 Solution

Here $D^2$ must denote the closed disc $\{(x,y):x^2+y^2\le 1\}$, as its boundary points participate in the identification. For $r=\sqrt{x^2+y^2}$, defineThe Pythagorean trigonometric identity gives $\Vert F(x,y){\Vert }^2=1$, so the image lies in the sphere $S^2$. Away from the origin the formula is continuous. At the origin, the first two coordinates have absolute value at most $|\sin (\pi r)|\to 0$ and the third tends to $1$, proving continuity there as well.

Every boundary point maps to $(0,0,-1)$, while the origin maps to $(0,0,1)$. For a point $(u,v,w)\in S^2$ with $-1<w<1$, there is a unique $r\in (0,1)$ with $\cos (\pi r)=w$, and its unique preimage isThis point has radius $r$, because $u^2+v^2=1-w^2={\sin }^2(\pi r)$. The two poles have the preimages already described. Thus $F$ is surjective and its fibers are exactly the prescribed equivalence classes.

The closed disc is compact by the Heine-Borel theorem, and $S^2$ is Hausdorff with the subspace topology from Euclidean space. The preceding result gives the homeomorphism