Nonexistence of a continuous injection $f:S^2 \rightarrow \mathbb{R^2}$

What is the "easiest" way to show that there is no continuous injection $f:S^2 \rightarrow \mathbb{R^2}$?

Sure the Borsuk-Ulam theorem implies that result, but this may be a "difficult" way.

• @Juho One can give an easy proof in case $f$ is an immersion – happymath Jul 5 '16 at 15:17
• Can Picard's theorem do this? – user334732 Jul 5 '16 at 15:30
• @RobertFrost I see no connection with Picard's theorem. – Hulkster Jul 5 '16 at 15:35
• What about if you project $\mathbb{R}^2$ onto a sphere the x-axis requires a point at infinity and so does the y-axis and so do every combination of the two. These can't be accommodated on a single point on the sphere. – user334732 Jul 5 '16 at 15:36
• @RobertFrost: Picard's Theorem, first of all, is about complex analytic mappings. If you're going to assume complex analyticity, it will then be far more elementary, even without the injectivity assumption. – Ted Shifrin Jul 5 '16 at 15:38

This is a non-obvious result. By invariance of domain, if there were such an $f$, its image would be open in $\Bbb R^2$. But it would, of course, also be compact. Oops.

• This is also very intuitive result, since it is impossible to draw a one "normal map" of the whole Earth. – Hulkster Jul 5 '16 at 15:38
• How does invariance of domain apply? – Christian Sievers Jul 5 '16 at 16:44
• @ChristianSievers: You get a basis for the topology on the sphere (or, indeed, any manifold) by taking sets homeomorphic to balls in Euclidean space. – Ted Shifrin Jul 5 '16 at 16:47
• Oh, I think you should link to en.wikipedia.org/wiki/Invariance_of_domain#Generalizations – Christian Sievers Jul 5 '16 at 16:56
• So the set $f(S^2)$ is always a countable union of open sets, and therefore open? – Hulkster Jul 6 '16 at 2:03

Consider two points $a,b \in S^2$ and a continuous closed curve $C\in S^2$ such that all coninuous transformations mapping $a\to b$ contain a value on $C$. The existtance of such a triplet is shown by example: The North and South poles, and the equator, on a sphere.
Now consider any continuous injection of that triplet onto some co-domain $D \in \Bbb{R}^2$. (Since the injection is also surjective onto $D$ it is a bijection.) Since every continuous curve line from $a$ to $b$ intersects $C$, and continuous injections preserve that property, exactly one of $a',b'$ must lie in the interior of $C'$. Without loss of generality, say $a'$ is not in the interior of $C'$.
Then $C'$ cannot be continuously deformed to lie completely in an arbitrarily small $\epsilon$-ball containing $a'$ in its interior. And a continous injection would preserve that property. Yet $C$ can be continuously deformed to lie completely in an arbitrarily small $\epsilon$-ball containing $a$ in its interior. This contradictoin shows that we could not have the desired continuous injection.
• I don't understand why exactly one of $a'$ and $b'$ has to lie on the interior of $C'$. – John Gowers Jul 5 '16 at 16:01
• If both are in the exterior or both in the interior, then there is a continuous curve not intersecting $C'$ that goes from $a'$ to $b'$. – Mark Fischler Jul 9 '16 at 4:06