When are polynomial maps open or closed? 1.$f:\mathbb{R} \rightarrow \mathbb{R}$, $f(x)=x^4+x^3$ is a closed map(or an open map)?
2.$f:\mathbb{R} \rightarrow \mathbb{R}$, $f(x)=x^5+x^4$ is a closed map?
3.$f:\mathbb{R}^2 \rightarrow \mathbb{R}$, $f(x,y)=x^2+y^2$ is a closed map?
Generally, $f:\mathbb{R} \rightarrow \mathbb{R}$, $f(x)=a_nx^n+...+a_1x+a_0$ is a closed map or an open map?
Generally, if $f:\mathbb{R}^n \rightarrow \mathbb{R}$ is a multivariable polynomials, $f$ is a closed map or an open map?
We have known that $x^2$ is a closed map but not an open map, $x^3+x$ is a homeomorphism, $x^3+2x$ is a closed map but not a homeomorphism.
Thank you very much.
 A: This is a small addition to Eric Wofsey's wonderful answer. Here is a useful and often easily checked criterion for properness of a polynomial map.
Proposition. Let $\Bbbk\in \left\{\mathbb R,\mathbb C \right\}$ and let  $\Bbbk^n\overset{f}{\to} \Bbbk^\ell$ be a polynomial mapping. Suppose the top degree homogeneous summand of each component has the property that its only root is the origin. Then $f$ is proper.
Here is a proof.
A: Any polynomial in one variable is a closed map.  Indeed, suppose $f:\mathbb{R}\to\mathbb{R}$ is a nonconstant polynomial and $C\subset\mathbb{R}$ is closed, and $y\in\overline{f(C)}$.  Then we can choose a sequence $(x_n)$ of points of $C$ such that $(f(x_n))$ converges to $y$.  In particular, the set $\{f(x_n)\}$ is bounded, which implies that the set $\{x_n\}$ is bounded (since as $|x|\to\infty$, $|f(x)|\to\infty$--this is the one place where we are using that $f$ is a nonconstant polynomial).  Passing to a subsequence, we may thus assume that $(x_n)$ converges to some $x\in\mathbb{R}$.  Since $C$ is closed, $x\in C$.  Since $f$ is continuous, $f(x)=\lim f(x_n)=y$.  Thus $y\in f(C)$.  Since $y\in\overline{f(C)}$ was arbitrary, this means $f(C)$ is closed.
This fails in general for polynomials in more than one variable.  For instance, let $f(x,y)=xy$ and let $C\subset\mathbb{R}^2$ be the set of points of the form $(x,1/x^2)$ for $x>0$.  Then $C$ is closed, but $f(C)=(0,\infty)$ is not closed.  I'm not aware of any general criterion for a polynomial in more than one variable to be a closed map, but I would expect that most of them aren't.
Very generally, if $X$ is a locally connected space, then a continuous map $f:X\to\mathbb{R}$ is open iff it has no local minima or maxima.  Indeed, if $x\in X$ is a local minimum of $X$, then there is some neighborhood $U$ of $x$ such that $f(y)\geq f(x)$ for all $y\in U$, and then clearly $f(U)$ contains no open interval around $f(x)$ (and similarly for local maxima).  Conversely, suppose $f$ has no local minima or maxima and let $U\subseteq X$ be open.  For any $x\in U$, let $V\subseteq U$ be a connected open neighborhood of $x$.  Since $x$ is neither a local minimum nor a local maximum of $f$, there exist $y,z\in V$ such that $f(y)<f(x)<f(z)$.  By connectedness of $V$ and continuity of $f$, it follows that $f$ must achieve every value in the interval $[f(y),f(z)]$ on $V$.  Thus $f(U)$ contains an open interval $(f(y),f(z))$ around $f(x)$.  Since $x\in U$ was arbitrary, this means $f(U)$ is open.
When $X=\mathbb{R}^n$ and $f$ is a polynomial, the existence of local minima or maxima can be tested using methods from calculus.  Note that in particular, for $n=1$, if $f$ has even degree, it must always have either a global minimum or a global maximum, so it is never open.  For $n=1$, if $f$ has no local minima or maxima, then it is monotone, and hence actually a homeomorphism $\mathbb{R}\to\mathbb{R}$.
