Definition 18.31.1. Let (f, f^\sharp ) : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) \longrightarrow (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \mathcal{O}') be a morphism of ringed topoi. We say (f, f^\sharp ) is flat if the ring map f^\sharp : f^{-1}\mathcal{O}' \to \mathcal{O} is flat. We say a morphism of ringed sites is flat if the associated morphism of ringed topoi is flat.
18.31 Flat morphisms
Lemma 18.31.2. Let f : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C}') be a morphism of ringed topoi. Then
is exact. If (f, f^\sharp ) : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \mathcal{O}') is a flat morphism of ringed topoi then
is exact.
Proof. Given an abelian sheaf \mathcal{G} on \mathcal{C}' the underlying sheaf of sets of f^{-1}\mathcal{G} is the same as f^{-1} of the underlying sheaf of sets of \mathcal{G}, see Sites, Section 7.44. Hence the exactness of f^{-1} for sheaves of sets (required in the definition of a morphism of topoi, see Sites, Definition 7.15.1) implies the exactness of f^{-1} as a functor on abelian sheaves.
To see the statement on modules recall that f^*\mathcal{F} is defined as the tensor product f^{-1}\mathcal{F} \otimes _{f^{-1}\mathcal{O}', f^\sharp } \mathcal{O}. Hence f^* is a composition of functors both of which are exact. \square
Definition 18.31.3. Let f : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}), \mathcal{O}') be a morphism of ringed topoi. Let \mathcal{F} be a sheaf of \mathcal{O}-modules. We say that \mathcal{F} is flat over (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}), \mathcal{O}') if \mathcal{F} is flat as an f^{-1}\mathcal{O}'-module.
This is compatible with the notion as defined for morphisms of ringed spaces, see Modules, Definition 17.20.3 and the discussion following.
Lemma 18.31.4. Let f : (\mathcal{C}, \mathcal{O}_\mathcal {C}) \to (\mathcal{D}, \mathcal{O}_\mathcal {D}) be a morphism of ringed sites. Let \mathcal{F}, \mathcal{G} be \mathcal{O}_\mathcal {D}-modules. If \mathcal{F} is finitely presented and f is flat, then the canonical map
of Remark 18.27.3 is an isomorphism.
Proof. Say f is given by the continuous functor u : \mathcal{D} \to \mathcal{C}. We have to show that the restriction of the map to \mathcal{C}/U for any U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}) is an isomorphism. We may replace U by the members of a covering of U. Hence by Sites, Lemma 7.14.10 we may assume there exists a morphism U \to u(V) for some V \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}). Of course, then we may replace U by u(V). Then since u is continuous, we may replace V by a covering and assume there is a presentation \mathcal{O}_ V^{\oplus m} \to \mathcal{O}_ V^{\oplus n} \to \mathcal{F}|_ V \to 0 over \mathcal{D}/V. Since formation of \mathop{\mathcal{H}\! \mathit{om}}\nolimits commutes with localization (Lemma 18.27.2) we may replace f by the morphism (\mathcal{C}/u(V), \mathcal{O}_{u(V)}) \to (\mathcal{D}/V, \mathcal{O}_ V) induced by f. Hence we reduce to the case where \mathcal{F} has a global presentation \mathcal{O}_\mathcal {D}^{\oplus m} \to \mathcal{O}_\mathcal {D}^{\oplus n} \to \mathcal{F} \to 0. Since f is flat and f^*\mathcal{O}_\mathcal {D} = \mathcal{O}_\mathcal {C} we obtain a corresponding presentation \mathcal{O}_\mathcal {C}^{\oplus m} \to \mathcal{O}_\mathcal {C}^{\oplus n} \to f^*\mathcal{F} \to 0, see Lemma 18.31.2. Using that \mathop{\mathcal{H}\! \mathit{om}}\nolimits commutes with finite direct sums in the first variable, using that both \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_\mathcal {C}}(\mathcal{O}_\mathcal {C}, -) and \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_\mathcal {D}}(\mathcal{O}_\mathcal {D}, -) are the identity functor, and using the functoriality of the construction of Remark 18.27.3 we obtain a commutative diagram
where the right two vertical arrows are isomorphisms. By Lemma 18.27.5 the rows are exact. We conclude by the 5 lemma. \square
Comments (0)