Two very short proofs, mostly topological, that a nonconstant polynomial map $f:{\bf C} \to \bf C$ is surjective (joint work with Robert Palais):

**(1)** Complex analysis shows that $f$ is an open map (images of open sets are open). A standard estimate, $|f(z)|\to\infty$ as $|z|\to\infty$, implies$f$ is also a closed map (images of closed sets are closed). Thus $f(\bf C)$ is an open, closed, nonempty subset of the connected space $\bf C$, therefore $ f(\bf C)=\bf C$.

The openness of $f$ is nontrivial, but it can be replaced by elementary algebra and topology:

**(2)** The set $K$ of roots of $f'$ is finite. The inverse function theorem shows that the set $A:=f({\bf C})\setminus f(K)$ is open, with finite boundary $A'=f(K) \setminus A$ because $f$ is closed. Thus $A$ has closure $\bar A = f({\bf C})$. Since a finite set cannot disconnect the plane, $\bar A = \bf C$.

A nice feature of these proofs is that they have straightforward (but not trivial) generalizations to higher dimensions:

**Theorem**: Every nonconstant, closed, holomorphic map between connected, complex *n*-dimensional manifolds, is surjective.