5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @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
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.