Suppose that $X$ is a sub affine variety of $Y$ , and let $φ : X \to Y$ be the inclusion. Prove that $φ^*$ is surjective...

Question

where $\varphi^*$ is the pullback homomorphism $k[Y] \to k[X]$

If you could also give an explanation for as to what the pullback morphism is, I'd really appreciate it; my understanding is somewhat shaky.

Answer 1

$k[X]$ is just the set of polynomial functions on $X$. I am assuming you are using affine varieties. So an element of $k[X]$ is represented by a polynomial function $p(x)$. The map $\varphi^*:k[Y]\to k[X]$ is defined by $$\varphi^*(p)(x)=p(\varphi(x))$$ So basically what it is saying is that any polynomial function on $X$ comes from some polynomial function on $Y$. This is obvious since polynomials are globally defined.

Another point of view is that $k[X]=k[x_1,\ldots ,x_n]/I(X)$ and you have $I(Y)\subseteq I(X)$, now the pullback is

$$k[x_1,\ldots ,x_n]/I(Y)\to k[x_1,\ldots ,x_n]/I(X)$$ and this is clearly surjective.