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In Hartshorne Proposition II.5.8, he shows, given a morphism $f \colon X \to Y$ where X and Y are schemes and $\mathcal{G}$ a quasi-coherent sheaf of $\mathcal{O}_{Y}$- modules, that $f^{*}\mathcal{G}$ is a quasi-coherent sheaf of $\mathcal{O}_{X}$-modules. Similarly, he shows that given $f$ quasicompact and separated, that the pushforward of a quasi-coherent sheaf is quasi-coherent.

My question concerns his reduction to the affine case. I understand that the quasi-coherent property is local, since it can be checked on an affine cover by definition, so for proving that the pullback of a quasi-coherent sheaf is quasi-coherent, one can assume $X$ to be affine. However, why is one allowed to also assume $Y$ to be affine as it is done in the proof, while for proving the pushforward, one can only assume $Y$ affine and not $X$?

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If $U\subseteq X$ is an open set and $V\subseteq Y$ is any open set containing $f(U)$, then the restriction of $f^*\mathcal{G}$ to $U$ depends only on the restriction of $\mathcal{G}$ to $V$ (more precisely, $(f^*\mathcal{G})|_U=g^*(\mathcal{G}|_V)$, where $g:U\to V$ is the restriction of the morphism $f$ to a morphism $U\to V$). To show $f^*\mathcal{G}$ is quasicoherent, it suffices to show that $X$ has an open cover by affine open sets $U\subseteq X$ on which $(f^*\mathcal{G})|_U$ is quasicoherent. For each affine open $V\subseteq Y$, $f^{-1}(V)\subseteq X$ can be covered by affine open sets, and these sets taken over all $V$ will cover all of $X$. So it suffices to show $(f^*\mathcal{G})|_U$ is quasicoherent if $U\subseteq X$ is affine open and there exists an affine open $V\subseteq Y$ such that $f(U)\subseteq V$. But in that case, $(f^*\mathcal{G})|_U=g^*(\mathcal{G}|_V)$ as above, so you may as well replace $f$ by $g$ and assume $X=U$ and $Y=V$ are affine.

For the pushforward, if $V\subseteq Y$ is open, $(f_*\mathcal{F})|_V$ depends on the restriction of $\mathcal{F}$ to the set $f^{-1}(V)\subseteq X$, so you can replace $Y$ by $V$ and $X$ by $f^{-1}(V)$ to assume that $Y$ is affine. To assume that $X$ is affine as well as in the pullback case, you would need to have an open cover of $Y$ by affine open sets $V$ such that $f^{-1}(V)$ is affine as well. Such an open cover need not exist (for instance, if $Y$ has only one point and $X$ is not affine).

So in short, the difference between the two cases is that there always exists a cover of $X$ by affine open subsets $U$ such that $f(U)$ is contained in an affine open set, but there need not exist a cover of $Y$ by affine open subsets $V$ such that $f^{-1}(V)$ is affine.

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  • $\begingroup$ Thanks. Just a quick elementary question. Is it obvious that given $U \subseteq X$ an affine open, that there necessarily exists an affine open $V \subseteq Y$ such that $f(U) \subseteq V$? Does this simply follow from the definition of morphism of schemes and the gluing construction? $\endgroup$ – combustion1925 Feb 8 '16 at 23:38
  • $\begingroup$ No, it's not obvious because it isn't even true. What is true is that there exists an open cover of $X$ by affine open sets $U$ which have this property (since you can take an affine open cover of $Y$, and then cover each of the sets $f^{-1}(V)$ by affine open sets). We're using here that to check that a sheaf is quasicoherent, it suffices to find an open cover such that its restriction to each set of the open cover is quasicoherent. $\endgroup$ – Eric Wofsey Feb 8 '16 at 23:40

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