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 

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 = 1_S 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 (1_X, f)_S: X ® X ×_S Y over S is called the graph morphism of f, denoted G_f, which is an X-section of X ×_S Y. If Y = X and f = 1_X, then the graph morphism of 1_X is called the diagonal morphism of X over S, denoted D_X/S (or simply D_X, or D_f, or D). 

Remark 5.1.4. Suppose X, Y are two objects over S. The map f ® G_f = (1_X, f)_S is a bijection: 

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, 1_X). Then E = S and i = j = 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: 

Proposition 5.1.10. (a) Any regular monomorphism is a mono. 
(b) Suppose f: X ® S is a morphism. Then D_f, G_f 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 f_S': X_S' ® 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 1_X 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 f_S' has the property P for any base extension S' to S. 

Proof. Suppose P satisfies the conditions (a) and (b). We have f_S' = f ×_S 1_S'. Therefore (i) implies (ii) by (a). On the other hand, f ×_S g is the composite morphism 

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 

where q is the second projection and q = (gf) ×_Z 1_Y: X ×_Z Y d Z ×_Z Y = Y. Then q has the property P by (b). Since G_f 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 D_g: 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 

where q is the second projection and q = (gf) ×_Z 1_Y as in (5.1.12). Then q has the property P by (b). By (5.1.7) G_f is the result of D_g under a base extension (see the diagram below), thus a closed regular monomorphism, since by assumption D_g is universally closed. Hence f has the property P by (a).