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Let $f: X\rightarrow Y$ be a morphism of affine varieties and $f^*: A(Y)\rightarrow A(X)$ the corresponding homomorphism of the coordinate rings. The question is whether this is true or false:

$f$ is injective if and only if $f^*$ is surjective.

The "only if" part is false. Here is a counterexample:

$$X=\mathbb{A^1}, Y=V(x^2-y^3)\\ f: X\rightarrow Y, t \rightarrow (t^3,t^2)\\ f^*: A(Y)\rightarrow A(X), (\bar{x},\bar{y})\rightarrow (t^3,t^2)$$ In this example $f$ is bijective, but $f^*$ is not surjective, since it does not map anything to $t$.

I cannot prove the "if" part or construct a counterexample of it. Thanks for any help!

KittyL
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    Well, without loss of generality, we may assume $Y = \mathbb{A}^n$ for some $n$. Do you agree? – Zhen Lin Mar 06 '15 at 16:04
  • @ZhenLin: Thank you for your reply! Are you saying the statement is false and use $Y=\mathbb{A}^n$ as a counterexample? If it is true, why can we just assume $Y=\mathbb{A}^n$? – KittyL Mar 06 '15 at 16:14
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    If $f^$ is surjective, then $f$ is a closed embedding ($X \cong \mathrm{Spec}(A(Y)/\ker f^)$), in particular injective. – Jake Levinson Mar 06 '15 at 17:26
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    @KittyL The claim is correct. Since $Y$ is affine, you may embed it in some $\mathbb{A}^n$. The claim is that $X$ embeds in the same $\mathbb{A}^n$, and moreover $X \subseteq Y$ under this embedding. – Zhen Lin Mar 06 '15 at 17:41

1 Answers1

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Assume $f^\ast$ is surjective and let us prove that $f$ is injective. Let $x_1,x_2\in X$ be two points with $y:=f(x_1)=f(x_2)$. Assume that $x_1\ne x_2$, then there is some function (the restriction of some coordinate function, say) $\varphi\in A(X)$ with $\varphi(x_1)\ne\varphi(x_2)$. However, $f^\ast$ is surjective, so there is some $\psi\in A(Y)$ with $f^\ast(\psi)=\varphi$. This implies $$\psi(y)=\psi(f(x_1))=f^\ast(\psi)(x_1)=\varphi(x_1)\ne\varphi(x_2)=f^\ast(\psi)(x_2)=\psi(f(x_2))=\psi(y),$$ which is a blatant contradiction. We must have had $x_1=x_2$.

Jesko Hüttenhain
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  • Of course, you can easily turn this into a direct proof, which is then more elegant. But it is late, and I will leave you with this. – Jesko Hüttenhain Mar 07 '15 at 00:36
  • Is it true also that $f$ must be an isomorphism? And that's the reason that the injective $f$ in the OP does not produce a surjective $f^*$? – JKEG Aug 31 '20 at 03:11
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    Dear @JKEG, I don't quite understand your follow-up question, but I suggest you ask it as an actual question (if you can't find an existing question here that is equivalent), and then leave a link here to that question. – Jesko Hüttenhain Sep 06 '20 at 16:14