1.6. Analytic Monos
A mono u^{c}: U^{c} > X is a complement
of a mono u: U > X if u and u^{c}
are disjoint, and any map v:
T > X such that u and v are disjoint factors through
u^{c} (uniquely). The complement u^{c} of
a mono u, if exists, is uniquely determined up to isomorphism.
Definition 1.6.1. (a) A mono f:
U > X is singular
if it is the complement of a strong mono
to X.
(b) A coflat singular mono is
an analytic mono.
(c) A subobject u: U > X of an object X is analytic
if u is an analytic mono.
(d) A strong mono is disjuctable
if it has a coflat complement.
Proposition 1.6.2. (a) The pullback
of any analytic mono is analytic.
(b) The pullback of any disjunctable strong mono is disjunctable.
(c) Any coflat complement of a mono is analytic.
Proof. Suppose u: U > X is a disjunctable strong
mono with the coflat complement u^{c}: U^{c}
> X. Suppose f: Y > X is any map. Then the pullback
f^{1}(u) of u along f is a strong mono. It
is easy to verify that f^{1}(u^{c}) =
(f^{1}(u))^{c}. Since the pullback
f^{1}(u^{c}) of coflat map is coflat by
(1.4.2), it is analytic; so f^{1}(u)
is disjunctable. This proves (a) and (b).
To see (c), suppose u: U > X is a mono with the coflat
complement u^{c}: U^{c} > X. By (1.5.2)
u^{c}: U^{c} > X is also disjoint with
u^{+1}(U). Since U
u^{+1}(U), clearly u^{c} is a complement of
the strong mono u^{+1}(U) > X, thus it is analytic.
Proposition 1.6.3. Composite of analytic
monos is analytic.
Proof. Let f: Y > X and g: Z > Y
be two analytic monos. Let t: T > X and s: S
> Y be two strong monos such that f = t^{c} and g
= s^{c}. Then f
t = 0 and g s = 0.
Since the composite fg of coflat maps is coflat by (1.4.2),
it suffices to prove that fg = (t
f^{+1}(s))^{c}. We have
(fg) (t
f^{+1}(s)) = (fg)^{1}(t
f^{+1}(s)) = (fg)^{1}(t)
(fg)^{1}(f^{+1}(s))
= g^{1}(f^{1}(t))
g^{1}(f^{1}(f^{+1}(s)))
= g^{1}(f^{1}(t))
g^{1}(s)
= (g (f
t)) (g
s) = 0
by (1.5.3) and (1.5.4).
Next consider a map r: W > X such that r _{X}
(t f^{+1}(s))
= 0. We have r _{X}
t = 0 and r _{X}
f^{+1}(s) = 0. Thus r factors through
f in a map k: W > Y. We have
k _{Y} s
= k^{1}(s) = k^{1}(f^{1}(f^{+1}(s)))
= (fk)^{1}(f^{+1}(s)))
= r^{1}(f^{+1}(s)) = r _{X}
(f^{+1}(s)) = 0.
Thus k factors through g in a map v: W > Z.
Hence r = fgv. This shows that fg = (t
f^{+1}(s))^{c}, i.e. fg is analytic.
Proposition 1.6.4. If u:
U > X and v: V > X are two disjunctable strong subobjects
of X. Then u^{c}
v^{c} = (u v)^{c}.
Proof. Let w = u
v: W > X be the join of u and v. Suppose u^{c}:
U^{c} > X and v^{c}: V^{c} >
X are the coflat complements of u and v respectively.
Let s: U^{c}
V^{c} > U^{c} and t: U^{c}
V^{c} > V^{c} be the pullback of u^{c}
and v^{c}. Then s and t are analytic monos
by (1.6.2), and r = u^{c}s = v^{c}t
= u^{c} v^{c}:
U^{c} V^{c}
> X is an analytic mono by (1.6.3). We prove
that r = w^{c}. We have
r w = r^{1}(W)
= r^{1}(u
v) = r^{1}(u)
r^{1}(v)
= s^{1}((u^{c})^{1}(u))
t^{1}((v^{c})^{1}(v))
= s^{1}(u^{c}
u) t^{1}(v^{c}
v)
= s^{1}(0)
t^{1}(0) = 0.
On the other hand, suppose z: Z > X is a map such that
z _{X} w = 0.
Then z _{X} u =
z _{X} v = 0. Thus
z factors through u^{c} and v^{c},
therefore also factors through r. This shows that r = w^{c}.
Proposition 1.6.5. Suppose u:
U > X and v: V > Y are two disjunctable strong monos
with coflat complements u^{c}: U^{c} >
X and v: V^{c} > Y respectively.
Then u + v: U + V > X + Y is a disjunctable strong mono
with coflat complement: u^{c} + v^{c}: U^{c}
+ V^{c} > X + Y.
Proof. We know that u^{c} + v^{c}
is coflat by (1.4.2.d) because u^{c}
and v^{c} are coflat. We have (u^{c} + v^{c})
(u + v) = u^{c}
u + v^{c} v = 0 +
0 = 0 by (1.3.4.a). Let t:
M > X + Y be a map such that t _{X+Y}
(u + v) = 0. Then t = t_{X} + t_{Y}
with t_{X} and t_{Y} being the pullback of
t along the injections X > X + Y and Y > X + Y
respectively. Thus (t_{X} + t_{Y}) _{X+Y}
(u + v) = 0 implies that t_{X} _{X}
u + t_{Y} _{Y}
v = 0. It follows that t_{X}
u = t_{Y} v = 0.
Thus t_{X} can be factored through u^{c}
and t_{Y} can be factored through v^{c}.
This implies that t can be factored through u^{c} + v^{c}.
Hence u^{c} + v^{c} is the complement of u +
v.
Proposition 1.6.6. (a) Isomorphisms
are analytic monos.
(b) Finite intersections of analytic monos are analytic monos.
(c) Finite sums of analytic monos are analytic monos.
(d) Injections of a sum are analytic monos.
Proof. (a) is obvious and (b) follows from (1.6.1)
and (1.6.2).
(c) follows from (1.6.5).
(d) The injections of a sum are coflat strong monos by (1.4.5),
and are complements of each other because finite sums are stable disjoint.
Therefore they are analytic monos.
Example 1.6.7. Consider the category
of affine schemes. Suppose A is a ring. From algebraic geometry
we know that a strong mono to Spec(A) is precisely a closed
immersion Spec(A/I) > Spec(A) determined
by an ideal I of A. Since by definition a singular mono is
a complement of a strong mono, any singular mono of affine schemes must
be an open embedding. By a result of Diers (cf. [D, p.42]) any singular
mono of affine schemes is always coflat, so analytic monos of affine schemes
are precisely open embedding of affine schemes. For instance, if a
is any element of A, then Spec(A/(a)) >
Spec(A) is a disjunctable strong mono with the open embedding
Spec(A_{a}) > Spec(A) as the coflat
complement, which is induced by the localization A > A_{a}
with respect to a.
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