7. Continuous Functors  Definition 7.1. Suppose C and D are two framed sites. A functor F: C ® D is continuous if all the objects and morphisms in C are geometric (therefore strongly geometric) over D (i.e., C = Geo(C/F)).  Remark 7.2. (a) Any bicontinuous functor of framed sites is continuous.  (b) A composition of two continuous functors is continuous.  (c) Suppose G: E ® C is a bicontinuous functor of framed sites and F: C ® D is any functor. Then FG: E ® D is continuous if and only if G(E) Í Geo(C/F).  (d) The restriction of any functor F: C ® D on the subsite Geo(C/F) of C is a continuous functor, denoted also by F.  Proposition 7.3. Suppose F: C ® D is a continuous functor. Suppose X is an object of C and W is an open  sieve on F(X). Then FX-1(W) Í FX*(W). If F is strict and W is open effective then  FX*(W) = FX-1(W).  Proof. Suppose f: Y ® X is a morphism in FX-1(W). Then F(f): F(Y) ® F(X) is in W, so 1D/F(Y) Í F(f)-1(W). Thus 1C/Y = FY*(1D/F(Y)) Í FY*(F(f)-1(W)) = f-1(FX*(W)), which implies that f is in FX*(W). Thus FX-1(W) Í FX*(W). If F is strict and W is open effective then FX*(W) Í FX-1(W) by (6.2.d), hence FX*(W) = FX-1(W). n  Theorem 7.4. Suppose F: C ® D is a strict functor. Then F is continuous if and only if for any object X of C and any open effective cover {Wi} of F(X), {FX-1(W)} is an open cover of X.  Proof. One direction comes from (7.3). For the other direction suppose the condition holds for F. Since any open effective sieve W on F(X) can be included in an open effective cover of F(X), FX-1(W) is open, and so FX*(W) = FX-1(W). It follows that X is geometric over D, and each morphism f: Y ® X is geometric because f-1(FX-1(W)) = FY-1(F(f)-1(W)) holds (unconditionally for any sieve W). n  An object X of a framed site C is called local if the joint M(X) of all the open sieves U ¹ 1C/X of X is not 1C/X. A morphism f: Y ® X of local objects of C is called a local morphism if f-1(M(X)) ¹ 1C/Y.  Remark 7.5. The following conditions for an object X of C are equivalent:  (a) X is a local object.  (b) G(X) is a local locale (i.e., a local object in Loc).  (c) Any open cover of X contains 1C/X.  Remark 7.6. Suppose X, Y and Z are local objects of C and f: Y ® X, g: Z ® Y are morphisms in C.  (a) If f and g are local morphisms, then fg is a local morphism.  (b) If fg is a local morphism then f is a local morphism.    Proposition 7.7. Suppose F: C ® D is a continuous functor of framed sites.  (a) If X is a local object of C, then F(X) is a local object of D and G(FX): G(X) ® G(F(X)) is a local morphism of local locales.  (b) If X and Y are two local objects of C and f: Y ® X is a local morphism, then F(f): F(X) ® F(Y) is a local morphism of local objects of D.  Proof. (a) First we show that if U is an open sieve on F(X) and FX*(U) = 1X, then U = 1F(X). The assumption  FX*(U) = 1X implies that there is an open cover {Vi} of X such that each restriction F(Vj) ® F(X) is in U. Since X is local there is some i such that Vi = 1X by (7.5.c). Thus F(X) ® F(X) is in U, hence U = 1F(X).  Now suppose {Ui} is an open cover of F(X). Since F is continuous, FX*: G(F(X)) ® G(X) is a morphism of frames by (6.6.a), thus {FX*(Ui)} is an open cover of X. Since X is a local object, there exits some i such that FX*(Ui) = 1X by (7.5.c). It follows from the above observation that Ui = 1F(X). We have proved that any open cover of F(X) contains 1X, thus F(X) is local by (7.5.c). Since M(F(X)) ¹ 1F(X), the above assertion also implies that FX*(M(F(X)) ¹ 1X. Thus G(FX): G(X) ® G(F(X)) is a local morphism of local locales.  (b) Since F is continuous we have G(FX)G(f) = G(F(f))G(FY). As G(FX) and G(f) are local morphisms, G(F(f)) is a local morphism by (7.6). n    [Next Section][Content][References][Notations][Home] 