By Akhil Mathew

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For the more general statement, consider instead the injection j! (OX |V ) → OX . (2) Let U ⊂ V . Then in the commutative diagram 0 G F (V ) G F(V ) G F (V ) G0 0 G F (U ) G F(U ) G F (U ) G0 both sequences are exact by flabbiness. Therefore the surjectivity of the left two downward arrows implies that of the rightmost downward arrow, hence proving that F is flabby. Following Grothendieck, we briefly indicate a more general approach to the above result. 14. Let C, C be two abelian categories and F : C → C a covariant left-exact functor.

So, first we look at how the ni /ti patch. The quotients ni /ti , nj /tj must become equal each of the localizations Mq for q ∈ D(ti tj ) = SpecAti tj . This means that, by the first part of the proof, the ti nj − tj ni become equal in Ati tj . In particular, there is a power (ti tj )kij such that (ti tj )kij (ti nj − tj ni ) = 0 ∈ M. ), we see that we have reduced our problem to the following: i 2The map M → p Mp is always injective. 2. SHEAVES OF MODULES ON SpecA 57 Suppose given quotients ni /ti ∈ Mti such that for each i, j, we have the gluing condition ti nj = tj ni .

Proof. We need to show that HomOX (M , N ) HomA (M, N ). Clearly, there is a map in the → direction, in view of the above proposition; given a sheap map, consider the associated homomorphism on the global sections. This is clearly surjective, since the association M → M is functorial. We have to prove that if M → N induces the zero map on global sections, then it is zero as a map of sheaves. This can be checked on the stalks. Fix a prime p. There is a diagram, which must be commutative: M Mp GN G Np If the map M → N is zero, then the localized map Mp → Np is zero.