Polynomial $g:\mathbb R^n \rightarrow\mathbb R^n$ with no critical point may have no root

Question

Version 1 (solved): If $g$ : $\mathbb R^n \rightarrow \mathbb R^n$ is a polynomial, $Dg(x)$ is non-degenerate for every $x$, then there exists $x$, such that $g(x)=0$.
Version 2: If $f$ : $\mathbb R^n \rightarrow \mathbb R$ is a polynomial, $D^2f(x)$ is non-degenerate for every $x$, then $f(x)$ has at least one critical point.
The problem is how to prove or disprove Version 1. The result of Version 2 is what I need to prove another problem.

Answer 1

The answer to question 1 appears to be "no". Pinchuk, in "A counterexample to the strong real Jacobian conjecture" found a polynomial map $g: \mathbb{R}^2 \to \mathbb{R}^2$ such that $\det(Dg)$ is nonvanishing, yet $g$ is not a diffemorphism. Campbell and Gwoździewicz (independently) computed the image of Pinchuk's map and found that it omitted $(0,0)$.

I say apparently because their computations don't seem to agree! Campbell finds that the omitted points are $(0,0)$ and $(-1,-163/4)$; Gwoździewicz gets $(0,0)$ and $(-1,0)$. It's probably just a notational issue, but one should probably check their work carefully before relying on it.

Still thinking about question 2. Given the complexity of Pinchuk's example, this seems likely to be hard.

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A few thoughts, no real progress, on the second question. If $D^2 f$ is nonsingular for all inputs, then the signature of $D^2 f$ is constant. The easiest case to think about is when $D^2 f$ is positive definite, in which case $f$ is convex. In this case, I will show that there is a point with $Df=0$.

It is well known that, when $f$ is convex, the image of $Df$ is a convex subset of $\mathbb{R}^n$. (So, if it is well known, why can't I find a reference? I'm sure I read this in a textbook on the multivariate Legendre transform, but I can't figure out which one.) So, if $0 \not \in (Df)(\mathbb{R}^n)$, then by Farkas' lemma there is a vector $\vec{v}$ such that $\langle \vec{v}, \cdot \rangle$ is positive on $(Df)(\mathbb{R}^n)$. So $f$ is strictly increasing on lines in direction $\vec{v}$. But the only convex strictly increasing polynomials are linear functions, so $f$ is linear on every line in direction $\vec{v}$.

Changing coordinates to assume that $\vec{v}$ is parallel to the $z_n$ axis, we deduce that $f$ is of the form $g(z_1, \ldots, z_{n-1}) z_n + h(z_1, \ldots, z_{n-1})$, for some $g$ which is always positive. If $D^2 g \neq 0$ at some point $(z_1,\ldots, z_{n-1})$, then either $D^2 f$ is non-positive-definite at $(z_1, \ldots, z_{n-1}, M)$ for $M$ very positive, or for $M$ very negative, contradicting our claim that $D^2 f$ is always positive definite. So $D^2 g=0$ and we conclude that $g$ is affine linear. But $g$ is always positive, so $g$ is a constant. In short, $f=z_n+h(z_1,\ldots, z_{n-1})$. But then the $n$-th column of $D^2 f$ vanishes, and $D^2 f$ is not invertible after all.

So there are no counter-examples with $D^2 f$ positive definite (or negative definite). But I never really thought there would be.

I'd love to see an analysis of the case of signature $+-$ on the plane. There definitely exist $f \in \mathbb{R}[x,y]$ with $D^2 f$ of signature $+-$ everywhere: Just take the real part of some $h(z) \in \mathbb{C}[z]$. But, by the fundamental theorem of algebra, $h'(z)$ has zeroes, so $D(\mathrm{Re}(h))$ vanishes in places.

Answer 2

Building on David Speyer's answer, and as the problem (version 2) seems hard, I venture a (nonconclusive) argument for the case where $D^2f(x)$ has signature $+-$, on the plane. (I was meaning this as a comment, but it was too long...)

We probably can select smooth vector fields $e^+(x)$ and $e^-(x)$ in the positive and negative (orthogonal) directions (eigenvectors of $D^2f(x)$). This defines $X(s,t)\in\mathbb R^2$, $s,t\in\mathbb R$, with $\partial_s X=e^+(X)$ and $\partial_t X=e^-(X)$ (plus $X(0,0)=0$ for definiteness). Now $f(X(s,t))$ is strictly convex in $s$ and strictly concave in $t$, isn't it?

This doesn't prove $f$ has a saddle point (example $\mathrm{Re} (e^z)$), but maybe polynomials have something more that allows a "mountain pass lemma" to apply ?

EDIT That seemed plausible, but no: maybe there exists $X(s,t)$ such that $f(X(s,t))$ is convex in $s$ and concave in $t$, but it cannot satisfy $\partial_s X=e^+(X)$ and $\partial_t X=e^-(X)$, because the necessary integrability condition $(e^-\nabla) e^+=(e^+\nabla) e^-$ is (generically) incompatible with the orthogonality $e^+\cdot e^-=0$ (if I'm not mistaken again...)