Homeomorphism between Sphere and $[0,1]^2$

Question

Following this question: Sphere homeomorphic to plane?

I understand that a sphere is not homeomorphic to the plane because the sphere is compact and the plane is not. But why is the sphere not homeomorphic to $[0,1]^2$?

This question interests me since if these two were homeomorphic, map projections would not be necessary. All previous mathematical answers I've seen to this question rely on the non-compactness of the plane, whereas the paper and screens maps are actually projected on in real-life are in fact compact.

Answer 1

Here are some ways to see that $S^2$ and $[0,1]^2$ aren't homeomorphic:

• $[0,1]^2$ is contractible while $S^2$ isn't. Indeed a contraction of $[0,1]^2$ is easy to write down, while $S^2$ is not contractible since it has a non-trivial second homology group.

• If there would exist a homeomorphism $\phi : [0,1]^2 \to S^2$ and if $x$ is an interior point of $[0,1]^2$ then $\phi$ would still be a homeomorphism $[0,1]^2 \setminus \{ x \} \to S^2 \setminus \{ \phi(x) \}$, where both spaces have the subspace topology. But now $[0,1]^2 \setminus \{ x \}$ has a nontrivial fundamental group, while the fundamental group of $S^2 \setminus \{ \phi(x) \}$ is still trivial.

• Take a look at this question if you want a proof which doesn't use algebraic topology (by showing that $S^2$ is not contractible without homology or cohomology).

As for your other point (if I understand it correctly), note that maps of the globe do not give homeomorphisms $[0,1]^2 \to S^2$ since on world maps there are no points corresponding to the north pole or the south pole. (That's one issue - another is that the inverse of a world map is not continuous.)

Answer 2

If they were homeomorphic, say by a homeomorphism by $f:S^2\rightarrow [0,1]^2$, then also are homeommorphic $S^2\setminus\{N\}$ and $[0,1]^2\setminus \{f(N)\}$. However, we know that $S^1\setminus\{N\}$ is homeomorphic to $\mathbb{R}^2$, so it is simply connected, but $[0,1]^2\setminus\{N\}$ is not simply connected, then they can not be homeomorphic.

Answer 3

$S^n$ is homogeneous, that is for $x, y\in S^n$ there is a homeomorphism $h:S^n\to S^n$ with $h(x) = y$. Indeed, from linear algebra we know that for each $x, y\in S^n$ there is an isometry $A$ sending $x$ to $y$ (for any two orthonormal bases $e_1, ..., e_n$ and $f_1, ..., f_n$ there is an isometry $A$ with $Ae_i = f_i$. Simply take $e_1 = x, f_1 = y$ and extend to orthonormal bases).

But $[0, 1]^n$ is not homogeneous, since points of the "interior" $(0, 1)^n$ and the "boundary" $[0, 1]^n\setminus (0, 1)^n$ can't be mapped to each other by a self-homeomorphism (here $n\geq 1$). But curiously, the Hilbert cube $[0, 1]^\omega$ is homogeneous.

Another reason why the two aren't homeomorphic is that $[0, 1]^n$ has the fixed point property, but $S^n$ doesn't, since the map $x\mapsto -x$ doesn't have a fixed point.

Answer 4

No, $[0,1]^2$ is contractible and $\mathbb{S}^2$ isn't.