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.

  • $\begingroup$ @Juho One can give an easy proof in case $f$ is an immersion $\endgroup$ – happymath Jul 5 '16 at 15:17
  • $\begingroup$ Can Picard's theorem do this? $\endgroup$ – user334732 Jul 5 '16 at 15:30
  • $\begingroup$ @RobertFrost I see no connection with Picard's theorem. $\endgroup$ – Hulkster Jul 5 '16 at 15:35
  • $\begingroup$ 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. $\endgroup$ – user334732 Jul 5 '16 at 15:36
  • $\begingroup$ @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. $\endgroup$ – 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.

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

A broad-strokes proof outline:

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.

  • 1
    $\begingroup$ I don't understand why exactly one of $a'$ and $b'$ has to lie on the interior of $C'$. $\endgroup$ – John Gowers Jul 5 '16 at 16:01
  • $\begingroup$ 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 $b'$. $\endgroup$ – Mark Fischler Jul 9 '16 at 4:06

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