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.

  • 1
    $\begingroup$ An open disc on $S^2$? Do you mean the intersection of an open ball with $S^2$? $\endgroup$
    – FShrike
    Jan 25 at 9:21
  • $\begingroup$ Which metric are you working with on $S^2$? $\endgroup$ Jan 25 at 9:31
  • $\begingroup$ @FShrike I think this is what was meant. $\endgroup$ Jan 25 at 11:02
  • $\begingroup$ 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. $\endgroup$
    – freakish
    Jan 25 at 13:07

1 Answer 1



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 $\frac{\sqrt{1-\rho^2}}{1- \rho}$.


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