# Is the closed hemisphere diffeomorphic to the closed disk?

This might be silly but it puzzles me. Let $M$ be the closed upper hemisphere of $\mathbb{S}^2$. This is a manifold with boundary. Is it diffeomorphic to a closed disk in $\mathbb{R}^2$?

These manifolds are definitely homeomorphic, via the projection map $M \to \mathbb{D}^2$: $(x,y,z)\mapsto (x,y)$.

However, this map is not a diffeomorphism. Its inverse $(x,y)\mapsto (x,y,\sqrt{1-x^2-y^2})$ is not differentiable at the boundary.

• Dear @IshanLevy: how is a diffeomorphism to the upper half plane an answer to the OP? Which diffeomorphism between the upper half plane and the upper hemisphere of $\mathbb{S}^2$ did you have in mind? – Peter Heinig Feb 26 '18 at 12:13

## 2 Answers

Whatever meaning you assign to diffeomorphism for manifolds with boundary, you are going to have to admit that the hemisphere and the disk are diffeomorphic: just use the stereographic projection instead of the coordinate projection you mentioned.

This is vertical projection; not differentiable at the edge.

Instead use stereographic projection. Consider the south pole $P=(0,0,-1)$. Project the northern hemisphere to the plane through the equator along the the lines through $P$, i.e., map $Q$ in the northern hemisphere to the point of intersection of $PQ$ with the equatorial plane.