Is the construction of Zariski topology from polynomial rings functorial? 
Given a polynomial ring $k[X_1,...,X_n]$ over a field $k$, we can consider the space $k^n$ equipped with Zariski topology whose closed sets are exactly the algebraic sets. Is this construction functorial ? 

I.e. given two polynomial rings over some fields $k[X_1,...,X_n]$ and $L[X_1,...,X_m]$, and a ring homomorphism $ \phi : k[X_1,...,X_n] \to L[X_1,...,X_m]$, does there exist a continuous map (w.r.t. Zariski topology)  $\phi^* : L^m \to k^n$ such that  the assignment $\mathcal F$ from the category of polynomial rings over some field to the category of topological spaces defined as $\mathcal F (k[X_1,...,X_n])=k^n$ and $\mathcal F(\phi)=\phi^*$ , is a contravariant functor ? If this cannot be done in general, then is there some modification of the underlying category or the topology on the space so that the construction is functorial ? (I want an answer for this particular case of polynomial rings, not any general algebraic geometric answer, as I don't know any general algebraic geometry yet.) 
 A: Not in the way requested, but yes with some mild changes. The easiest mild change is to require that $k = L$ and that we only consider $k$-algebra homomorphisms. Then, given such a homomorphism
$$\phi : k[x_1, \dots x_n] \to k[x_1, \dots x_m]$$
the corresponding map
$$\phi^{\ast} : k^m \to k^n$$
is obtained by precomposing evaluation homomorphisms. That is, you can think of $k^n$ as the set of $k$-algebra homomorphisms $k[x_1, \dots x_n] \to k$, so the functor I'm proposing here is just the functor (co)represented by $k$. This is clearly a functor into sets; it's an exercise to show that with Zariski topologies it's also a functor into topological spaces. 
Alternatively, $\phi$ determines and is determined by $n$ polynomials $\phi(x_1), \dots \phi(x_n) \in k[x_1, \dots x_m]$, and $\phi^{\ast}$ is given very explicitly by applying each of these polynomials in turn to a point in $k^m$. 
If you don't require $k = L$ this recipe no longer works. For example, consider the homomorphism
$$\varphi : \mathbb{R}[x] \ni f(x) \mapsto f(i) \in \mathbb{C}[x].$$
If we attempt to write down a map $\mathbb{C} \to \mathbb{R}$ going the other way, we run into the problem that if we precompose by an evaluation homomorphism $\mathbb{C}[x] \to \mathbb{C}$, we get a homomorphism $\mathbb{R}[x] \to \mathbb{C}$ which is not guaranteed to (and won't, in general) land in $\mathbb{R}$. 
The fix to this is to consider a more complicated object called the spectrum. The spectrum of $k[x_1, \dots x_n]$ is more complicated than $k^n$ (but strictly contains it) as soon as $k$ fails to be algebraically closed, but is necessary to consider for doing modern algebraic geometry in the standard style. The upside is that the construction of the spectrum is functorial with respect to arbitrary homomorphisms between commutative rings. 
In particular, $\varphi^{\ast}$ acts as follows on spectra. $\text{Spec } \mathbb{C}[x]$ consists of the points in $\mathbb{C}$ together with a "generic point" corresponding to the prime ideal $(0)$, which you can think about as associated to the homomorphism $\mathbb{C}[x] \to \mathbb{C}(x)$, and which won't really concern us here. $\varphi^{\ast}$ sends every point $z \in \mathbb{C}$ to a point in $\text{Spec } \mathbb{R}[x]$ corresponding to the prime ideal $(x^2 + 1)$, which you can think about as associated to the homomorphism 
$$\mathbb{R}[x] \ni f(x) \mapsto f(i) \in \mathbb{C}.$$
Geometrically you can think of $\text{Spec } \mathbb{R}[x]$ as, not $\mathbb{R}$, but the orbits of $\mathbb{C}$ under complex conjugation. So it has points corresponding to the fixed points of complex conjugation, which is the expected copy of $\mathbb{R}$, but it also has other points corresponding to the nontrivial orbits; in the above case the orbit is $\{ i, -i \}$. 
