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Let $X,Y$ be two irreducible, projective $k$-schemes. $k$ is assumed to be algeraically closed. Consider a dominant morphism $f: X \to Y$ between them which is not an isomorphism!

Let $\mathcal{L}$ be an invertible sheaf on $Y$ and $\mathcal{G}:=f^*\mathcal{L}$ it's pullback. Assume that $\mathcal{G}$ is very ample; that is $\mathcal{G}$ induces a closed embedding $g: X \to \mathbb{P}^n= \operatorname{Proj} \ H^0(X, \mathcal{G})$.

$f^*$ induces a $k$-map $f^*: H^0(Y,\mathcal{L}) \to H^0(X,f^*\mathcal{L})=H^0(X,\mathcal{G})$.

Question: Under assumption that $\mathcal{G}$ is very ample: why the $k$-map $f^*: H^0(X,\mathcal{L}) \to H^0(Y,\mathcal{G})$ cannot be surjective?

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  • $\begingroup$ @wnx: Are you sure? The point that leaded me to this conjecture on non surjectivity is the following: $f$ is assumed not to be a isomorphism. That is we can find an open affine subset $\operatorname{Spec} R = U \subset Y$ such that the restriction of $f$ to $f^{-1}(U) \to U =\operatorname{Spec} R$ is also a non isomorphism. The advantage is now that $U$ is affine and that is the restriction $f$ to $f^{-1}(U) \to U$ is conpletly determined by corresponding ring map $r_U: R \to O_X(f^{-1}(U))=f_*O_X(U)$. As $f$ is dominant and $Y$ is reduced, $r_U$ is injective. $\endgroup$
    – user267839
    Commented Mar 16, 2020 at 13:06
  • $\begingroup$ As $f$ is dominant and $Y$ is reduced, $r_U$ is injective. By assumption $r$ is not an isomorphism and thus not surjective. Now we can tensor $r_U$ by $\mathcal{L}(U)$ and obtain $r_U \otimes id_{\mathcal{L}(U)}: R \otimes_R \mathcal{L}(U) = \mathcal{L}(U) \to f_*O_X(U) \otimes \mathcal{L}(U)= \mathcal{G}(f^{-1}(U))$ (last one is projection formula). This map is cannot also be sujective. Take into account that $r_U \otimes id_{\mathcal{L}(U)}= f^*_U$. Fetch a local section $s_U \in \mathcal{G}(f^{-1}(U))$ which is not in image of $r_U \otimes id_L$. $\endgroup$
    – user267839
    Commented Mar 16, 2020 at 13:06
  • $\begingroup$ As by assumption $\mathcal{G}$ is ample we can lift $s_U$ to a global section which is by construction not in the image of $f^*: H^0(X,\mathcal{L}) \to H^0(Y,\mathcal{G})$. Two questions occure now: Is my approach correct? If not what I have done wrong? Since if it would be correct that would imply that the map isn't surjective and it would contradict your statement. Otherwise I have done somewhere in my approach an error. $\endgroup$
    – user267839
    Commented Mar 16, 2020 at 13:07

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If the map $f^* \colon H^0(Y,\mathcal{L}) \to H^0(X,f^*\mathcal{L})$ is surjective then there is a commutative diagram $$ \require{AMScd} \begin{CD} X @>>> \mathbb{P}(H^0(X,f^*\mathcal{L})^\vee) \\ @VfVV @VVV \\ Y @>>> \mathbb{P}(H^0(Y,\mathcal{L})^\vee), \end{CD} $$ where the right vertical arrow is an embedding and the bottom arrow is a priori rational. But since $f$ is proper and dominant, it is surjective, hence the bottom arrow is regular. Moreover, it follows that $f$ is a closed embedding. But then $f$ is an isomorphism.

EDIT. If you want to check the surjectivity of $H^0(Y,\mathcal{L}) \otimes \mathcal{O}_Y \to \mathcal{L}$ at point $y \in Y$, choose $x \in X$ such that $y = f(x)$, and consider the commutative diagram $$ \require{AMScd} \begin{CD} H^0(Y,\mathcal{L}) @>>> \mathcal{L}_y \\ @V f^* VV @| \\ H^0(X,f^*\mathcal{L}) @>>> (f^*\mathcal{L})_x. \end{CD} $$ The left vertical arrow is surjective by assumption, and the bottom arrow is surjective, because $f^*\mathcal{L}$ is very ample. Therefore, the top arrow is surjective as well.

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  • $\begingroup$ One question on your argument why rational $Y \to \mathbb{P}(H^0(Y,\mathcal{L})^\vee)$ becomes regular: that's clear pure set theoretically: $f$ is surjective, so we can assign to every $y \in Y$ the image of arbitrary $x \in f^{-1}(y)$ under composition $X \to \mathbb{P}(H^0(X,f^*\mathcal{L})^\vee) \to \mathbb{P}(H^0(Y,\mathcal{L})^\vee)$ simply by "going another way around". Why that (at first glace) pure set theoretical assignment gives rise for regular $Y \to \mathbb{P}(H^0(Y,\mathcal{L})^\vee)$ mapping as morphism or category of schemes? $\endgroup$
    – user267839
    Commented Mar 16, 2020 at 13:56
  • $\begingroup$ Because regularity is equivalent to surjectivity of the morphism $H^0(Y,\mathcal{L}) \otimes \mathcal{O}_Y \to \mathcal{L}$, which, by Nakayama lemma, is enough to check only at closed points. $\endgroup$
    – Sasha
    Commented Mar 16, 2020 at 15:43
  • $\begingroup$ Yes, I understand. But why the surjectivity of $H^0(Y,\mathcal{L}) \otimes \mathcal{O}_Y \to \mathcal{L}$ (=regularity, as you said) follows from surjectivity of $f$? I not see it. $\endgroup$
    – user267839
    Commented Mar 17, 2020 at 10:12
  • $\begingroup$ @user7391733: I included an explanation to this question in the answer. $\endgroup$
    – Sasha
    Commented Mar 17, 2020 at 10:43
  • $\begingroup$ Thank you very much. A side remark: with same conditions as above for $X,Y$ and $f$ we can moreover say that $f^* \colon H^0(Y,\mathcal{L}) \to H^0(X,f^*\mathcal{L})$ is always injective. what we need I think is that $Y$ is reduced & $f$ dominant $\endgroup$
    – user267839
    Commented Mar 18, 2020 at 15:58

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