Prove that the helicoid is homeomorphic to the plane.
Good!
Here are a few abstract thoughts.
We know the universal covering space of the punctured plane is the plane , with the projection .
On the other hand, the universal cover of is also the Riemann surface of the (multi-valued) inverse function , 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 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.)