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Let $X$ be an affine Noetherian scheme, $Y$ a separated Noetherian scheme, $f:X\rightarrow Y$ a morphism of schemes inducing a homeomorphism on the underlying topological spaces.

Let $F$ be a coherent sheaf on $Y$; treat it as a sheaf of abelian groups and pull it back to $X$. It is possible that $H^1(X, f^{-1}(F))\neq 0$?

This question may be confusing because we are essentially computing the sheaf cohomology of the exact same abelian sheaf on the exact same topological space but the non-trivial piece of information here is the existence of the morphism $f$ (remember, the forgetful functor from schemes to spaces is not full). Thus a random coherent sheaf on a separated scheme with non-vanishing $H^1$ is not going to do the trick.

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    $\begingroup$ I do not see the relevance of the "may be confusing" part here. If $f:X\to Y$ is a homeomorphism then $H^1(Y,F)\cong H^1(X,f^{-1}(F))$ for every sheaf $F$, coherent or not. $\endgroup$
    – S. carmeli
    Commented Apr 13, 2019 at 22:41
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    $\begingroup$ I don't see anything stopping us from taking $Y = X$ and $f = \mathrm{id}$. Then Certainly $H^1(X, f^{-1}\mathcal{F}) = H^1(X, \mathcal{F}) = 0$ since $\mathcal{F}$ is a quasi-coherent sheaf on an affine scheme. $\endgroup$
    – cgodfrey
    Commented Apr 14, 2019 at 5:36
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    $\begingroup$ What's more relevant than the forgetful functor from schemes to spaces not being full is that the forgetful functor from quasi-coherent sheaves to sheaves of abelian groups is exact. So you can compute cohomology of a quasi-coherent sheaf $\mathcal{F}$ as the cohomology of the underlying sheaf of abelian groups. $\endgroup$
    – cgodfrey
    Commented Apr 14, 2019 at 5:40
  • $\begingroup$ @S.carmeli the question asked for an example where the cohomology of $f^{-1}(F)$ does not vanish. Do you have such an example? $\endgroup$
    – user137767
    Commented Apr 14, 2019 at 9:50
  • $\begingroup$ @cgodfrey $f^{-1}(F)$ is not necessarily a quasi-coherent sheaf (the quasi-coherent sheaf is $f^{-1}(F)\otimes_{f^{-1}(O_Y)}O_X$). The question explicitly states that we are taking the pullback of an abelian sheaf. The cohomology of an arbitrary abelian sheaf on an affine Noetherian scheme does not necessarily vanish. $\endgroup$
    – user137767
    Commented Apr 14, 2019 at 9:55

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