The equalizer of a pair of morphisms, if exists, is uniquely determined up to an isomorphism. 

Example 3.1.1. Suppose u, v: Y --> X is a pair of functions of sets. Let K = ker(u, v) = {a Î Y | u(a) = v(a)}. Then the inclusion function k: K --> Y is an equalizer of u, v in the category of sets. 

Proposition 3.2. Suppose f, g: Y --> X is a pair of morphisms. Let K = ker(f, g) = {a Î Y | f(a) = g(a)} be the equalizer of f, g in the category of sets. Then K is a closed subset of Y. The subobject K with the inclusion morphism is a equalizer of f, g in A. 

Proof. Let d: Z[K] --> Y be the morphism induced by inclusion of K --> Y. Then fd = gd. So d(Z[K]) = K as K is the equalizer of f, g in the category of sets. Thus K = G(K) implies that K is a closed subset of Y. Any morphism t: T --> Y such that ft = gt factors through K in a function m, which must be a morphism by (2.8.c). Thus K is the equalizer of f, g in A. 

Definition 3.3. Suppose {X_i} is a set of objects in a category. A product of {X_i} is an object X, together with a collection of morphisms p_i: X --> X_i, with the universal property that, if Y is another object together with a collection of morphisms t_i: Y --> X_i, there is a unique morphism s: Y --> X such that t_i = p₁s for each i. 

Example 3.3.1. If {X_i} is a set of sets, the product set Õ_i X_i together with the projections Õ_i X_i --> X_i is the product of {X_i} in the category of sets. 

Proposition 3.4. Suppose {X_i} is a set of objects. Then there is unique algebra structure on the product set Õ_i X_i such that the projections Õ_i X_i --> X_i are morphisms, which is the product of objects {X_i} in A. 

Proof. For simplicity we give the proof for the product of two objects X and Y. The general cases is similar. The projections p₁, p₂ of X × Y determines two projections q₁: Z[X × Y] --> X and q₂: Z[X × Y] --> Y extending p₁, p₂ , which then induces a function t: Z[X × Y] --> X × Y extending the identity function of X × Y such that q₁ = p₁t and q₂ = p₂t. Let s: X × Y --> Z be the surjective generic extension of t. Then there are morphisms u: Z --> X and v: Z --> Y such that us = p₁ and vs = p₂. Also u and v induces a function s': Z --> X × Y such that p₁s' = u and p₂s' = v. Thus we have p₁s's = p₁ and p₂s's = p₂, which implies that s is injective, therefore bijective. By (1.2) X × Y carries an algebraic structure such that s is an isomorphism. Then p₁ = us and p₂ = vs are morphisms. 

Next consider two morphisms u: W --> X and v: W --> Y. They induces a function r: W --> X × Y such that u = p₁r and v = p₂r. Let t': Z[W] --> W be the canonical morphism and r': Z[W] --> Z[X × Y] be the morphism induced by r. Then p₁rt' = q₁t' = p₁tr' and p₂rt' = q₂t' =p₂tr' implies that rt' = tr'. Since t' is surjective and tr' is a morphism of objects, by (2.3.b) r is a morphism. This shows that the object X × Y is the product of X with Y. The uniqueness of the algebraic structure on X × Y also follows from this and (1.2). 

Definition 3.5. Let f: Y --> S and g: X --> S be two morphisms of objects. A fibred product of X and Y over S is an object X × _S Y, together with two morphisms p₁: X × _S Y --> X and p₂: X × _S Y --> Y such that fp₁ = gp₂, which has the universal property that, if u: Z --> X and v: Z --> Y are two morphisms of objects such that fu = gv, there is a unique morphism t: Z --> X × _S Y such that u = p₁t and v = p₂t. 

Example 3.5.1. Let f: Y --> S and g: X --> X be two functions of sets. Let X × _S Y = {(a, b)| a in X and b in Y such that f(a) = g(b). Then X × _S Y together with the projection p₁: X × _S Y --> X sending (a, b) to a, and the projection p₂: X × _S Y --> Y sending (a, b) to b, is the fibre product of X and Y over S in the category of sets. 

Proposition 3.6. Suppose f: Y --> X is a morphism of objects. Then Y × _X Y is a closed subset of Y × Y. The subobject Y × _X Y, together witht the projections p₁, p₂: Y × _X Y is the fibre product of Y and X over S. 

Proof. Consider the morphism h: Z[Y × _X Y] --> Y × Y determined by the inclusion Y × _X Y --> Y × Y. Let p₁, p₂ be the projections of Y × _X Y. Then p₁h = p₂h, which implies that h(Z[Y × _X Y]) \subseteq Y × _X Y. Thus Y × _X Y is a closed subset of Y × Y. It is straightforward to verify that Y × _S X is the fibre product of Y and X over S. 

Definition 3.7. Suppose f_i, g_i: Y --> X is a set of pairs of morphisms. A common coequalizer of {f_i, g_i} is a morphism k: X --> K of objects such that kf_i = kg_i for all i, with the universal property that, if m: X --> M is another morphism of object with mf_i = mg_i for all i, then there is a unique morphism n: K --> M such that m = nk. 

Proposition 3.8. Any set f_i, g_i: Y --> X of pairs of morphisms has a common coequalizer. 

Proof. Let Õ_i X_i be the products of X = X_i with itself indexed by i. Let f: Y --> Õ X_i be the morphism determined by f_i: Y --> X = X_i. Let g: Y --> Õ X_i be the morphism determined by g_i: Y --> X = X_i. Then the equalizer of f, g is the common equalizer of f_i, g_i. 

Definition 3.9. Suppose {X_i} is a set of objects. A coproduct of {X_i} is an object X, together with a collection of morphisms q_i: X₁ --> X, with the universal property that, if Y is another object together with a collection of morphisms t_i: X_i --> Y, there is a unique morphism s: X --> Y such that t_i = sp₁ for each i. 

Proposition 3.10. Suppose {X_i} is a set of objects. Let r_i: X_i --> Z[È_i X_i] be the inclusions. Then there is a unique surjective morphism t: Z[È_i X] --> B such that each tr_i is a morphism and (B, {tr_i}) is the coproduct of {X_i}. 

Proof. For simplicity we give the proof for the coproduct of two objects X and Y. The general case is similar. Let r: X --> Z[X È Y] and s: Y --> Z[X È Y] be the inclusions. Consider the two morphisms u: Z[X] --> Z[X È Y] and v: Z[Y] --> Z[X È Y] determined by the inclusion X --> X È Y and Y --> X È Y. Let a: Z[X] --> X and b: Z[Y] --> Y be the canonical morphisms. Let t: Z[X È Y] --> B be the common coequalizer of (u, ra) and (v, sb). First tra = tu and a is a surjective morphism, and tu is a morphism implies that tr is a morphism. Similarly ts is a morphism. Next assume f: X --> B' and g: Y --> B' are any two morphisms. Let t': Z[X È Y] --> B' be the morphism induced by the function (f, g): X \cup Y --> T. Then hu = fa. But f = hr. So hu =hra. Similarly hv = hsb. Since t is the common coequalizer of u, ra and v, sb, there is a unique morphism k: B --> T such that h = kt. Then f = hr = ktr and g = hs = kts. The uniqueness of k is obvious. This shows that (B, tr, ts) is a sum of X, Y.