5.2 Universal Morphisms  In this section we assume C is a metric site with finite limits (i.e., C has fibre products and a finite object).  Definition 5.2.1. We say a morphism f: X d Y is universally open (resp. closed, resp. bicontinuous, resp. surjective, resp. injective, resp. effective, etc) if, for any base extension Y' ® Y, the morphism fY': XY' ® Y' is open (resp. closed, resp. bicontinuous, resp. surjective, resp. injective, resp. effective, etc).  Remark 5.2.2. More generally, if a morphism f: X ® Y has certain property P, and if for any base extension Y' ® Y the morphism fY': XY' ® Y' also has the property P, then we say that f has the property P universally. We say the property P is stable under base extension if for any morphism f having the property P, f has P universally (in which case we also say that the property P is an absolute property).  Example 5.2.3. The following properties for a morphism f: X ® Y in a site C are absolute:  (a) f is an open effective morphism (1.2.1);  (b) f is a regular monomorphism (5.1.10.d);  (c) f is an injective monomorphism (1.3.6);  (d) f is a surjective morphism (1.3.8.c).  (e) f(X) is a closed subset of Y (1.3.8.b);  (f) f(X) is an open subset of Y (1.3.8.b);  (g) f(X) is a locally closed subset of Y (1.3.8.b).  Example 5.2.4. Suppose C is a site such that t: C ® Top preserves finite limits. Then the following properties for a morphism f: X ® Y are absolute:  (a) f is injective.  (b) f is bicontinuous.  (c) f is homeomorphic.  Proposition 5.2.5. Suppose a monomorphism f: X ® Y is universally closed (resp. open), then f is universally bicontinuous.  Proof. Since f is a monomorphism, for any base extension g: Y' ® Y, fY': X ×Y Y' ® Y' is a monomorphism, therefore fY' is injective (1.3.6). But fY' is closed (resp. open), therefore fY' is bicontinuous.  Proposition 5.2.6. Suppose f: X ® Y is a universally bicontinuous morphism. Then f is universally closed (resp. open, resp. homeomorphism) if and only if f(X) is closed (resp. open, resp. the whole set X)   Proof. Since by assumption f is universally bicontinuous, for any base extension g: Y' ® Y, the morphism fY': XY' ® Y' is closed (resp. open, resp. homeomorphism) if and only if the image |fY'(XY')| = g-1(|f(X)|) is closed (resp. open, resp. surjective) (1.3.8), i.e. if and only if |f(X)| in X is closed (resp. open, resp. the whole set X).  The following three propositions (5.2.7) - (5.2.9) hold for universally open, universally closed, universally bicontinuous, universally injective and universally homeomorphic morphisms. For simplicity we only treat the case of universally closed morphisms.  Proposition 5.2.7. (a) The composition of two universally closed morphisms is universally closed.  (b) If f: X ® X', g: Y ® Y' are universally closed, then so is f ×S g.  (c) If f: X ® Y is a universally closed S-morphism, so is fS' : XS' ® YS' for any base extension S' ® S.  (d) Suppose f: X ® Y, g: Y ® Z are two morphisms such that f is surjective. If gf is universally closed, so is the morphism g.  (e) Suppose f: X ® Y is a morphism. Suppose {Ui} is an open effective cover of |Y|. Then f is universally closed if and only for any i, the morphism f-1(Ui) ® Ui is so.  Proof. (a) and (c) follows directly from the definitions and (4.2.6.d). (b) follows from (a) and (c) by (5.1.11).  To prove (d), suppose f is surjective, then for any morphism Z' ® Z, fZ': XZ' ® YZ' is surjective (1.3.8.c). Thus we only need to prove that if gf is closed, so is g, which can be verified easily.  The " only if " part of (e) follows from (c). Conversely, suppose the condition is satisfied and g: Y' ® Y is a morphism. Then g-1(Ui) = U'i forms an open covering of |Y'|. If we denote by fi the restriction f-1(Ui) ® Ui of f and by f' the morphism fY', the restriction f'-1(U'i) ® U'i of f' is precisely (fi)Ui'. Thus we are reduced to prove that the hypotheses implies f is closed, which is obvious.  Proposition 5.2.8. Suppose f: X ® S is a morphism. The following conditions are equivalent:  (a) The diagonal morphism Df: X ® X ×S X is universally closed.  (b) Suppose g, h: T ® X are two S-morphisms with the kernel p: ker(g, h) ® T. Then p is universally closed.  (c) Suppose g: Y ® S is a morphism. Then any Y-section h of Y ×S X over Y (i.e., the graph of any S-morphism from Y to X) is universally closed.   Proof. First we assume (a) holds. Using the notation of (5.1.5) we have p = (DX)T. Since by assumption DX: X ® W is universally closed, p is universally closed by (5.2.7.c). Thus (a) implies (b).  Next we assume (b) holds. Suppose h: Y ® X ×S Y is an Y-section of X ×S Y over Y. Then h is the kernel of the morphisms p, phq : X ×S Y ® X, where p, q are the projections of X ×S Y to X and Y. Applying (b) to p and phq we see that h is universally closed. This proves (c).  Finally (a) is a special case of (c) with g = f and h = Df .  Proposition 5.2.9. The following assertions for a site C are equivalent:  (a) For any morphism f: X ® S, the diagonal morphism Df: X ® X ×S X is universally closed.  (b) For any morphism f: X ® S, any S-section of X is universally closed.  (d) Any special morphism is universally closed.  Proof. By (5.2.8).  Proposition 5.2.10. (a) Suppose in C any spacial morphism is universally bicontinuous. If f: X ® Y and g: Y ® Z are morphisms and gf is universally bicontinuous, then so is f.  (b) If gf is universally closed and g is separated (see (5.3)), then f is universally closed.  (c) If f: X ® Y and g: Y ® Z are morphisms and gf is universally injective, then so is f.  Proof. These assertions follow from (5.1.12) and (5.1.13).  Proposition 5.2.11. Suppose f: X ® Y and h: Y' ® Y are two morphisms in a site C. If h is surjective and fY': X ×Y Y' ® Y' is universally injective. Then f is universally injective.  Proof. Consider a base extension Y1 ® Y, f1 = fY1. Let Y1' = Y1 ×Y Y', X1' = X' ×Y' Y1' = X1 ×Y1 Y1', f1': X1' ® Y1'. Consider the diagram  Since f ' is universally injective, so is f1'. Since h is surjective, so is h1. Since X1' is the fibre products of X1 and Y1', for any two points x, x' Î |X1| over a point y Î |Y1|, there is a point y' Î |Y1'|, and two points z, z' Î |X1'| over y' and over x, x' respectively. Since f1' is injective, z = z', it follows that x = x', thus f1 is injective. This shows that f is universally injective.  Proposition 5.2.12. Suppose C is a site with fibre products. Then the regular monomorphisms in C are universally bicontinuous if and only if the topology t: C ® Top preserves the fibre products of the form f ×X g where f: U ® X is a section and g: Y ® X is any morphism.   Proof. In any site a section f: U ® X is bicontinuous. Since C is separable, the projection p: U ×X Y ® Y is injective with the image g-1(U). Thus the regular monomorphism p is bicontinuous if and only if t preserves the fibre products of this type.    [Next Section][Content][References][Notations][Home]