{C} \mid |z|<1 \ \ \text{ and } \ \ \operatorname{Im}(z)>0\}$, i.e. the upper half disk.

Consider the conformal mapping $f_1(z) = i(z^{-1}+z)$, then we see \begin{align} f_1(G) = \{ z \in \mathbb{C} \mid \operatorname{Re}(z)>0\} \end{align} which is the right half plane. Now, consider the conformal mapping given by \begin{align} f_2(z) = \frac{z-1}{z+1} \end{align} then we see \begin{align} f_2(f_1(G)) = \{ z \in \mathbb{C} \mid |z|<1\}. \end{align}

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

The map $f(re^{it}) = [r/(1-r)]e^{it}$ takes the half disc to the upper half plane. The map $g(x,y) = (x,\ln y)$ takes the upper half plane to the whole plane. The map $h(re^{it}) = [r/(1+r)]e^{it}$ takes the whole plane to the disc. So the map $h\circ g \circ f$ maps the half disc to the disc. All of the component maps are diffeormorphisms, hence so is $h\circ g \circ f.$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.