5.1 Regular Monomorphism Suppose C is a category with fibre products.  Definition 5.1.1. Suppose X, S are two objects and i, j are two morphisms from X to S. The kernel (or equalizer) of i and j is an object ker(i,j), together with a map p: ker(i,j) ® X such that ip = jp, which has the following universal property:  for any object W and any map u: W ® X such that iu = ju, there exists a unique morphism h: W ® ker(i,j) such that ph = u.  Note that p is a monomorphism representing the subfunctor  W ® {f Î hom (W, X)| if = jf} of hX = hom (~, X).  Definition 5.1.2. Suppose X is an object over S with the structure morphism f: X ® S. A morphism s: S ® X such that fs = 1S is called an S-section of X. (Thus an S-section of X is an S-morphism s: S ® X.) The set of S-sections of X is denoted by G(X/S).  Definition 5.1.3. Suppose X and Y are objects over S. Suppose f: X ® Y is an S-morphism. The morphism (1X, f)S: X ® X ×S Y over S is called the graph morphism of f, denoted Gf, which is an X-section of X ×S Y. If Y = X and f = 1X, then the graph morphism of 1X is called the diagonal morphism of X over S, denoted DX/S (or simply DX, or Df, or D).  Remark 5.1.4. Suppose X, Y are two objects over S. The map f ® Gf = (1X, f)S is a bijection:  homS (X, Y) = homX (X, X ×S Y) = G(X ×S Y/X) The relationship of these concepts are exhibited in the following lemmas. (The proofs are straightforward which will be omitted.)  Suppose u: X ® S is a morphism. Consider an S-object T and an S-morphism t: T ® X ×S X = W. Form the fibre product (X ×W T, i, j) as below:  Lemma 5.1.5. Suppose g, h: T ® X are two S-morphisms with t = (g, h)S. Then E = ker (g, h) is the kernel of g and h:  Lemma 5.1.6. Suppose T = X, and s: S ® X is an S-section of u: X ® S with t = (su, 1X). Then E = S and i = j = s canonically:  Lemma 5.1.7. Suppose f: Y ® X is an S-morphism with T = Y ×S X and t = (f ×S 1X). Then (E, j) is the graph of f. We have E = Y, i = f and j = (1Y, f)S canonically:    Lemma 5.1.8. Suppose f: Y ® X, g: Z ® X are two S-morphisms with T = Y ×S Z and t = f ×S g. Then E = Y ×X Z canonically. We have j = (p, q)S and i = fp = gq, where p: Y ×X Z ® Y and q: Y ×X Z ® Z are projections:  Definition 5.1.9. A morphism is called regular monomorphism if it is the kernel of some pair of morphisms. A morphism is called a section if it is a section of a morphism.  Proposition 5.1.10. (a) Any regular monomorphism is a mono.  (b) Suppose f: X ® S is a morphism. Then Df, Gf and any S-section of X are regular monomorphisms.  (c) A morphism is a regular monomorphism if and only if it is the result of making a base extension of a diagonal morphism.  (d) If f: X ® S is a regular monomorphism, then fS': XS' ® S' is a regular monomorphism for any base extension S' ® S.  Proof. (a) is obvious and (b) follows from (5.1.6, 5.1.7, 5.1.8); (c) comes from (5.1.5); (d) follows from (c) and (4.2.6d).  Proposition 5.1.11. Suppose P is a property of morphisms satisfying the following conditions:  (a) For any object X, the identity 1X has the property P.  (b) The composition of two morphisms has the property P if each of them has the property P.  Then the following assertions are equivalent:  (i) Suppose f: X ® X', g: Y ® Y' are S-morphisms having the property P, then f ×S g has the property P.  (ii) If f: X ® Y is an S-morphism having the property P, then fS' has the property P for any base extension S' to S.  Proof. Suppose P satisfies the conditions (a) and (b). We have fS' = f ×S 1S'. Therefore (i) implies (ii) by (a). On the other hand, f ×S g is the composite morphism  f × 1Y                                 1X' × g X ×S Y -----------------------> X' ×S Y ----------------------> X' ×S Y' Hence (ii) implies (i) by (b).  Proposition 5.1.12. Suppose P is a property of morphisms having the following properties:  (a) If f: X ® Y is any regular monomorphism and g: Y ® Z is a morphism having the property (P), then gf has the property P.  (b) If f: X ® X' and g: Y ® Y' are two S-morphisms having the property P, then f ×S g has the property P.  Then if the composition gf of two morphisms f: X ® Y, g: Y ® Z have the property P, f has the property P.  Proof. Note that f can be factored as  Gf                                                               q f: X --------------------------> X ×Z Y -----------------------------> Y where q is the second projection and q = (gf) ×Z 1Y: X ×Z Y d Z ×Z Y = Y. Then q has the property P by (b). Since Gf is a special morphism, f has the property P by (a).  Proposition 5.1.13. Suppose C is a site. Suppose P is a property of morphisms having the following properties:  (a) If f: X ® Y is any closed regular mono and g: Y ® Z is a morphism having the property (P), then gf has the property P.  (b) If f: X ® X' and g: Y ® Y' are two S-morphisms having the property P, then f ×S g has the property P.  Then if the composition g.f of two morphisms f: X ® Y, g: Y ® Z have the property P, and Dg: Y ® Y ×Z Y is universally closed (i.e., g is separated, see 5.3) then f has the property P.  Proof. Note that f can be factored as  Gf                                                 q f: X ----------------------------> X ×Z Y --------------------------------> Y where q is the second projection and q = (gf) ×Z 1Y as in (5.1.12). Then q has the property P by (b). By (5.1.7) Gf is the result of Dg under a base extension (see the diagram below), thus a closed regular monomorphism, since by assumption Dg is universally closed. Hence f has the property P by (a).          [Next Section][Content][References][Notations][Home]