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Suppose $F$ is an morphism between algebraic variety $V$ and $W$. Prove that the pull back $F^\#$ between the coordinate ring $C[W]$ and $C[V]$ is surjective if and only if the morphism $F$ is an isomorphism between $V$ and some algebraic subvariety of $W$.

This is an exercise from Karen.E.Smith 'An invitation to algebraic geometry.'

I can show without difficulty that if $F^\#$ is surjective, then $F$ is injective. But how to prove that the image of $F$ is a subvariety of $W$? Generally it's not true that the image of an injective morphism is an algebraic variety.

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I will assume varieties are irreducible. Let $I$ be the kernel of $f^\sharp : \mathbb{C}[W] \to \mathbb{C}[V]$. Since $f^\sharp$ is surjective, $I$ must be a prime ideal, so corresponds to some closed subvariety $Y \subseteq W$. Thus the morphism $f : V \to W$ must factor through the inclusion $Y \hookrightarrow W$. We may assume without loss of generality that $Y = W$. But then $f^\sharp$ is an isomorphism, say with two-sided inverse $g^\sharp : \mathbb{C}[V] \to \mathbb{C}[W]$, and the fundamental theorem regarding morphisms of varieties implies the morphism $g : W \to V$ corresponding to $g^\sharp$ must be a two-sided inverse for $f : V \to W$.

You are right that the image of a morphism of affine varieties need not be closed: for example, if $V = \{ (x, y) \in \mathbb{C}^2 : x y = 1 \}$, $W = \mathbb{C}$, and $f(x, y) = x$, then the image of $f$ is a not closed in $W$. But what about $f^\sharp$? Well, $\mathbb{C}[V] = \mathbb{C}[x, y] / (x y - 1)$ and $\mathbb{C}[W] = \mathbb{C}[x]$, so $f^\sharp : \mathbb{C}[V] \to \mathbb{C}[W]$ is not surjective. (In fact, it is injective!) Thus the argument in the previous paragraph does not apply.

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Dear Zhen Lin, I was wondering: if $f:V\rightarrow W$ is a closed morphism can we then conclude that the pullback is surjective? thanks – Heitor Fontana May 31 '13 at 15:16
No. Let $V = \mathbb{A}^1$ and let $W$ be a point. – Zhen Lin May 31 '13 at 15:50

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