By Andre Joyal, Myles Tierney

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Proof: Necessity is clear by Proposition 1 §1. Conversely, to calculate the supremum of an Ef-subset of M, just calculate its supremum as an E-subset. Proposition 1. equivalence Let A be a locale in S. Then there is a natural sA(sh(A)} * Mod(A). Proof: N: A 0p Let N + s£(5) be an A-'module. We have to show that the functor of §2 is a sheaf. So let a = \/ a. iel 1 be a supremum in A, 50 A. JOYAL § M. TIERNEY (xi^ieT and let for ^e a ^am^^Y °f elements x. )«x. )'X. 1 . iel a. -x . «x . J i i J Then a*x = a* V/ x .

Next, let A e Loc(S) and consider the pullback square (Proposition 4 §1) sh(f A) -* sh(A) Pf P By Proposition 1 §3, s£(sh(A)) = Mod(A), Making the identification, # A-modules to f A-modules. regarded as a sup-lattice, and w-e have proved Theorem 5• The spatial reflection over Let S. # # f becomes simply f itself, which takes If M is an A-module, p*M is just M so the Beck-Chevalley condition becomes trivial, 1. Top/S denote the category of topoi and geometric morphisms Taking sheaves on a space defines a functor sh: Sp(S) -^ Top/S.

Then We have to show: enough to show is open, and x u e 0 (X) u = satisfies \/ u1-- ieJ Vi e I, u. £ V / u . But u. <_ p~31u - , so it is 1 1 jeJ 3 u i A p"3 U i <_ X / u . J j eJ 3 u . < p*(u. ) 1 x jeJ 3 By the usual argument, it is enough to show that if the left hand side of the bottom line equals 1, then the right hand side equals 1, but this is obvious. Proof of Theorem 1: An atom of X is an open subspace a c — > X such that a*a CZ A and 3 a = 1- Let A be the set of atoms. Each atom a defines a point of X, since a = 1 by Lemma 1.

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