Hey there! Let's work through this problem step by step—first I'll clear up what the pullback morphism actually is (since you said you're fuzzy on it), then we'll prove the surjectivity for the inclusion map of an affine subvariety.

First, let's ground this in basics. Suppose we have two affine varieties (X \subseteq \mathbb{A}^n) and (Y \subseteq \mathbb{A}^m), plus a morphism (\phi: X \to Y). Remember: a morphism of affine varieties is a map defined by polynomial coordinate functions. So if (Y) uses coordinates (y_1,...,y_m), then (\phi(x_1,...,x_n) = (f_1(x_1,...,x_n), ..., f_m(x_1,...,x_n))) where each (f_i) is a polynomial in (k[x_1,...,x_n]).

The pullback homomorphism (\phi^*: k[Y] \to k[X]) is all about "pulling back" functions from (Y) to (X) using the map (\phi). Here's the precise, intuitive definition:

• For any regular function (g \in k[Y]) (which is just a polynomial function that acts on (Y)), (\phi^(g)) is the function on (X) where (\phi^(g)(x) = g(\phi(x))) for every (x \in X).

To make this concrete: if (g) is a polynomial in (k[y_1,...,y_m]), substituting the coordinate functions of (\phi) into (g) gives a polynomial in (k[x_1,...,x_n]). Since (X) is an affine variety, we restrict this polynomial to (X) to get an element of (k[X]) (which is (k[x_1,...,x_n]/I(X)), where (I(X)) is the ideal of polynomials that vanish everywhere on (X)). So (\phi^*(g)) is just the equivalence class of (g(f_1,...,f_m)) in (k[X]).

Now let's tackle the proof. We have (X) as an affine subvariety of (Y), so (X \subseteq Y) and (\phi: X \to Y) is just the inclusion map—it sends every (x \in X) to itself in (Y). We need to show (\phi^: k[Y] \to k[X]) is surjective (meaning every element of (k[X]) is hit by (\phi^)).

First, recall the ring definitions:

• (k[Y] = k[y_1,...,y_m]/I(Y)), where (I(Y)) is the ideal of polynomials vanishing on all of (Y).
• (k[X] = k[y_1,...,y_m]/I(X)), since (X) lives inside (Y), we can use the same coordinate system for both.

What does (\phi^) do here? For any element (g + I(Y) \in k[Y]), (\phi^(g + I(Y)) = g|_X + I(X))—this is just the restriction of the polynomial (g) to the subvariety (X), treated as an element of (k[X]).

To prove surjectivity, we need to show: for every element (f \in k[X]), there exists some (g \in k[Y]) such that (\phi^*(g) = f).

Here's how we do it:

1. Take any (f \in k[X]). By definition, (f) is a regular function on (X), which means there's a polynomial (h \in k[y_1,...,y_m]) where (f(x) = h(x)) for every (x \in X).
2. Now consider (g = h + I(Y) \in k[Y]). When we apply (\phi^) to (g), we get (\phi^(g)(x) = g(\phi(x)) = g(x) = h(x) = f(x)) for all (x \in X).
3. This means (\phi^*(g) = f) in (k[X]).

In short: every regular function on (X) is just the restriction of some polynomial (which defines a regular function on (Y)) to (X). So the pullback map hits every element of (k[X])—it's surjective.

Quick Example to Solidify This #

Let's take a simple case:

• Let (Y = \mathbb{A}^2) (the affine plane), so (k[Y] = k[x,y]).
• Let (X) be the line (x=0), an affine subvariety of (Y). Then (k[X] = k[y]) (since (I(X) = (x)), the ideal of polynomials vanishing on the line (x=0)).
• The inclusion map (\phi: X \to Y) sends ((y) \mapsto (0,y)).
• The pullback (\phi^: k[x,y] \to k[y]) takes any polynomial (f(x,y)) to (f(0,y))—exactly the restriction to (X). Clearly, every polynomial in (k[y]) is the image of (f(x,y) = 0 \cdot x + f(y)) under (\phi^), so it's surjective. Perfect, this matches our proof!

内容的提问来源于stack exchange，提问作者m. lekk