Sketched Solutions to Exercises

(ii) Denote this map by φ, which is induced by the compositions with εi : Xi ↪→ ⊔ j Xj , i ∈ I. If φ(fi) = φ(gj) for some fi ∈ Hom(Y,Xi) and gj ∈ Hom(Y,Xj), then we must have i = j because φ(fi) and φ(gj) have the same image. Since φ(fi) = εi ◦ fi, φ(gi) = εi ◦ gi and εi is an injective map, we deduce that fi(y) = gi(y), ∀y ∈ Y , i.e. fi = gi. (iii) This map is induced by the composition with the projections pi : ∏ j Xj → Xi. As an example where this map is not injective, take Y = {pt}. Then Hom( ∏ iXi, Y ) = {pt} while ⊔ iHom(Xi, Y ) = ⊔ i{pt} has the same cardinality as that of I. So the map is not injective as long as #I ≥ 2. 2.2 A morphism from (Z ′, f ′, g′) to (Z, f, g) in D is a morphism φ : Z ′ → Z in C such that f ◦φ = f ′ and g ◦ φ = g′. Then (Z, f, g) is a terminal object in D if and only if the map HomC(Z ′, Z)→ HomC(Z ′, X)×HomC(Z ′, Y ), φ 7→ (f ◦ φ, g ◦ φ) is a bijection for each Z ′ ∈ C, in which case the inverse map takes (f ′, g′) to the unique morphism from (Z ′, f ′, g′) to (Z, f, g) in D. This is equivalent to Z ∼= X × Y in C, and f : Z → X, g : Z → Y are the natural morphisms.