Question : Does there exist a surjective function $F$ that maps $\mathbb{R}^n\_+$ to $\mathbb{R}^n$ (where $\mathbb{R}^n\_+$ denotes the set of vectors of length $n$ with only positive entries). The answer is yes by considering the function $F$:
$$(x\_1,\ldots,x\_n)\to(\log x\_1,\ldots,\log x\_n)$$
It is easy to see that $F$ is surjective.


Now my real question is if there is such a function $F$ whose components are restricted to be polynomials of degree at most $n$ (and $F$ maps $\mathbb{R}^n\_+$ to $\mathbb{R}^n$ and is surjective). For example:
$$(x\_1,x\_2,x\_3,x\_4)\to(1-x\_1,x\_1\cdot x\_2+x\_3,x\_3^2+x\_2-x\_4,5x\_4)$$
has polynomial components, but is not surjective as the restriction $x\_i>0$ implies we cannot get a $1$ in the first coordinate.


Any help or advice is greatly appreciated. I think the answer is no but am not sure how to prove it.



Vector valued functions
Complete Version
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This is motivated by the complex cubing map described below, which generalizes to the cubing map on quaternions: 


For $n > 1$, the map analogous to cubing a complex number or a quaternion is
f\_3: $ (x\_1, x\_2, \dots, x\_n) \rightarrow 
(x\_1^3 -3 x\_1( x\_2^2 + x\_3^2 + \dots + x\_n^2),
-x\_2^3 + 3 x\_2 x\_1^2, 
 \dots, - x\_n^3 + 3 x\_n x\_1^2)$.
This map takes any plane through the $x\_1$ axis to itself, and acts like the complex cubing map in such a plane.


Therefore, the image of the cone $C\_{60}$ within a 60 degree angle of the $x\_1$ axis is all of $\mathbb R^n$.


There are linear transformations that take the positive orthant to nearly all of a half-space, in particular, there are linear images that contain $C\_{60}$.


Composing, we get a surjective homogeneous degree 3 polynomial map from the positive orthant to all of $\mathbb R^n$.


The dimension 1 case is impossible with any degree polynomial, as discussed in Richard Palais's response, and the degree 2 cases cannot be done with degree < 3, as discussed below.


NEW VERSION
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For n = 2, if we consider a homogeneous quadratic polynomial, it acts on lines through the origin as a rational map of degree 2. This is either a double cover map of $S^1$ to itself, equivalent to the complex map $z \rightarrow z^2$ , or it folds the circle in half to an interval. The only chance it has to be surjective is the double cover case. The image of any interval under the complex squaring map may be surjective on the circle of lines, but when lifted to its double cover, the circle of rays, its image is an interval, since there are lines which are only hit once. Therefore, no homogeneous degree 2 polynomial is surjective.


Adding a linear or constant term doesn't change the limiting action on rays unless the homogeneous part is degenerate and is 0 on some line, as for instance 
$(x,y) \rightarrow (0, y^2)$ (up to linear transformations in the domain and range).
Adding lower degree terms obviously can't make this surjective. So, for $n=1$ or $2$, the answer is **NO**.


There are degree 3 polynomial maps of $\mathbb R^2$ to itself that take the positive quadrant to the whole plane: first, compress the plane to fatten the quadrant to make an angle more than 120 degrees, then cube it as a complex number.


In even dimension $2m \ge 4$, a solution in complex coordinates, as noted by fedja in comments, is 
$(z\_1, \dots, z\_m) \rightarrow (z\_1^4, \dots, z\_m^4)$, and a solution of degree 3 is obtained by performing the degree 3 map above coordinatewise.


For odd dimensions $2m+1 \ge 7$, a solution of degree 6 can be obtained as the degree 3 solution on $2m$ coordinates followed by the complex squaring map on the last coordinate with any of the first $2m$. The degree of the composition is 6.


This leaves dimensions 3 and 5 unanswered. Maybe with a little more cleverness?...


Earlier Version
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> Note, added: I was hasty reading the question, and didn't pay attention to degree restrictions. The comment of fedja to the question took this into account, and answered it as well as this. The missing cases at the moment seems to be $n = 3, 5, 7$.
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In the complex plane, the map $z \rightarrow z^4$ maps the positive quadrant surjectively to $\mathbb C$. Expressed in terms of real and imaginary parts, this satisfies your question for dimension $n = 2$.


You can get a positive answer for any $n \ge 2$ by composing copies of this function applied to pairs of coordinates (the pairs can overlap).
For dimension $n = 1$, it is impossible


