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I would have to ask for apology for following question by everybody who is familar with algebraic geometry since this might be a quite trivial problem that cause some irritations for me: we consider a morphism $f: Z \to Y$ between schemes. then the induced direct image functor $f_*$ is claimed to be left exact. the proofs I found used always following argument: We consider an exact sequence of sheaves on $Z$

$$0 \to \mathcal{F} \to \mathcal{G} \to \mathcal{H} \to 0$$

choose arbitrary open subset $V \subset Y$ and apply firstly $f_*$ and then the $\Gamma(V, -)$ functor. Since $\Gamma(V, -)$ is left exact for sheaves we obtain exact sequence

$$0 \to \mathcal{F}(f^{-1}(V)) \to \mathcal{G}(f^{-1}(V)) \to \mathcal{H}(f^{-1}(V))$$

at this point the proves end. why now we are done? why is this sufficient? I thought that a sequence of sheaves if exact if and only if the induced sequence at every stalk is exact namely we have to verify that

$$0 \to (f_*\mathcal{F})_y \to (f_*\mathcal{G})_y \to (f_*\mathcal{H})_y$$

is exact in every $y \in Y$. at that moment I encounter the problem that I don't know how explicitely calculate the stalk $(f_*\mathcal{F})_y$ of the direct image sheaf. on the other hand the exactness for all sequences of second type seems to be much weaker as the conditions for exactness on stalks as in the last one. or are these two criterions for exactness equivalent?

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    $\begingroup$ Taking stalks is an exact functor hence the composition of pushforward and taking stalks will give you a left exact functor $\endgroup$
    – leibnewtz
    Mar 2 '19 at 4:09
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Left exact on all open sets implies left exact on stalks. This follows from exactness of direct limits for categories of modules, cf. Why do direct limits preserve exactness?.

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  • $\begingroup$ But stalks are direct limits (it doesn't change your argument, because direct limits are exact on abelian groups) $\endgroup$ Mar 2 '19 at 10:04
  • $\begingroup$ @Max Thanks, fixed. $\endgroup$
    – Ben
    Mar 2 '19 at 10:27
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A sequence of sheaves $$0 \to \mathcal{F} \to \mathcal{G} \to \mathcal{H} \to 0$$ is exact if and only if the induced sequence on stalks is exact for every $x \in X$. It's easy to show this fact is equivalent to the statement that taking stalks takes any exact sequence to an exact sequence. This means that a sequence $$0 \to \mathcal{F} \to \mathcal{G} \to \mathcal{H}$$ is exact if and only if it's exact on the level of stalks.

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  • $\begingroup$ This does not answer the question - the OP is asking why the given sequence of sections over $f^{-1}(U)$ being exact implies the sequence of direct image sheaves is exact. $\endgroup$
    – KReiser
    Mar 2 '19 at 4:26
  • $\begingroup$ Why not? If all sections over $f^{-1}(U)$ are exact then the sequence of sheaves is exact. This is equivalent, by my argument, to every sequence of stalks of the direct image sheaves being exact $\endgroup$
    – leibnewtz
    Mar 2 '19 at 5:33
  • $\begingroup$ If anything it's missing it's the detail that sheafification preserves finite (co)limits. This gives us that sheafification preserves kernels and cokernels, which is necessary to show that exactness on the level of sections actually gives us exactness on the level of sheaves $\endgroup$
    – leibnewtz
    Mar 2 '19 at 5:55
  • $\begingroup$ Well, for one, you didn't mention that in your answer. The OP is looking for an argument that shows "$0\to \mathcal{F}(f^{-1}(U)) \to \mathcal{G}(f^{-1}(U))\to \mathcal{H}(f^{-1}(U))$ exact" implies "$0\to f_*\mathcal{F} \to f_*\mathcal{G} \to f_*\mathcal{H}$ is exact". Your post contains "exactness can be checked on the level of stalks" which is already present in the OP's question and does not connect to taking sections, which is the thing the OP professes difficulty with. $\endgroup$
    – KReiser
    Mar 2 '19 at 6:56

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