Open map D⁴ → S² 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  $D^4\hookrightarrow S^2\times\mathbb R^2$.
 A: 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. 
