Induced morphim on Spec

Classical Case

Finding pullback

Question: Suppose $X \subset k^m$ and $Y \subset k^n$ are irreducible algebraic sets (with Zarisky topology). Let $f: X \to Y$ be given by
      $(x_1, \ldots, x_m) \mapsto (f_1(x), \ldots, f_n(x)).$

What is the induced map on coordinate rings?

Answer: The induced maps $f^{\sharp}: A(Y) = k[y_1, \ldots, y_n]/ I  \to A(X) = k[x_1. \ldots, x_m]/ J$ is given by pre-composition with $f$ so it is
   $y_i \mapsto f_i$.

Recovering from Pullback

Question: On the other hand, suppose we have a map $\varphi: A(Y) \to A(X)$. How do we recover $f$ such that $\varphi = f^{\sharp}$?

Answer: $f: X \to Y$ can be defined by
 $(x_1, \ldots, x_m) \mapsto \left(\varphi (y_1)(x), \ldots, \varphi(y_n)(x)\right).$

Affine Scheme Case.      

In the examples below, we first define the maps on points on $\mathbb{C}^n$. Then we interpret the map as the maps on coordinate rings as before. Afterwards, we recover the map on $\text{Spec} \mathbb{C}^n$.

Example. Parabola

Let $f: \mathbb{C} \to \mathbb{C}$ be given by $x \mapsto y = x^2$. Then the induced map on coordinate rings is $f^{\sharp}: \mathbb{C}[y] \to \mathbb{C}[x]$ given by $y \mapsto x^2$.

We claim that the induced map on Spec is

$ f: \text{Spec} \mathbb{C} \to \text{Spec} \mathbb{C}$
$ (x - a) \mapsto (y - a^2)$. 

Indeed $f(x-a) = (f^{\sharp})^{-1} (x-a)$. Now $f^{\sharp}(y - a^2) = (x^2 - a^2) \in (x-a)$ so $(f^{\sharp})^{-1} (x-a) \ni (y- a^2)$, but the latter is maximal. 

 Generalization

Suppose $f: k^m \to k^n$ is given by $x \mapsto (f_1(x), \ldots, f_n(x))$, where $f_1, \ldots, f_n \in k[x_1, \ldots, x_m]$. Then as before, it induces $f^{\sharp}: k[y_1, \ldots, y_n] \to k[x_1, \ldots, x_m]$ be given by

$y_i \mapsto f_i$. 

Note: We could have also just started with $f^{\sharp}.$

For any ideal $I \subset k[x_1, \ldots, x_m]$ and $J \subset k[y_1, \ldots, y_n]$ such that $I \supset \phi(J)$ we have a morphism 

$f: \text{Spec}  k[x_1, \ldots, x_m]/ I \to k[y_1, \ldots, y_n]/J$. 

Claim: $f$ sends points $(x_1 - a_1, \ldots, x_m - a_m)$ to $(y_1 - f_1(a), \ldots, y_n - f_n(a)).$

Proof: As $(y - f_1(a), \ldots, y - f_n(a))$ is a maximal ideal, it suffices to show that it is in the preimage under $f^{\sharp}$ of $(x_1 - a_1, \ldots, x_m - a_m)$. Indeed, 

$$f^{\sharp}(y_1 - f_1(a), \ldots, y_n - f_n(a)) = (f_1(x) - f_1(a), \ldots, f_n(x) - f_n(a)) \in (x_1 - a_1, \ldots, x_n - a_n) $$  as the latter is the kernel of the evaluation at $a$ map. 

Thus we see that the induced map on Spec generalizes the induced map on classical algebraic varieties.