Is $S^2$ homeomorphic to $D^2$?

Question

I'm trying to figure out whether $S^2 := \{ v \in \mathbb R^3, |v| = 1 \} $ is homeomorphic to $D^2 := \{ v \in \mathbb R^2, |v| \leq 1 \}$, the closed unit disk. Intuitively I would say no, since $S^2$ has empty interior as a subset of $\mathbb R^3$ but $D^2$ doesn't as a subset of $\mathbb R^2$. However, I couldn't find anything about the boundary or interior of homeomorphic sets.

Here's an actual attempt to a line of reasoning : a homeomorphism maps open sets to open sets, however $S^2$ does not have any open subset, while $D^2$ does (such as any open ball of radius smaller than 1 centered at the origin), so the two are not homeomorphic.

The part I'm unsure about is that technically, $D^2$ has empty interior as a subset of $\mathbb R^3$ as well. I'm not sure how to address this "is $D^2$ a subset of $\mathbb R^2$ or $\mathbb R^3$" problem.

Answer 1

I'm assuming you don't know about homotopy groups otherwise this problem would be easy. Here is a proof using a result that might show up in a first topology class.

The Borsuk-Ulam theorem states that any for any map $f: S^n \to \mathbb{R}^n$ there exists $x\in\mathbb{R}^n$ so that $f(x) =f(-x)$. Using $n=2$ and replacing the plane with the disk (since the disk is a subset of the plane), the theorem still holds; hence, there are no injective maps from the 2-sphere to the disk, so they cannot be homeomorphic.

Answer 2

To decide whether or not $S^2$ is homeomorphic to $D^2$ you have to regard both sets as separate topological spaces, carrying the subspace topology of $\mathbb{R}^3$ and $\mathbb{R}^2$, respectively. For instance, if $S^2$ carries the subspace topology then it has open sets, although these sets are not open if you consider them as sets in the topological space $\mathbb{R}^3$. Hence your statement "$S^2$ does not have any open subset, while $D^2$ does" is not correct in this setting.

Consider a point $(x,y,z) \in S^2$. If you project this point down onto the $x$-$y$-plane via $(x,y,z) \mapsto (x,y,0)$ then you end up in $D^2$ (by identifying the $x$-$y$-plane in $\mathbb{R}^3$ with $\mathbb{R}^2$). If you now use the equivalence relation $$ (x,y,z) \sim (x,y,-z) $$ then you can see that $S^2 /\sim$ is homeomorphic to $D^2$. See for instance here for a similar discussion.

EDIT:

$S^2$ is a manifold without boundary, while $D^2$ is a manifold with boundary. Hence they are not homeomorphic.

Answer 3

Upon further inspection, it turns out that contractibility is a topological property. $D^2$ is trivially contractible (with homotopy $f : (t, x) \mapsto (1-t)x$) but $S^2$ is not contractible (not trivially ...). As such, $D^2$ and $S^2$ are not homeomorphic.

Answer 4

$\mathbb{S}^2$ is an ANR that is not an AR, while $\mathbb{D}^2$ is an AR.