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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.
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