5.3 Separated Morphisms  In this section we assume C is a metric site with finite limits. Let Z be a final object of C.  Definition 5.3.1. A morphism f: X ® S is called separated if the diagonal morphism D = (1X, 1X)S: X ® X ×S X is a universally closed morphism. If f is separated we also say that X is separated over S. An object X is called separated if it is separated over the final object Z.  Proposition 5.3.2. Suppose f: X ® S is a morphism. The following conditions are equivalent:  (a) f is separated.  (b) The kernel p: ker (i,j) ® Y of any two S-morphisms i, j: Y ® X is universally closed.  (c) Any Y-section of Y ×S X over Y for any S-object Y ® S is universally closed.   (c') The graph Gf = (1, f): Y ® Y ×S X of any S-morphism f: Y ® X is universally closed.  Proof. By (5.2.8).  Corollary 5.3.3. Suppose C is a site such that any special morphism is bicontinuous. Suppose f: X ® S is a morphism. Then the following conditions are equivalent:  (a) f is separated.   (a') The image |D(X)| of X in X ×S X is a closed subset of X ×S Y.  (b) Suppose i, j: Y ® X are two S-morphisms with the kernel p: ker (i,j) d Y. Then |p(ker(i,j))| is a closed subset of Y.  (c) Suppose g: Y ® S is a base extension and h is a Y-section of X ×S Y over Y. Then |h(Y)| is a closed subset of X ×S Y.   (c') Suppose f: Y ® X is an S-morphism with the graph morphism Gf = (1, f): Y ® Y ×S X. Then |Gf(Y)| is a closed subset of Y ×S X.  Proof. By (5.2.6).  Proposition 5.3.4. (a) Any monomorphism is separated.  (b) Any effective morphism is separated.  (c) The composition of two separated morphisms is separated.  (d) Suppose f: X ® Y is a separated S-morphism and j: S' ® S is a morphism. Then the S'-morphism fS' is separated.  (e) If f: X' ® X and g: Y' ® Y are separated S-morphisms, then the product morphism f × g: X' ×S Y' ® X ×S Y is also separated.  (f) If the composite gf of two morphisms is separated, f is separated.  Proof. (a) Suppose f: X ® S is a monomorphism. Then X = X ×S X (4.2.6.h). Thus |D(X)| = |X ×S X| is a closed subset of |X ×S X| and f is separated.  (b) Any effective morphism is a monomorphism.  (c) Suppose f: X ® Y , g: Y ® Z are two morphisms. It is easy to see that DX/Z = ((p, q)Y)DX/Y  If f and g are separated, then DX/Y and (p, q)Y are universally closed monomorphisms (5.1.8). Since the composition of two universally closed morphisms is universally closed, DX/Z is a universally closed morphism; therefore gf is separated.  (d). We have XS' × YS' XS' = (X ×Y X) ×Y YS' and DXS' = DX ×Y 1YS'. Then apply (5.2.7.c).  (e) is equivalent to (d) by (5.1.11).  (f) By (5.1.12) in view of (a) and (e).  Corollary 5.3.5. Let f: X ® S be a separated morphism. Suppose U Í |X| and V Í |S| are effective subsets and f(U) Í V. Then U is separated over V.  Proof. The composite g: U ® S of the effective morphism U ® X and the separated morphism f: X ® S is separated (5.3.4.c). Let h: U ® V be the induced morphism and i: V ® S the inclusion. Then g = ih is separated implies that h is separated (5.3.4.f).  Corollary 5.3.6. (a) An object X is separated if and only for any object S, any morphism f: X ® S is separated.   (b) If f: X ® S is a separated morphism and S is separated, then X is separated.  Proof. (a) The condition is sufficient because we may take f to be the canonical morphism from X to the final object Z. For the necessity let i: X ® Z and j: S ® Z be the natural morphisms to the final object Z. Then we have i = jf. Thus f is separated if i is so by (5.3.4.f).  (b) If f: X ® S and j: S ® Z are separated, then i = jf is separated by (5.3.4.c).  Corollary 5.3.7. Suppose X is a separated. Then any effective subobject of X is a separated object.  Definition 5.3.8. Suppose f: X ® S is a morphism. A diagonal cover of X over S is an effective open cover {Ui} of |X| such that {|Ui ×S Ui|} is an open cover of |X ×S X|.  Remark 5.3.9. An effective open cover {Ui} is a diagonal cover if any of the following conditions are satisfied.  (a) Ui = f-1(Vi) for an effective open cover {Vi} of |S|.  (b) Each pair x, y Î |X| lie in some Ui.  Proposition 5.3.10. Suppose {Ui} is a diagonal cover of X such that each Ui is separated over S. Then f is separated.  Proof. Suppose each Ui is separated over S. Then Ui = DX-1(Ui ×S Ui) ® Ui ×S Ui is a universally closed morphism. Since {|Ui ×S Ui|} is an open cover of |X ×S X|, Df is a universally closed morphism by (5.2.7.e). Thus f is separated.  Proposition 5.3.11. Suppose C is a site such that the topology t: C ® Top preserves finite limits. An object X is separated if and only if the underlying space of X is Hausdorff.   Proof. An object X in such a site C is separated if and only if the space of X is separated in the site Top. Thus it suffices to prove the assertion for Top. First we assume X is a Hausdorff topological space. For any (x, y) Î X × Y and (x, y) Ï D(X), we have x ¹ y, therefore we can find two neighborhoods U and V of x and y respectively such that U Ç V = Æ. Then W = p-1(U) Ç q-1(V) = U × V is a neighborhood of (x, y) and W Ç D(X) = Æ. This means that D(X) is closed, hence X is separated since D is universally bicontinuous. Conversely, suppose X is separated. Then D(X) is a closed subset of X × X. If x, y are two points of X and x ¹ y, then (x, y) - D(X). Thus we can find a neighborhood W of (x, y) such that W Ç D(X) = Û. By the definition of product topology on X × Y we can find neighborhoods U and V of x and y respectively such that U × V is a neighborhood of (x, y) contained in W. Then we must have U Ç V = Æ. This shows that X is Hausdorff.  Example 5.3.12. Suppose X is a Hausdorff space. Suppose f: X ® S is a continuous map and g: S ® X is an S-section of X. Then g(S) must be a closed subset (5.3.3.c). This fact can also be checked directly. Regarding S as a subset of X, if x Î X - S then g(x) ¹ x. Since X is Hausdorff, we can find two neighborhoods V and V' of x and g(x) respectively such that V Ç V' = Æ. Let W = V Ç g-1(V'). Then W is a neighborhood of x and W Ç S = Æ. This means that S is closed.       [Next Section][Content][References][Notations][Home]