52

Is it possible to construct an embedding $D^4\hookrightarrow S^2\times \mathbb R^2$ such that the projection $D^4\to S^2$ is an open map?

Here $D^n$ denotes closed $n$-ball.

An open map D⁴ → S². It is easy to construct an embedding $D^3\hookrightarrow S^3$ such that its composition with Hopf fibration $f_3:D^3\to S^2$ is open.

Composing $f_3$ with any open map $D^4\to D^3$, one gets an open map $f_4:D^4\to S^2$.

The map $f_3$ is not a projection of embedding $D^3\hookrightarrow S^2\times\mathbb R$. (We have $f_3^{-1}(p)=S^1$ for some $p\in S^2$ and $S^1$ can not be embedded in $\mathbb R$.)

I still do not understand if one can present $f_4$ as a projection of an embedding $D^4\hookrightarrow S^2\times\mathbb R^2$.

ε-δ
  • 1,785
  • Related: http://mathoverflow.net/questions/24748/hopf-fibration-inside-the-retraction-of-r4-minus-line-s2 – Ryan Budney Feb 06 '11 at 22:33
  • @Ryan: The induced map $S^3\to S^2$ is contractible; in particular it can not be Hopf fibration. – ε-δ Feb 06 '11 at 22:49
  • 11
    Do you know any open map from $D^4$ to $S^2$? – Sergei Ivanov Feb 07 '11 at 13:31
  • How do you construct this embedding of $D^3$ in $S^3$? The boundary 2-sphere of $D^3$ cannot be transverse to fibers everywhere, so at some point it must be tangent, and it seems to me that the composition is not open there. – Bruno Martelli Feb 11 '11 at 15:21
  • 2
    @Bruno: just take a hemi-sphere as your $D^3$. The composition of inclusion $D^3 \to S^3$ with the Hopf fibration $S^3 \to S^2$ is an open map in this case. – Ryan Budney Feb 11 '11 at 16:21
  • @Ryan: I disagree. If you "just take a hemi-sphere as your $D^3$", it's not an open map (see Bruno's argument). – André Henriques Mar 06 '11 at 13:06
  • 3
    Here's an aside that might be instructive: There is no open map from $D^2$ to $S^1$. The reason is that every map $D^2\to S^1$ lifts to $\mathbb R$. Take the point in $D^2$ that realizes the maximum. The map isn't open around that point. Moral: the non-existence argument uses Morse theory. – André Henriques Mar 06 '11 at 13:16
  • 4
    @André: you are wrong, it is open map for any disc of radius $\ge\pi/2$ in $S^3$. the projection map $D^3\to S^2$ can not be open if orbit is tangent to the boundary from outside, but it is OK to be tangent from inside. – ε-δ Mar 11 '11 at 00:11
  • 3
    I stand corrected. – André Henriques Mar 17 '11 at 17:54
  • 1
    It's quite a question! I reduce it to "if a map $S^3\to S^2$ is null-homotopic (i.e. extends to D^4), then can it be open?" Maybe the regularity that one asks should be made precise: does "map" mean continuous? or smooth ($C^\infty$)? Let us say "smooth". Do you admit certain generecity hypotheses? An underlying problem is that "open" is less than "rank 2". I've arrived at the following preliminary question. Take a connected open subset $U\subset\R^n$ and a smooth open map $f:U\to\R^n$. Can the Jacobian of $f$ be $>0$ at some point and $<0$ at some other point? (I can answer only for $n=1$!) – Gael Meigniez Mar 25 '21 at 16:18

1 Answers1

2

This is too long for a comment but maybe it'll help me clarify what you're looking for. Interpret $D^4$ as the unit compact ball in $\mathbb C^2$.

$$ D^4 = \{ (z_1,z_2) \in \mathbb C^2 : |z_1|^2+|z_2|^2 \leq 1 \} $$

There is a function $f : D^4 \setminus \{(0,0)\} \to S^2$ given by $f(z_1,z_2) = z_2/z_1$, where we're thinking of $S^2$ as the Riemann sphere. $f$ is an open map. But $f$ is also a composite:

$$ D^4 \setminus \{(0,0)\} \to S^2 \times \mathbb C \to S^2 $$

the 1st map $D^4 \setminus \{(0,0) \} \to S^2$ being $(z_1,z_2) \longmapsto (z_2/z_1, z_2)$ and the second map being $(z_1,z_2) \longmapsto z_1$.

The first map is an embedding, and the 2nd map is a projection. My original post had the domain as $D^4$ but that makes no sense. Okay, maybe I'm starting to wrap my head around the question you're asking. This map above has the property that the restriction $S^3 \to S^2$ is the Hopf fibration, which as a map $S^3 \to S^2$ is not null-homotopic. So it's impossible to extend the above construction to a map $D^4 \to S^2 \times \mathbb C$. If you don't leave the world of submersions this means your map $S^3 \to S^2$ has to be a null-homotopic submersion, but such things do not exist -- a submersion would have to be a circle bundle over $S^2$ and those are only Hopf fibrations.

So if there is a positive answer to your question, the map $S^3 \to S^2$ has to have have some degeneracies.

Ryan Budney
  • 43,013
  • 5
    What is f(0,0)? – Sergei Ivanov Feb 11 '11 at 00:34
  • 9
    That's problematic, isn't it... – Ryan Budney Feb 11 '11 at 01:06
  • Thank you. I was trying to modify your construction to get the embedding I want. The map $(z_1,z_2)\mapsto(z_2/z_1,z_2)$ is a locally embedding only for $z_2\not=0$. So one might look at embedding of $D^4$ in $\mathbb C^2\backslash \mathbb C$, but the projection does not cover a point in $S^2$, so it can not be open... – ε-δ Feb 14 '11 at 19:09