Are surjective homogeneous maps open at zero?

Question

I'm asking this question as a follow-up inspired by this one: An open mapping theorem for homogeneous functions?

I'm actually wondering whether there exists an homogeneous map $f:\mathbb R^n\to\mathbb R^m$ that is surjective but not open at zero.

The question came up to my mind since the converse is true, i.e. an homogeneous map open at zero is clearly surjective, and I cannot find any counterexample to the counterpart of this statement up to now. I posted some time ago a related question Boundedness of the preimage of sphere via homogeneous polynomials that actually constituted the main hole towards a proof of this fact.

As a matter of fact it is fairly easy to prove that an homogeneous map open at zero that has non trivial zeros is surjective, but I feel that this assumption is too restrictive: at least in the case of quadratic maps it is proved here in Lemma 2 that if $n$ is big enough then surjectivity implies the existence of nontrivial zeros.

Thanks in advance for any kind of help.

Gil

Answer 1

Consider the set $$Q:=\Big(B\big((0,0),1\big)\setminus(0,\infty)^2\Big)\cup B\big((\frac12,0),\frac12\big)\subset \mathbb R^2.$$ (Three quarters of an Euclidean disk of radius one, plus one half of an Euclidean disk of radius one half). Note that it is a star-shaped set, it is not a neighbourhood of the origin, but its trace on every line through the origin, is a relative neighbourhood of the origin, so that $\bigcup_{t>0}tQ=\mathbb R^2$. The boundary of $Q$ is a simple closed curve, so we can choose a parametrisation $\partial B(0,1)\to \partial Q$ and extend it to a homogeneous map $f:\mathbb R^2\to \mathbb R^2$. Then $f$ is continuous, positively homogeneous, surjective, not open at the origin.

For a fully homogeneous map, one may replace $Q$ with the symmetric set $$R :=\{(x,y)\in B((0,0),1), xy\le0\}\cup B((1/2,0),1/2)\cup B((-1/2,0),1/2)), $$ whose boundary is a symmetric closed figure-eight curve, and extend an odd parametrisation of the boundary it to a fully $1$-homogeneous map.

Hints for a possible polynomial example

Consider a symmetric curve $S$ like this

That is the set $[0,1] \cdot S$, like the above set $R$, is not a nbd of the origin, yet $\mathbb R \cdot S$ is the whole plane $\mathbb R^2$. If we can make such a symmetric curve $S$ by a Cartesian parametrisation on $[0,2\pi]$ given by two trigonometric polynomials, $X(t)=P(\cos t,\sin t ),$ $Y(t)=Q(\cos t,\sin t)$, where $P(x,y)$, $Q(x,y)$ are homogeneous polynomials of the same degree (in fact it would be sufficient they have all their terms of even degree, or all terms of odd degree, since we may multiply each term by some power of $x^2+y^2$), then the map $(x,y)\mapsto (P(x,y), Q(x,y))$ would be a surjective non-open homogeneous polynomial map $f$ such that $f(\partial B(0,1)=S$. Note that for a mop $f$ of even degree, it is not required that $S$ be symmetric.