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Let $X$ and $Y$ be smooth projective surfaces over an alg. closed field. Let $f:X\longrightarrow Y$ be a finite morphism of degree 2. Let $C$ be a smooth curve which maps to a smooth $C'$ under $f$. Is it obvious that $\mathcal{O}_X(C)=f^*\mathcal{O}_Y(C')$?

Conversely if $L'$ is a line bundle on $Y$, and if $C\in |f^*L'|$, then is $f(C)\in |L'|$?

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  • $\begingroup$ Dear npna: $k$ is a notation for your algebraically closed base field. But now that you have added indices to $\mathcal O$ your question is perfectly clear. I have deleted my previous comment which is no longer relevant. $\endgroup$ – Georges Elencwajg Nov 4 '15 at 18:52
  • $\begingroup$ Considering these are false even for curves (and hence for surfaces), why don't you first check the curve case to see what goes wrong. $\endgroup$ – Mohan Nov 4 '15 at 20:46
  • $\begingroup$ @Mohan, in case of curves, there is a degree constraint. That is $deg{f^*L'}=2degL'$. I was not sure how to translate it to general surfaces. $\endgroup$ – gradstudent Nov 5 '15 at 7:18
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a) No, it is not true that $\mathcal{O}_X(C)=f^*\mathcal{O}_Y(C')$.
For example take $X=\mathbb P^2_{x:y:z}, Y=\mathbb P^2_{u:v:w}, C=V(x)$ and let $$f(x:y:z)=(u:v:w)=(x^2:yz:z^2)$$ so that $C'= V(u)$.
Then $\mathcal O_X(C)=\mathcal O_{\mathbb P^2}(1)$, whereas $f^*(\mathcal O_Y(C'))=f^*(\mathcal O_{\mathbb P^2}(1))=\mathcal O_{\mathbb P^2}(2)$

b) The same set-up also gives a negative answer to your second question:
Take for $L'$ the line bundle $\mathcal O_{\mathbb P^2}(1)$. Then $f^*(L')=\mathcal O_{\mathbb P^2}(2)$ and you may take for $C$ the conic $z^2=xy$.
Its image $f(C)$ is the curve $C'$ given by $ w^3=uv^2$ which nor all thy Piety nor Wit will lure to $|L'|$ .

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  • $\begingroup$ Since we have $f(0:1:0) = (0:0:0)$, how do you define $f$ at $(0:1:0) \in \mathbb{P}^2$? Does the same argument work if I just use $f(x:y:z) = (x^2:y^2:z^2)$? $\endgroup$ – user113988 Dec 17 '18 at 2:21

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