B3 Exercise 4A (p.19)

To show that (2) ⇒ (3) : Proof. Given R-modules N, B and f ∈ Hom_{R}(N, B), we have a short exact sequence of R-modules 0 ⟶ N ⟶^{f} B ⟶ cokerf ⟶ 0. Applying (2) to this exact sequence with A = N gives an exact sequence of abelian groups/R-modules Hom_{R}(B, N) ⟶^{f_{∗}} Hom_{R}(N, N) ⟶ 0 . That is ∀g ∈ Hom_{R}(N, N) ∃h ∈ Hom_{R}(B, N) : g = f_{∗}(h) = h ∘ f. Taking g = 1_{N} gives (3). □

B6 Exercise 6A (p.30)

(a) Let X, Y ∈ Obj(C), then the zero map 0_{XY} : X ⟶ 0 ⟶ Y does not depend on the zero object chosen. Proof. Let 0 and 0′ be two zero objects of C. Let 0′ ⟶^{φ} 0 be the unique element of Hom_{C}(0′, 0). Similarly, we denote by f ∈ Hom_{C}(X, 0), g ∈ Hom_{C}(0, Y), f′ ∈ Hom_{C}(X, 0′), and g′ ∈ Hom_{C}(0′, Y), the unique elements of each of thoses sets. We then have a commutative diagram :

X ⟶^{f} 0 ⟶^{g} Y
f′ ↘  ^{φ}↑  ↗_{g′}
0′

(For example, φ∘f′ ∈ Hom_{C}(X, 0) by composition, and since it contains only one element, we must have f = φ ∘ f′; similarly for the others compositions). Hence, 0_{XY} = g ∘ f = g ∘ (φ ∘ f′) = (g ∘ φ) ∘ f′ = g′ ∘ f′ = 0′_{XY}, using the associativity of ∘. □

(b) Let C be a category containing a zero object 0. Then it is unique up to isomorphism. Proof. Let 0 and 0′ be two zero objects of C. We denote by f the unique element in Hom_{C}(0, 0′) and g the unique element in Hom_{C}(0′, 0). Then g ∘ f ∈ Hom_{C}(0, 0) and since it has only one element, namely the identity, we must have g ∘ f = 1_{0}. Similarly f ∘ g = 1_{0′} □