Math Help - Helicoid

1. Helicoid

Prove that the helicoid is homeomorphic to the plane.

2. Originally Posted by Bruno J.
Prove that the helicoid is homeomorphic to the plane.
We have $\begin{cases} x=r\cos(a\theta)\\y=r\sin(a\theta)\\z=\theta \end{cases}$.

Can't we just let $a\to0$ (in a continuous manner to make it a homeomorphism)?

3. Originally Posted by chiph588@
We have $\begin{cases} x=r\cos(a\theta)\\y=r\sin(a\theta)\\z=\theta \end{cases}$.

Can't we just let $a\to0$ (in a continuous manner to make it a homeomorphism)?
So what is the image of $(x,y,z)$ under this map? *

*corrected

4. Originally Posted by Bruno J.
So what is the image of $(x,y,z)$ under this map? *

*corrected
$(x,y,z)\mapsto(r,0,\theta)$ where $r$ and $\theta$ are free variables, so this map is the x-z plane.

5. Originally Posted by chiph588@
$(x,y,z)\mapsto(r,0,\theta)$ where $r$ and $\theta$ are free variables, so this map is the x-z plane.
Good!

Here are a few abstract thoughts.

We know the universal covering space of the punctured plane $\mathbb{C}-\{0\}$ is the plane $\mathbb{C}$, with the projection $\mbox{exp }: \mathbb{C} \to \mathbb{C}-\{0\}$.

On the other hand, the universal cover of $\mathbb{C}-\{0\}$ is also the Riemann surface of the (multi-valued) inverse function $\mbox{Log}$, of which the helicoid is the most obvious model. By the universal property of the universal cover, there exists a homeomorphism between the helicoid and the plane.

(Informally : the slit plane $\mathbb{C}-\mathbb{R}_{\geq 0}$ is easily seen to be homeomorphic to a strip without its boundary, (any branch of the logarithm doing the job). Now lift the cut to the helicoid, and map each slice of the helicoid to a strip, identifying the appropriate boundaries of two strips if the corresponding boundaries of the slices are identified on the helicoid. In this way, the helicoid is mapped homeomorphically on the plane.)