1.3. Distributive Properties In Section 3 - 6 we consider a fixed analytic
category Denote by
x):_{2} X --> X +
X be the injections. Then (x _{1},x)
is the cokernel pair of _{2}0 --> X. Since 0 is the pullback
of (x _{1},x), the kernel of (_{2}x
_{1},x) factors through _{2}0. But 0 is strict.
Thus the kernel of (x _{1},x) is _{2}0.
This shows that 0 --> X is a regular mono.
(c) Since 0 --> 1 is a regular mono it is the kernel of its
cokernel pair (t _{1},t):_{2} 1 --> 1
+ 1. If 0 is not isomorphic to 1, then t
and _{1}t are not isomorphism, which implies that _{2}1
is not strict.
Remark 1.3.2. It follows from (1.3.1.a)
and (c) that a category and its opposite are both analytic (or lextensive)
iff it is equivalent to the terminal category 1 .
g: Z --> S be
two maps. Then
X) _{1} + X_{2} = _{S}
ZX._{1} _{S}
Z + X_{2} _{S}
Zf: X and _{1} + X_{2} --> Sg:
Y be two maps. Then
_{1} + Y_{2} --> SX) _{1} + X_{2}
(_{S}Y) =_{1} + Y_{2} X_{1} _{S}
Y_{1} + X_{1} _{S}
Y_{2} + X_{2} _{S}
Y_{1} + X_{2} _{S}
Y_{2}.
s: T --> S) be the pullback of (f, g). Since
X is stable, we have _{1} + X_{2}r = r , where _{1}
+ r_{2}r is the pullback of (_{1}r:
T --> X _{1} + X_{2},x:_{1} X)_{1}
--> X_{1} + X_{2}, and r
is the pullback of (_{2}r: T --> X
_{1} + X_{2},x:_{2} X)_{2} --> X_{1} + X_{2}.
But r is also the pullback of (_{1}fx:_{1}
X _{1} --> S,g: Z --> S), and r
is also the pullback of (_{2}fx:_{2} X
_{2} --> S,g: Z --> S).
(b) follows from (a).
X _{1} --> T_{1},v:_{1} Y _{1}
--> T_{1},u:_{2} X
and _{2} --> T_{2},v:_{2} Y be four
maps. Then
_{2} --> T_{2}X(_{1} _{T1}
Y_{1} + X_{2} _{T2}
Y_{2} = X)_{1} + X_{2} (_{T1}
+ T_{2}Y)._{1} + Y_{2}u:_{1} X _{1} --> T_{1},x:_{1}
X) is the pullback of
_{1} --> X_{1} + X_{2}t:_{1} T:_{1} --> T_{1} + T_{2},
u_{1} + u_{2} X)._{1} + X_{2} -->
T_{1} + T_{2}
X)_{1} + X_{2} (_{S}
Y)_{1} + Y_{2} = X_{1} _{S}
Y_{1} + X_{1} _{S}
Y_{2} + X_{2} _{S}
Y_{1} + X_{2} _{S}
Y_{2}.X
_{1} _{S}
Y_{1} = X_{1} _{T1}Y_{1} and X
_{2} _{S}
Y_{2} = X_{2} _{T2}Y. Also _{2}X as _{2} _{S}
Y_{1} = X_{1} _{S}
Y_{2} = 0S = T is disjoint.
Thus _{1} + T_{2}X
_{1} _{T1}Y + _{1}X _{2}_{T2}Y is the pullback of _{2}X
and _{1} + X_{2}Y over _{1} + Y_{2}S.
(b) is a special case of (a) with Y,
_{1} = T_{1}v: _{1} = 1_{T1}T, _{1}
--> T_{1}Y and _{2} = 0v:_{2}
0 --> T
_{2}.
g:_{1} Y and _{1} --> X_{1}f
_{2},g:_{2} Y be fours maps.
If _{2} --> X_{2}f
then _{1} + f_{2} = g_{1} + g_{2},f and _{1} = g_{1}f
_{2} = g_{2.}
X and _{1} + X_{2}y
be the injections of _{1}, y_{2}Y . Then _{1} + Y_{2}x(_{1}f_{1}
= f)_{1} + f_{2}y and _{1}x(_{1}g_{1}
= g)_{1} + g_{2}y. Since
_{1}f and
_{1} + f_{2} = g_{1} + g_{2}x is monic, we have _{1}f.
Similarly we obtain _{1} = g_{1}f.
_{2} = g_{2}Proposition 1.3.6. Finite sums commute
with equalizers.
T
be the equalizer of a pair of maps (_{1} --> X_{1}g _{1},h):_{1}
X and let _{1} --> Z_{1}t:_{2} T be the equalizer of a pair of maps (_{2}
--> X_{2}g
_{2},h):_{2} X . Since _{2} --> Z_{2}X is stable, any map _{1}
+ X_{2}m: M --> X has the form _{1}
+ X_{2}m = m
with _{1} + m_{2}m:_{1} M -->
X_{X1}_{1} and m:_{2} M
--> _{X2}X . Then by (1.3.5) _{2}m
equalizes (g _{1} + g_{2},h)
iff _{1} + h_{2}m equalizes (_{1}g)
and _{1}, h_{1}m equalizes (_{2}g).
Thus _{2}, h_{2}t is the equalizer of (_{1} + t_{2}g _{1}
+ g_{2},h).
_{1} + h_{2}
Yand _{1} --> X_{1} f:_{2} Y be two maps. Then _{2}
--> X_{2}f
is epic iff _{1} + f_{2}f and _{1}f are epic.
_{2}
m, n):
X --> M be two maps such that mf
Consider the maps _{1} = nf_{1}.m + 1 and _{X2}n
+ 1 from _{X2}X
to _{1} + X_{2}M + X. We have
_{2}n + 1)(_{X2}f)_{1}
+ f_{2} = nf_{1} + f_{2} = m_{°}f(_{1}
+ f_{2} = m + 1)(_{X2}f)_{1}
+ f_{2}.f is epic, we have _{1} + f_{2}n + 1
= _{X2}m + 1. Thus _{X2}n = m
by (1.3.5), which means that f
is epic. Similarly _{1}f is epic. n
_{2}Proposition 1.3.8. (a) f:_{1}
+ f_{2} Y is a mono (resp. strong mono, resp. regular mono) if
and only _{1} + Y_{2} --> X_{1}
+ X_{2}f:_{1} Y and
_{1} --> X_{1}f:_{2} Y are so.
_{2} --> X_{2}{b} The injections x: X --> X + Y and y: Y
--> X + Y are regular monos.
Y is a mono (resp.
strong mono, resp. regular mono) then each of _{1}
+ Y_{2} --> X_{1} + X_{2}f and
_{1}f is so by (1.3.4)(b) because monos
(resp. strong monos, resp. regular monos) are stable under base extension.
Next suppose _{2}f and _{1}f are monos. Then
(_{2}1, _{Y1}1)
is the pullback of (_{Y1}f _{1},f)_{1},
and (1, _{Y2}1)
is the pullback of (_{Y2}f _{2},f)_{2}.
It follows that (1 + _{Y1}1,
_{Y2}1 + _{Y1}1)
is the pullbacks of (_{Y2}f _{1} + f_{2},f) by (1.3.4.a). Thus _{1}
+ f_{2}f is monic. If _{1}
+ f_{2}f and _{1}f
are strong monos, then any pullback of _{2}f
is not non-isomorphic epic by (1.3.4) and (1.3.7),
thus _{1} + f_{2}f is a strong mono. Finally assume
that _{1} + f_{2}f and _{1}f are regular monos. Suppose
_{2}f is the equalizer of a pair of maps (_{1}u):_{1},
u_{2} X and _{1} --> Uf
is the equalizer of (_{2}v):_{1}, v_{2} X. Applying (1.3.6) we see that _{1}
--> Vf is the equalizer of (_{1}
+ f_{2}u). Thus _{1} + v_{1},
u_{2} + v_{2}f
is a regular mono.
_{1} + f_{2}(b) x is the sum of the identity map X --> X and the
regular mono 0 --> Y . Thus it is regular by (a). Similarly y
is regular.
U} is a finite family of morphisms. Denote by _{i} --> Xf
= f:_{i} U
= U
the morphism induced by _{i} --> Xf. The following conditions
are equivalent:
_{i}(a) f is a monomorphism.
(b) Each f for is a monomorphism and _{i}U for any pair _{i}
U_{j} = 0i
j .
If f: U --> X is a monomorphism then U is the
smallest subobject of X containing each subobject U.
_{i}
U and _{1} --> Xf:_{2}
U. Let _{2} --> Xe:_{1} U and _{1} -->
U_{1} + U_{2}e:_{2} U be the injections.
_{2}
--> U_{1} + U_{2}Suppose f: U is a monomorphism.
The injections _{1} + U_{2} --> Xe and _{1}e are monomorphisms.
Thus _{2}f and _{1} = fe_{1}f
are monomorphisms. Let (_{2} = fe_{2}m: M --> U, _{1}n:
M --> U) be the pullback of (_{2}f _{1},f)
. Then _{2}fe.
Since _{1}m = f_{1}m = f_{2}n = fe_{2}nf is a monomorphism, we have e.
Since the sum _{1}m = e_{2}nU is disjoint, we have
_{1} + U_{2}M = 0 . Thus U. Hence (a) implies (b).
_{1}
U_{2} = 0Conversely assume (b) holds. Since f and _{1}f
are monomorphisms, _{2}U and _{1} _{X}
U_{1} = U_{1}U . Also _{2} _{X}
U_{2} = U_{2}U. Applying (1.3.3.b) we see that
the pullback of _{1}
U_{2} = U_{1} _{X}
U_{2} = U_{2} _{X}
U_{1} = U_{2}
U_{2} = 0f: U with
itself over _{1} + U_{2} --> XX is U. This means that _{1} _{X}
U_{1} + U_{1} _{X}
U_{2} + U_{2} _{X}
U_{1} + U_{2} _{X}
U_{2} = U_{1} + U_{2}f
is a monomorphism.
Finally we assume f: U --> X is a monomorphism. Suppose
g: V --> X is a monomorphism such that f and _{1}
gf . Then there are monomorphisms _{2}
gu:_{1} U and _{1}
--> Vu:_{2} U such that
_{2} --> Vgu and _{1} = f_{1}gu.
The induced morphism _{2} = f_{2}u = (u _{1},u):_{2}
U is then a monomorphism because
it satisfies (b), and we have _{1} + U_{2} --> Vgu = f. Thus f
g. This shows that U is the join of {U}.
_{i} |