# Is the open half-disc homeomorphic to an open disk?

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 $\mathbb{Z}$.

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.
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}
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 $h\circ g \circ f.$