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. n 
 
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 

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 (f.x₁: X₁ ® S, g: Z ® S), and r₂ is also the pullback of (f_°x₂: X₂ ® S, g: Z ® S). 
(b) follows from (a). n 

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

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

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 = g_2. 

Proof. Let x₁, x₂ be the injections of X₁ + X₂ and y₁, y₂ be the injections of Y₁ + Y₂ . Then x_1°f₁ = (f₁ + f₂)_°y₁ and x_1°g₁ = (g₁ + g₂)_°y₁. Since f₁ + f₂ = g₁ + g₂ and x₁ is monic, we have f₁ = g₁. Similarly we obtain f₂ = g₂. n 
 
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: M_X1 ® X₁ and m₂: M_X2 ® 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). n 

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 m_°f₁ = n_°f₁. Consider the maps m + 1_X2 and n + 1_X2 from X₁ + X₂ to M + X₂. We have 

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 (1_Y1, 1_Y1) is the pullback of (f₁, f₁), and (1_Y2, 1_Y2) is the pullback of (f₂, f₂). It follows that (1_Y1 + 1_Y2, 1_Y1 + 1_Y2) 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. n 

Proposition 1.3.9. Suppose {f_i: U_i ® X} is a finite family of morphisms. Denote by f = å f_i: U = å U_i ® X the morphism induced by f_i. The following conditions are equivalent: 
(a) f is a monomorphism. 
(b) Each f_i for is a monomorphism and U_i Ç U_j = 0 for any pair i ¹ j . 
If f: U ® X is a monomorphism then U is the smallest subobject of X containing each subobject U_i. 

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 = f_°e₁ and f₂ = f_°e₂ are monomorphisms. Let (m: M ® U₁, n: M ® U₂) be the pullback of (f₁, f₂) . Then f_°e_1°m = f_1°m = f_2°n = f_°e_2°n. Since f is a monomorphism, we have e_1°m = e_2°n. Since the sum U₁ + U₂ is disjoint, we have M = 0 . Thus U₁ Ç 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 Ç U₂ = U₁ × _X U₂ = U₂ × _X U₁ = U₂ Ç 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 g_°u₁ = f₁ and g_°u₂ = f₂. The induced morphism u = (u₁, u₂): U₁ + U₂ ® V is then a monomorphism because it satisfies (b), and we have g_°u = f. Thus f £ g. This shows that U is the join of {U_i}. n