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If $\pi :C\to C'$ is an affine $\mathscr O$-connected morphism,then by definition,pull back map $\mathscr O_{C'}(U)\to \mathscr O_C(\pi^{-1}(U))$ is an isomorphism for every affine $U$ of $C'$,since $\pi^{-1}(U)$ is affine,this means $\pi|_{\pi^{-1}(U)}$ is an isomorphism,so $\pi$ is and isomorphism,right?But it seems that we need some big machines to do this,like:

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coming from vakil's FOAG,Page736.Here $\pi$ is finite,hence affine by definition.So what I miss?

schuming
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    I think perhaps Vakil made a mistake. If $\pi:X\to Y$ being $\mathcal{O}$-connected means that $\mathcal{O}Y\to \pi\ast\mathcal{O}X$ is an isomorphism then if $\pi$ is affine we know that $X$ is the relative spectrum $\underline{\mathrm{Spec}}(\pi\ast\mathcal{O}_X)$ which is equal to $\underline{\mathrm{Spec}}(\mathcal{O}_Y)$ which is just $Y$. Then again, I haven't really looked into the context of the exercise so maybe we're missing something. – Alex Youcis Sep 12 '20 at 06:04
  • You can also prove this from just a gluing argument. Can you provide more context about the exercise? For instance, if Zariski's connectedness lemma has just been proven, it may be the case that the exercise is intended to give you practice with this. – KReiser Sep 12 '20 at 07:33

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