1.4 Exact Presites Definition 1.4.1. Suppose C is a presite and X an object of C. A subset U of |X| is called exact if the following conditions are satisfied:  (a) For any point x Î U there is a morphism f: Y ® X such that |f(Y)| Í U and x Î |f(Y)|.  (b) For any x Î U the category C*/(x,U) of pointed objects f: (y, Y) ® (x, X) over (x, X) such that |f(Y)| Í U is connected.  (c) A subset V of U is open in the subspace U if for any morphism f: Y ® X such that |f(Y)| Í U, f-1(V) is an open subset of |Y|.  Remark 1.4.2. For any subset U of |X| let C/U be the subcategory of C/X consisting of objects (Y, f) over X such that |f(Y)| Í U. We denote by |hU| the colimit of the functor tp: C/U ® C ® Top where t is the metric functor of C. The maps |f|: |Y| ® U for all (Y, f) Î C/U determine a continuous map G: |hU| ® U. Then (1.4.1 a,b,c) are equivalent to the assertions that G is surjective, injective, and open respectively. Thus we have  (a) U is active if and only if the map G: |hU| ® U is a homeomorphism.  (b) If U is active and hU is represented by a morphism g: Z ® X, then we may identify |hU| with the space |Z| of Z since (Z, g) is a final object of C/U.  (c) Combining (a) and (b) we see that a subset U of |X| is effective if and only if U is active and exact.  Remark 1.4.3. If C is separable, then (1.4.1b) may be simplified to  (1.4.1b') For any x Î U and any f: (y, Y) ® (x, X) and g: (z, Z) ® (x, X) in C*/(x,U), there is a pointed object (w, W) with two morphisms (w, W) ® (y, Y) and (w, W) ® (z, Z).  Definition 1.4.4. A presite C is called exact if any open subset of the space |X| of any object X Î C is exact.  Proposition 1.4.5. Suppose for any object X Î C the open exact subsets of |X| form a basis for |X|. Then C is exact.  Proof. Suppose U is an open subset of |X|. We verify the conditions of (1.4.1) for U:  (a) (1.4.1.a) follows from the fact that U has an exact open cover {Vi} and (1.4.1a) holds for each active open subset Vi.  (b) Suppose x Î U. We have to prove that the category C*/(x,U) defined in (1.4.1a) is connected. Let V be an exact open neighborhood of x contained in U. Then C*/(x,V) is a connected subcategory of C*/(x,U). It suffices to prove that any object of C*/(x,U) is connected to an object of C*/(x,V) by a morphism in C*/(x,U). Suppose f: (y, Y) ® (x, X) is an object of C*/(x,U). Then f-1(V) is an open subset of Y containing y. By (a) there is a g: (z, Z) ® (y, Y) such that g(Z) Í f-1(V). Since fg: (z, Z) ® (x, X) is an object of C*/(x,V), we see that the object f Î C*/(x,U) is connected by g to an object fg e C*/(x,V).  (c) Suppose V is a subset of U such that (1.4.1.c) holds for V. Take an exact open cover {Vi} of U. Then (1.4.1c) holds for the subset V Ç Vi of Vi for each i. Thus each V Ç Vi is open in Vi. Hence V is open in U.  Corollary 1.4.6. Suppose for any object X Î C the open effective subsets of |X| form a basis for |X|. Then C is exact. Any locally effective presite is exact.  Remark 1.4.7. (1.4.6) yields another proof for (1.2.15) since active, quasi-effective subset is effective.  Suppose (C, t) is a presite and C' a category containing C as a full subcategory. Since colimits exist in Top, the metric functor t: C ® Top has a Kan extension t': C' ® Top. We obtain a presite (C', t').  Remark 1.4.8. Suppose S is an object of C'. The underlying space |S| of S is determined by the following properties:  (a) For any s Î S there is a morphism f: (x, X) ® (s, S) in C'* with X Î C.  (b) The category C*/(s,S) of pointed objects f: (x, X) ® (s, S) over (s, S) with X Î C is connected.  (c) The topology on |S| is the finest one such that for each f: X ® S with X Î C, the map |f|: |X| ® |S| is continuous. Thus a subset V of |S| is open if and only for any X Î C and f: X ® S, f-1(V) is open in |X|.  Remark 1.4.9. If (C', t') is a presite and C a full subcategory of C' such that the above (a) - (c) hold, then t' is the Kan extension of the restriction t'|C of t' on C.  Remark 1.4.10. The underlying set of the space |S| may be identified with the set of connected components of the category C*/S of the triples (x, X, f) where X Î C, x Î |X|, and f: X ® S is a morphism in C'; a morphism from (x, X, f) to (y, Y, g) is a morphism h: (x, X) ® (y, Y) such that gh = f.  Remark 1.4.11. If C is separated, then (1.4.8.b) may be simplified to:  (1.4.8.b') For any s Î |S| and any morphisms f: (x, X) ® (s, S) and g: (y, Y) ® (s, S) with X, Y Î C, there exists a pointed object (z, Z) such that Z Î C, with two morphisms p: (z, Z) ® (x, X) and q: (z, Z) ® (y, Y) such that fp = gq.  Example 1.4.12. (a) If (C, t) is representable with t* = h'T = homC (T, ~) for some object T Î C, then (C', t') is also representable with t'* = h'T = homC' (T, ~) in C'.  (b) Suppose C has a generic subsite D, then D is also a generic subsite of (C', t').  Proposition 1.4.13. Suppose (C, t) is a presite and C' a category containing C as a full subcategory. Suppose t': C' ® Top is the Kan extension of t on C'. Then (C, t) is separable if and only if (C', t') is separable.  Proof. Suppose (C, t) is separable. Suppose f': (x', X') ® (s, S) and g': (y', Y') ® (s, S) are two morphisms in C'*. Let f": (x, X) ® (x', X') and g": (y, Y) ® (y', Y') be morphisms in C' with X, Y Î C (1.4.8.a). We obtain two morphisms f = f'f": (x, X) ® (s, S) and g = g'g": (y, Y) ® (s, S). Applying (1.4.8.b') we find two morphisms p: (z, Z) ® (x, X) and q: (z, Z) ® (y, Y) as required. Thus (C', t') is separable.  Next assume that (C', t') is separable. Suppose f: (x, X) ® (s, S) and g: (y, Y) ® (s, S) are two morphisms in C*. There are two morphisms p': (z', Z') ® (x, X), x) and q': (z', Z') ® (y, Z) in C'*. There is a morphism h: (z, Z) ® (z', Z') with Z Î C (1.4.8.a). Then p'h: (z, Z) ® (x, X) and q'h: (z, Z) ® (y, Y) are what we are looking for. This proves that (C, t) is separable.  1.4.14 Suppose D is a category and E a full subcategory of D.  (a) We say E is a sieve of D if any object X Î D such that there is a morphism from X to an object of E is in E.  (b) We say E is a final subcategory of D if for any object X Î D, the subcategory E/X of C/X consisting of X-objects Y Î E is non-empty and is connected.  Remark 1.4.15. Suppose E is a final subcategory of D.  (a) If F is a subcategory of E, then F is a final subcategory of D. (The proof is similar to the following (1.4.15.a).)  (b) If F is a full subcategory of D containing E, then E is final in F.  Lemma 1.4.16. Suppose D is a category and E a final subcategory of D.  (a) D is connected if and only if E is connected.  (b) Suppose F is a sieve of D and F Ç E is a final subcategory of E. Then E is connected if and only if F is connected.  Proof. (a) If E is connected then D is so since any object of D is connected to an object of E. Conversely assume D is connected. For any two objects X and Y in E there are finitely many objects X0 = X, X1,...,X2n = Y in C such that there are morphisms X2j-2 ® X2j-1 and X2j ® X2j-1 for j = 1,...,n. Since E is final, there exist morphisms fi: Zi ® Xi with Zi Î E; for i = 0 and n we let Zi = Xi. Now Z2j-2 and Z2j-1 are objects over X2j-1, thus they are connected in E because C/X2j-1 is connected by (1.4.14b). Similarly Z2j and Z2j-1 are connected in E. Thus X = Z0 and Y = Z2n is connected in E, which shows that E is connected.  (b) Suppose E is connected. Since E is final, for any X Î F there is a morphism Y ® X with Y Î E. But F is a sieve, so Y Î F, thus Y Î F Ç E. Hence any object of F is connected to an object of E Ç F. Since E Ç F is final in E and E is connected by assumption, E Ç F is connected by (a).  Conversely, suppose F is connected. Since E is final in D and F Ç E is final in E, F Ç E is final in D by (1.4.15.a), thus is final in F by (1.4.15.b). Applying (a) we see that F Ç E and E are connected because F is so.  Proposition 1.4.17. Suppose (C, t) is a presite and C' a category containing C as a full subcategory. Suppose t': C' ® Top is the Kan extension of t on C'. Then (C, t) is exact if and only if (C', t') is exact.  Proof. First suppose (C, t) is exact. Suppose U is an open subset of the space |X| of an object X Î C'. We prove that U is exact by verifying the conditions of (1.4.1).  (a) For any point x Î U there is a morphism f: (y, Y) ® (x, X) such that Y Î C (1.4.8.a). Since C is exact, we can find an active open neighborhood V of y contained in f-1(U). Let g: (z, Z) ® (y, Y ) be a morphism with Z Î C and g(Z) Í V. Then fg: (z, Z) ® (x, X) is a morphism such that |fg(Z)| Í U. This proves (1.4.1.a) for U.  (b) We need to prove that for any x Î U, the category F = C'*/(x,U) of pointed objects f: (y, Y) ® (x, X) with f(Y) Í U is connected. Consider the category D = C'*/(x,X). Then E = C*/(x,X) is connected and is final in C'*/(x,X) by (1.4.8), and F is a sieve in D. Since C is active, F Ç E = C*/(x,U) is final in E. Thus we may apply (1.4.16.b) to see that F = C'*/(x,U) is connected.  (c) Suppose V is a subset of U such that for any morphism f: Y ® X with |f(Y)| Í U, |f-1(V)| is an open subset of |Y|. We have to prove that V is open in U, or equivalently, V is an open subset of X. It suffices to prove that for any morphism g: Z ® X with Z Î C, g-1(V) Í |Z| is open (1.4.8.c). Since g-1(U) is open we only need to show that g-1(V) is open in g-1(U). Since g-1(U) is exact, by (1.4.1.c) it suffices to prove that for any h: W ® Z with |h(W)| Í g-1(U), h-1(g-1(V)) is open. But |gh(W)| Í U. Thus (gh)-1(V) = h-1(g-1(V)) is open by assumption.  Conversely suppose (C', t') is exact. Suppose U is an open subset of the space |X| of an object X Î C. We prove that U is exact in C by verifying the conditions of (1.4.1).  (a') Since U is exact in C', for any point x Î U there is a morphism f: (y, Y) ® (x, X) with f(Y) Í U (1.4.1.a). Let g: (z, Z) ® (y, Y ) be a morphism with Z Î C. Then fg: (z, Z) ® (x, X) is a morphism in C such that |fg(Z)| Í U. This proves (1.4.1.a) for U.  (b') We need to prove that for any x Î U, the category E = C*/(x,U) of pointed objects f: (y, Y) ® (x, X) with f(Y) Í U and Y Î C is connected. Since U is exact in C', D = C'*/(x,U) is connected by (1.4.1.b), and E is final in D by (1.4.8). Thus we may apply (1.4.16.a) to see that E = C*/(x,U) is connected.  (c') Suppose V is a subset of U such that for any morphism f: Y ® X with |f(Y)| Í U and Y Î C, |f-1(V)| is an open subset of |Y|. We have to prove that V is open in U. Since U is exact in C', it suffices to prove that for any morphism g: Z ® X with g(Z) Í U, g-1(V) Í |Z| is open (1.4.1c). By (1.4.8c) it suffices to prove that for any h: Y ® Z with Y Î C, h-1(g-1(V)) is open. But |gh(Y)| Í U. Thus (gh)-1(V) = h-1(g-1(V)) is open by assumption.