# Homeomorphism of open disks on a sphere

Are open disks defined on the unit sphere $$S^2$$ homeomorphic to open disks on $$\mathbb{R}^2$$? I know that the unit sphere is a 2D manifold, but that tells me the (seemingly weaker?) point that any point on the sphere has an open neighbourhood that is homeomorphic to a Euclidean open disk. Can we force that neighbourhood to be an open disk on $$S^2$$?

Context: I am trying to understand a proof that $$\mathbb{R}\mathbb{P}^2$$, defined as the quotient of a unit sphere by antipodal images, is a manifold. It involves taking a closed disk around a point $$p \in S^2$$, noting that $$S^2$$ is compact, and so the quotient map restricted to this disk is actually a homeomorphism. Then we restrict it to the open disk that is the interior.

• An open disc on $S^2$? Do you mean the intersection of an open ball with $S^2$? Jan 25 at 9:21
• Which metric are you working with on $S^2$? Jan 25 at 9:31
• @FShrike I think this is what was meant. Jan 25 at 11:02
• You don't need this neighbourhood to be a disc (whatever that means on $S^2$). The conclusion for $\mathbb{R}\mathbb{P}^2$ holds if you take an open neighbourhood $U$ of $p$ such that $U$ is homeomorphic to $\mathbb{R}^2$ and $\overline{U}$ doesn't contain antipodal points. Such neighbourhood always exists. Jan 25 at 13:07

## 1 Answer

Yes.

An open disk in $$S^2$$ is a set of the form $$D(\xi,r) = S^2 \cap B(\xi,r)$$, where $$\xi \in S^2$$ and $$r \le 2$$. Here $$B(\xi,r) = \{\eta \in \mathbb R^3 \mid \lVert \eta - \xi \rVert < r\}$$ is the open ball with center $$\xi$$ and radius $$r > 0$$. Of course $$\lVert - \rVert$$ denotes the Euclidean norm.

Note that $$D(\xi,r) \subset S^2 \setminus \{-\xi\}$$. Equality holds for $$r = 2$$.

W.l.o.g. we may assume that $$\xi$$ is the south pole $$S = (0,0,-1)$$. This is possible because we can find an orthogonal linear automorphism $$\phi$$ on $$\mathbb R^3$$ mapping $$\xi$$ to $$S$$. By orthgonality we get $$\phi(S^2) = S^2$$ and $$\phi(B(\xi,r)) = B(S,r)$$, thus $$\phi(D(\xi,r)) = D(S,r)$$.

Now take the stereographic projection from the north pole $$N = (0,0,1)$$

$$s : S^2 \setminus \{N\} \to \mathbb R^2, s(x_1,x_2,x_3) = \frac{1}{1-x_3}(x_1,x_2) .$$

It is well-known that $$s$$ is a homeomorphism. You can easily check that $$\phi(D(S,r))$$ is an open disk . It is $$= \mathbb R^2$$ for $$r = 2$$; for $$r < 2$$ just observe that the boundary of $$D(S,r)$$ has the form $$S^2 \cap E(\rho)$$, where $$E(\rho) = \{ (x_1,x_2, \rho) \mid x_1, x_2 \in \mathbb R\}$$ is the hyperplane $$x_3 = \rho$$ for some $$\rho \in (-1,1)$$. The set $$S^2 \cap E(\rho)$$ is a circle in $$E(\rho)$$ with center $$(0,0,\rho)$$ and radius $$\sqrt{1-\rho^2}$$. It is mapped by $$s$$ to a circle in $$\mathbb R^2$$ with center $$(0,0)$$ and radius $$\frac{\sqrt{1-\rho^2}}{1- \rho}$$. Thus $$D(S,r)$$ is mapped by $$\phi$$ to the open disk with center $$(0,0)$$ and radius $$\frac{\sqrt{1-\rho^2}}{1- \rho}$$.