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 $\mathbb{Z}$.
Best Answer
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.