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I'm trying to understand what does inverse exceptional image $f^!$ of a coherent sheaf look like.

Let's say for the sake of this question that $f:X\rightarrow Y$ is a finite flat morphism between schemes. In this case, my understanding is that $f^!$ is a functor from coherent sheaves on $Y$ to coherent sheaves on $X$, and that if $\mathcal{F}$ is a coherent sheaf on $Y$ then $f^!\mathcal{F}$ can be described as $\underline{Hom}_{\mathcal{O}_Y}(f_*(\mathcal{O}_X),\mathcal{F})$.

If $X$ and $Y$ are affine then it's not hard to work out what this does for a coherent sheaf on Y.

As another basic case, I'm trying to understand what happens if $\mathcal{L}$ is a line bundle on $Y$, in terms of divisors. Say $D=\sum{P_i}$ is a divisor on $Y$, and $\mathcal{L}(D)$ is the corresponding line bundle. My understanding is that if $f$ is etale then $f^! = f^*$, so that in that case $f^!(\mathcal{L}(D))=\mathcal{L}(D')$ where $D'=\sum_i\sum_{f(Q)=P_i}{Q}$.

Can we understand $f^!(\mathcal{L}(D))$ in general (by which I mean, when $f$ is assumed finite flat but not necessarily etale)? (can we say something more specific than $f^!\mathcal{L}(D)=\underline{Hom}_{\mathcal{O}_Y}(f_*(\mathcal{O}_X),\mathcal{L}(D))$, in terms of the divisor $D$)? And how should I think of the functor $f^!$ for general coherent sheaves, in this setting?

Thanks!

xlord
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  • You're aware that this is not actually a functor, right? Only the derived version exists. – KReiser Aug 18 '20 at 06:08
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    @KReiser I thought that if $f:X \rightarrow Y$ is a finite flat morphism, then $f^!$ is a functor from coherent sheaves on $Y$ to coherent sheaves on $X$, was I wrong about this? – xlord Aug 18 '20 at 09:13
  • Although I don't know what will happen in general, but if $f^!=f^$ when $f$ is etale, then I think it should also be true at least for branched covering between curves. For example, if you can show $\mathcal{Hom}{\mathcal{O}_Y}(f\mathcal{O}_X,\mathcal{F})$ is $f^*\mathcal{F}$ when $f$ is a covering between curves, then I don't see anything goes wrong when branching points appear. – AG learner Aug 18 '20 at 17:36
  • @xlord this remark was mostly meant for your "in general" question. Unless you restrict to a situation where the thing exists, you can't say anything "in general" because the functor you want to talk about doesn't exist most of the time. – KReiser Aug 18 '20 at 18:23
  • @KReiser, I suppose I meant for the set up for this question to be when $f$ is finite and flat. I'll edit the question to make this more clear. – xlord Aug 18 '20 at 18:40
  • @AGlearner is there a reason to believe this will also hold for branched coverings which are not etale? – xlord Aug 18 '20 at 18:46

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