Are there any known criteria for quadratic mapping from R^n to R^n being surjective? Let quadratic mapping be the function from $\mathbb{R}^n$ to $\mathbb{R}^n$, where each coordinate is a quadratic form of $n$ variables. Are there any known criteria for it being surjective? May somebody give relevant references too?
 A: If $n=2$ the quadratic map $\mathbb{R}^2\to\mathbb{R}^2$ with $(x_1,x_2)\mapsto (x_1^2-x_2^2,2x_1x_2)$ is surjective. This follows because the map $\mathbb{C}\to\mathbb{C}$ with $z\mapsto z^2$ is surjective. Hence there exist
real quadratic maps $\mathbb{R}^{2m}\to\mathbb{R}^{2m}$ for all even values
of $n$. (Identify $\mathbb{R}^{2m}$ and $\mathbb{C}^m$ and consider $(z_1,\dots,z_m)\mapsto(z_1^2,\dots,z_m^2)$.) For $n=1$ a quadratic
map $\mathbb{R}\to\mathbb{R}$ is not surjective. The question can thus be rephrased:
for which values of $m\geq1$ is there a real quadratic surjective map $\mathbb{R}^{2m+1}\to\mathbb{R}^{2m+1}$?
A: Some comments.
Check this question and the comments.
A paper from the answer

Proposition 6. No algorithm is possible that, given a polynomial mapping
  $f : \mathbb{R}^n \to \mathbb{R}^n$ with computable coefficients, decides whether this mapping is
  surjective.

On the other hand another answer gives relatively efficient criterion
for deciding if the map is bijective over $\mathbb{Q}$ (this implies surjective).
A: I do not know the answer in general but, just for fun, I tried to analyze the small dimensions. The results call for an obvious conjecture. This may be known, but I haven't seen this in the literature.
A quadratic map (given by homogeneous quadratic forms)
$\varphi\colon\mathbb{R}^{n+1}\to\mathbb{R}^{n+1}$ is essentially the same as an
$n$-dimensional linear system $L_\varphi$ of quadrics in $\mathbb{R}\mathrm{p}^n$.
We also have the projectivization, which is a rational map
$\bar\varphi\colon\mathbb{R}\mathrm{p}^n\to\mathbb{R}\mathrm{p}^n$. For the sake
of simplicity, let us assume that $L_\varphi$ has no basepoints, so that
$\bar\varphi$ is regular (we do not need to blow up anything).
The first observation is that, assuming $n\ge1$, the original map $\varphi$ is
surjective if and only if so is $\bar\varphi$. Indeed, we can restrict/normalize
$\varphi$ to the unit spheres, and the key is the fact that the double covering
$S^n\to\mathbb{R}\mathrm{p}^n$ is nontrivial.
Here is the common approach to the description of the topology of $L_\varphi$
(after Dixon, Agrachev, etc.) The spectral variety is the set
$C\subset L_\varphi\cong\mathbb{R}\mathrm{p}^n$ of singular quadrics. Typically,
it is a hypersurface of degree $n+1$ (possibly nonreduced),
although in some very degenerate cases it may coincide with $L_\varphi$.
Passing from quadrics to quadratic forms (and normalizing), we
get a nontrivial double
covering $S^n\to L_\varphi$ and the pull-back $\tilde C\subset S^n$ of $C$. On
the complement $S^n\setminus\tilde C$ one has the index function, sending a
point to the negative inertia index of the corresponding nondegenerate quadratic form. It
takes values between $0$ and $n+1$, is symmetric in an appropriate sense, and has a lot of
other nice properties that I will skip here.

Theorem.
  A necessary condition for
  a quadratic map $\varphi$ to be surjective is that the index
  function $\iota\colon S^n\setminus\tilde C\to\mathbb{Z}$
  should not take value $0$. If $n\le2$ and $\varphi$ is generic, this
  condition is also sufficient.

Proof
The necessity follows from the fact that the members of $L_\varphi$ are the
pull-backs of hyperplanes in $\mathbb{R}\mathrm{p}^n$, and quadrics of
index $0$ are empty.
The case $n=0$ is trivial: $\varphi$ is never surjective and $\iota$
always takes value $0$.
The case $n=1$ is also easy. Indeed, there are two projective classes of
generic pencils, with the spectral curve $C\subset L_\varphi$ empty or not.
In the former case, $\varphi$ is surjective; in the latter case, it is not
and $\iota$ does take value $0$. (I can refer to my paper, but the fact is
trivial and well known.)
Let $n=2$. A point $p$ in the target $\mathbb{R}\mathrm{p}^n$ defines a
pencil $\ell\subset L_\varphi$, and the pull-back $\varphi^{-1}(p)$ is the
base locus of $\ell$. Generic pencils of plane conics admit a simple
projective classification (see my paper again); they are characterized by
their basepoints. The index function of
the only pencil with all four basepoints non-real does take value $0$; hence,
such pencils do not occur if $\iota>0$.
QED

Remark.
  Taking care of a few technicalities, I believe that the proof can be adjusted
  to non-generic linear systems. The big question is whether the sufficiency
  holds as well in higher dimensions.

A: I'll consider the question posed in Glasby's answer.   Take $n=3$ and $\pi(x,y,z)=( (x-1)^2 - y^2, 2(x-1)y, xz )$.  To  see that $(u,v,w)$ lies in the image, note that there is always at least one pair $(x_0,y_0)$ such that $((x_0-1)^2-y_0^2,2(x_0-1)y_0) = (u,v)$ and $x_0 \neq 0$.   To see this note that $(0,y_0) \mapsto (1-y_0^2,-2y_0)$ and $(2,-y_0) \mapsto (1-y_0^2,-2y_0)$.
Thus $\pi(x_0,y_0,w/x_0) = (u,v,w)$.
Thus there exist surjective quadratic maps for all $n \neq 1$.  This doesn't really answer the original request for a criterion though. 
A: Similar questions were studied in the paper
A V Arutyunov, S E Zhukovskii, "Properties of real surjective quadratic mappings", SB MATH, 2016, 207 (9), DOI:10.1070/SM8611 .
In particular, for $n=3$ there are necessary and sufficient conditions for a quadratic mapping to be surjective. The paper is written in Russian. Probably it will be translated this year. If it is interesting, I can write a short summary.
A: A necessary condition is that if the mapping is $Q:x\mapsto(q_1(x),\cdots,q_n(x))$ where $x=(x_1,\cdots,x_n)\in\mathbb{R}^n$, then $q_1,\cdots,q_n$ are not simultaneously diagonalizable. 
In fact if not after a change base, we may assume that $Q(x_1,\cdots,x_n)=A\begin{bmatrix}x_1^2\\ \vdots\\ x_n^2\end{bmatrix}$, where $A\in M_n(\mathbb{R})$. But this map is obviously not surjective.
Using above we get the following necessary and sufficient condition for the case $n=2$.

$Q=(q_1,q_2)$ is surjective iff $q_1$ and $q_2$ are both regular and not simultaneously diagonalizable. 

(The example of Glasby manifests this).
