Show that the sphere, S, and $\mathbb{R}^2$ is not homeomorphic

Question

I am trying to show that the sphere $S^2$ and $\mathbb{R}^2$ are not homeomorphic.I understand that you can't 'compress' a 3D shape into a 2D plane but I don't know how I would express this formally.

$S^2 = \{(x, y, z) ∈ \mathbb{R}^3: x^2 + y^2 + z^2 = 1\}$

As always, any help is appreciated!

Answer 1

Homeomorphism will preserve any "topological" property of spaces - in particular, $S^2$ is compact and $\mathbb R^2$ is not, so they can't be homeomorphic.

In fact, the image of a compact space under a continuous map is compact, so there is not even a surjective continuous map $S^2 \to \mathbb R^2$.

Answer 2

If $\;f:S^2\to\Bbb R^2\;$ is a homeomorphism then $\;f:S^2\setminus\{x_0\}\to\Bbb R\setminus\{f(x_0)\}\;$ also is, yet the sphere minus any point is simply connected whereas the plane minus any point isn't.

Or as above but with "contractible" instead of simply connected.

Answer 3

If they were homeomorphic, then they have the same homotopy groups $\pi_k$ for every $k\ge0$. Notice that $\mathbb{R}^2$ is contractible whereas $S^2$ is not. In particular $\pi_2(\mathbb{R}^2)$ is trivial whereas $\pi_2(S^2)\cong\mathbb{Z}$. So they are not homeomorphic.