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 $\mathbb{Z}$.
 A: 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.
A: 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}
A: 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 $h\circ g \circ f.$
