Are there Zariski-continuous maps between algebraic sets that are not polynomial maps? Suppose that $S_1$ and $S_2$ are the vanishing sets of a system of polynomial equations in n variables over a field $\mathbb{k}$ (ideal in $\mathbb{k}[X_1,\dots,X_n]$) and a system of polynomial equations in m variables over a field $\mathbb{k}$ (ideal in $\mathbb{k}[Y_1,\dots,Y_m]$). We can give $S_1$ and $S_2$ the Zariski topology by letting all algebraic subsets of $S_1$ and $S_2$ to be closed.
Surely, if $Y_j\circ f\colon S_1\to\mathbb{k}$ is a polynomial in $\mathbb{k}[X_1,\dots,X_n]/I(S_1)$, for every $j$, then $f$ is continuous in the Zariski topology.
The condition that $Y_j\circ f$ be polynomial, however, seems too strict: as long as the vanishing set of $Y_j\circ f$ were equal to the vanishing set of a polynomial function, $f$ would be continuous, which means that the usual morphisms from $S_1$ to $S_2$ are not the same as the continuous maps between them as (Zariski-)topological spaces.
My question then is of two parts.

  
*
  
*Is it true that there exist Zarsiki-continuous maps that are not polynomial? 
  
*If yes, then what is the geometric way of thinking about the extra structure that polynomial maps preserve, i.e. continuous maps squash subsets of $S_1$ in such a way that non-algebraic subsets never get squashed into algebraic ones, but some algebraic ones may get squashed into non-algebraic ones, but polynomials also give what features to the geometric squashing?
  

 A: Just an easy remark. If $k = \mathbb{F}_q$ is the finite field with $q$ elements, then every function $k^n \to k$ is a polynomial map. In fact, for every point $p = (a_1, \dots, a_n) \in k^n$ consider the polynomial
$$
f(x_1, \dots, x_n) = (-1)^n \prod_{i=1}^n \prod_{b \in \mathbb{F}_q \setminus \{ a_i \} } (x_i - b);
$$
$f(p) = 1$ and $f$ is zero in $k^n \setminus \{ p \}$.
A: Yes, there exists Zariski continuous maps between algebraic varieties that are not polynomials. For example all  bijection between $\mathbb A ^1$ and itself is Zariski continuous but not polynomial in general: think of bijection that permutes $0$ and $1$ and leaves other points fixed.
Geometric way of thinking is algebraic subsets must be preserved. Only use polynomials and rational functions.
A: Consider the affine line and Look for an example there yourself.
A: If I understand your question correctly, then (1) yes, there are going to be lots of Zariski continuous maps that are not polynomial (although, if one asks that all higher products of the map preserve the Zariski topology on the corresponding sources and targets, this is a much stronger restriction, I think --- does anyone know the details?); and (2) the usual extra structure that one adds is the sheaf of regular functions on source and target; polynomial maps pullback regular functions on the target to regular functions on the source.  Thus varieties are naturally thought of as being objects
in the category of locally ringed spaces (i.e. spaces equipped with a certain structure sheaf of rings) rather than just in the category of topological spaces.
A: The complex conjugation $\bar \cdot : \Bbb C \to \Bbb C$ is a concrete example of a Zariski-continuous function that is not a polynomial function.
It is not a polynomial function because if it were, since this property is independent of the underlying topology, $\bar \cdot$ would also be a polynomial function in the transcendental topology (the "usual" topology on $\Bbb C$), therefore it would also be holomorphic - which it clearly is not, since it doesn't satisfy the Cauchy-Riemann relations.
To see that it is continuous, notice that if $Z = f^{-1}(\{0\})$ with $f \in \Bbb C[x]$, and if $\bar f$ is the polynomial obtained by conjugating the coefficients of $f$, then
$$\bar \cdot ^{-1} (Z) = \bar \cdot (Z) = \{ \bar x \in \Bbb C \mid f(x) = 0\} = \{ \bar x \in \Bbb C \mid \overline{f(x)} = 0\} = \{ \bar x \in \Bbb C \mid \bar f(\bar x) = 0\} = \\
\{ z \in \Bbb C \mid \bar f(z) = 0\} = \bar f ^{-1} (\{0\})$$
which is Zariski-closed.
If $C$ is an arbitrary Zariski-closed set, then there exist polynomials $f_1, \dots, f_k$ such that
$$C = f_1 ^{-1} (\{0\}) \cap \dots \cap f_k ^{-1} (\{0\}) ,$$
whence it follows that
$$\bar \cdot ^{-1} (C) = \bar \cdot ^{-1} \Big( f_1 ^{-1} (\{0\}) \cap \dots \cap f_k ^{-1} (\{0\}) \Big) = \bar f_1 ^{-1} (\{0\}) \cap \dots \cap \bar f_k ^{-1} (\{0\})$$ which, being an intersection of Zariski-closed sets, is itself Zariski-closed.
Since the preimage of every Zariski-closed set is Zariski-closed, it follows that $\bar \cdot$ is Zariski-continuous.
A: There are continuous functions between algebraic sets that are not polynomial functions, the simpliest case that I have in mind is regular maps. For instance, take two algebraic sets $V$ and $W$ and a regular map globally defined as a quotient of polynomials. This is not always the case for a general regular map between algebraic sets on $\mathbb{C}$ but if you think real every regular function between real algebbraic sets $V$ and $W$ can be written as a global quotient of two polynomials such that the denominator never vanishes on $V$, as an application of the sum of squares. Thus, in the real world the previous setting is completely general for regular functions. Finally, it is very simple to verify that such functions are continuous with respect to the Zariski topology. If $f^{-1}(0)$ is an algebraic subset of $W$ (every real algebraic set is given by a unique equation by the sum of squares) then 
\begin{equation*}
\left(f\circ\frac{p}{q}\right)^{-1}(0)=(q^n(f\circ p))^{-1}(0)
\end{equation*} 
where $n$ is the degree of $f$.
