1.3. Distributive Properties 

In Section 3 - 6 we consider a fixed analytic category A. The results obtained in Section 3 and 4 also hold for any lextensive categories

Denote by 0 an initial and 1 a terminal object. Recall that in any category an initial 0 is called strict if any map X --> 0 is an isomorphism. The dual notion is a strict terminal object. 

Proposition 1.3.1. (a) The initial 0 is strict. 
{b} Any map 0 --> X is regular. 
{c} If the terminal object 1 is strict then the category is equivalent to the terminal category 1 (i.e. 0 = 1 ). 

Proof. (a) Consider a map f: X --> 0. The pullback of an injection 0 --> 0 + 0 = 0 along f is X. Since 0 + 0 = 0 is stable, X + X is naturally isomorphic to X and the injections X --> X + X are isomorphisms. It follows that X is initial because X + X is disjoint. 
(b) Let (x1, x2): X --> X + X be the injections. Then (x1, x2) is the cokernel pair of 0 --> X. Since 0 is the pullback of (x1, x2), the kernel of (x1, x2) factors through 0. But 0 is strict. Thus the kernel of (x1, x2) is 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 (t1, t2): 1 --> 1 + 1. If 0 is not isomorphic to 1, then t1 and t2 are not isomorphism, which implies that 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

Proposition 1.3.3. (a) Let f: X1 + X2 --> S and g: Z --> S be two maps. Then 

(X1 + X2S Z = X1 S Z + X2 S Z.
{b} Let f: X1 + X2 --> S and g: Y1 + Y2 --> S be two maps. Then 
(X1 + X2S (Y1 + Y2) = X1 S Y1 + X1 S Y2 + X2 S Y1 + X2 S Y2.

Proof. (a) Let (r: T --> X1 + X2, s: T --> S) be the pullback of (f, g). Since X1 + X2 is stable, we have r = r1 + r2 , where r1 is the pullback of (r: T --> X1 + X2, x1: X1 --> X1 + X2), and r2 is the pullback of (r: T --> X1 + X2, x2: X2 --> X1 + X2). But r1 is also the pullback of (fx1: X1 --> S, g: Z --> S), and r2 is also the pullback of (fx2: X2 --> S, g: Z --> S). 
(b) follows from (a). 

Proposition 1.3.4. (a) Let u1: X1 --> T1, v1: Y1 --> T1, u2: X2 --> T2, and v2: Y2 --> T2 be four maps. Then 

X1 T1 Y1 + X2 T2 Y2 = (X1 + X2) T1 + T2 (Y1 + Y2).
(b) (u1: X1 --> T1, x1: X1 --> X1 + X2) is the pullback of 
(t1: T1 --> T1 + T2,   u1 + u2: X1 + X2 --> T1 + T2).

Proof. (a) Let S = T1 + T2. Applying (1.3.3.b) we have 

(X1 + X2) S (Y1 + Y2) = X1 S Y1 + X1 S Y2 + X2 S Y1 + X2 S Y2.
But X1 S Y1 = X1 T1 Y1 and X2 S Y2 = X2 T2 Y2. Also X2 S Y1 = X1 S Y2 = 0 as S = T1 + T2 is disjoint. Thus X1 T1 Y1 + X2 T2 Y2 is the pullback of X1 + X2 and Y1 + Y2 over S. 
(b) is a special case of (a) with Y1 = T1, v1 = 1T1: T1 --> T1, Y2 = 0 and v2: 0 --> T2. 

Proposition 1.3.5. Let f1, g1: Y1 --> X1 and f2, g2: Y2 --> X2 be fours maps. If f1 + f2 = g1 + g2, then f1 = g1 and f2 = g2. 

Proof. Let x1, x2 be the injections of X1 + X2 and y1, y2 be the injections of Y1 + Y2 . Then x1f1 = (f1 + f2)y1 and x1g1 = (g1 + g2)y1. Since f1 + f2 = g1 + g2 and x1 is monic, we have f1 = g1. Similarly we obtain f2 = g2
Proposition 1.3.6. Finite sums commute with equalizers. 

Proof. Let t1: T1 --> X1 be the equalizer of a pair of maps (g1, h1): X1 --> Z1 and let t2: T2 --> X2 be the equalizer of a pair of maps (g2, h2): X2 --> Z2 . Since X1 + X2 is stable, any map m: M --> X1 + X2 has the form m = m1 + m2 with m1: MX1 --> X1 and m2: MX2 --> X2 . Then by (1.3.5) m equalizes (g1 + g2, h1 + h2) iff m1 equalizes (g1, h1) and m2 equalizes (g2, h2). Thus t1 + t2 is the equalizer of (g1 + g2, h1 + h2). 

Proposition 1.3.7. Let f1: Y1 --> X1 and f2: Y2 --> X2 be two maps. Then f1 + f2 is epic iff f1 and f2 are epic. 

Proof. One direction is obvious because any sum of epis is epic. Suppose f1 + f2 is epic. Let (m, n): X --> M be two maps such that mf1 = nf1. Consider the maps m + 1X2 and n + 1X2 from X1 + X2 to M + X2. We have 

(n + 1X2)(f1 + f2) = nf1 + f2 = m°f1 + f2 = (m + 1X2)(f1 + f2).
Since f1 + f2 is epic, we have n + 1X2 = m + 1X2. Thus n = m by (1.3.5), which means that f1 is epic. Similarly f2 is epic. n 
Proposition 1.3.8.  (a) f1 + f2: Y1 + Y2 --> X1 + X2 is a mono (resp. strong mono, resp. regular mono) if and only f1: Y1 --> X1 and f2: Y2 --> X2 are so. 
{b} The injections x: X --> X + Y and y: Y --> X + Y are regular monos

Proof. (a) If f1 + f2: Y1 + Y2 --> X1 + X2 is a mono (resp. strong mono, resp. regular mono) then each of f1 and f2 is so by (1.3.4)(b) because monos (resp. strong monos, resp. regular monos) are stable under base extension. Next suppose f1 and f2 are monos. Then (1Y1, 1Y1) is the pullback of (f1, f1), and (1Y2, 1Y2) is the pullback of (f2, f2). It follows that (1Y1 + 1Y2, 1Y1 + 1Y2) is the pullbacks of (f1 + f2, f1 + f2) by (1.3.4.a). Thus f1 + f2 is monic. If f1 and f2 are strong monos, then any pullback of f1 + f2 is not non-isomorphic epic by (1.3.4) and (1.3.7), thus f1 + f2 is a strong mono. Finally assume that f1 and f2 are regular monos. Suppose f1 is the equalizer of a pair of maps (u1, u2): X1 --> U and f2 is the equalizer of (v1, v2): X1 --> V. Applying (1.3.6) we see that f1 + f2 is the equalizer of (u1 + v1, u2 + v2). Thus f1 + f2 is a regular mono. 
(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. 

Proposition 1.3.9. Suppose {fi: Ui --> X} is a finite family of morphisms. Denote by f =  fi: U =  Ui --> X the morphism induced by fi. The following conditions are equivalent: 
(a) f is a monomorphism. 
(b) Each fi for is a monomorphism and Ui  Uj = 0 for any pair j
If f: U --> X is a monomorphism then U is the smallest subobject of X containing each subobject Ui

Proof. It suffices to prove the assertion for a pair of morphisms f1: U1 --> X and f2: U2 --> X. Let e1: U1 --> U1 + U2 and e2: U2 --> U1 + U2 be the injections. 
Suppose f: U1 + U2 --> X is a monomorphism. The injections e1 and e2 are monomorphisms. Thus f1 = fe1 and f2 = fe2 are monomorphisms. Let (m: M --> U1, n: M --> U2) be the pullback of (f1, f2) . Then fe1m = f1m = f2n = fe2n. Since f is a monomorphism, we have e1m = e2n. Since the sum U1 + U2 is disjoint, we have M = 0 . Thus U1  U2 = 0. Hence (a) implies (b). 
Conversely assume (b) holds. Since f1 and f2 are monomorphisms, U1 X U1 = U1 and U2 X U2 = U2 . Also U1  U2 = U1 X U2 = U2 X U1 = U2  U2 = 0. Applying (1.3.3.b) we see that the pullback of f: U1 + U2 --> X with itself over X is U1 X U1 + U1 X U2 + U2 X U1 + U2 X U2 = U1 + U2. This means that f is a monomorphism. 
Finally we assume f: U --> X is a monomorphism. Suppose g: V --> X is a monomorphism such that f1  g and f2  g . Then there are monomorphisms u1: U1 --> V and u2: U2 --> V such that gu1 = f1 and gu2 = f2. The induced morphism u = (u1, u2): U1 + U2 --> V is then a monomorphism because it satisfies (b), and we have gu = f. Thus g. This shows that U is the join of {Ui}. 

 [Next Section][Content][References][Notations][Home]