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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 $D^4\hookrightarrow S^2\times\mathbb R^2$.

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Related: mathoverflow.net/questions/24748/… –  Ryan Budney Feb 6 '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 6 '11 at 22:49
Do you know any open map from $D^4$ to $S^2$? –  Sergei Ivanov Feb 7 '11 at 13:31
@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
@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 0:11
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1 Answer

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.

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What is f(0,0)? –  Sergei Ivanov Feb 11 '11 at 0:34
That's problematic, isn't it... –  Ryan Budney Feb 11 '11 at 1:06
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
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