Definition 3.1. Suppose f, g: Y --> X is a pair of morphisms in a category. An equalizer of f, g is a morphism k: K --> Y of objects such that fk = gk, which has the universal property that, if m: M --> Y is another morphism of objects with fm = gm, then there is a unique morphism n: M --> K such that m = kn. 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 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 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} = p_{1}s for each i. Example 3.3.1. If {X_{i}} is a set of sets, the product set _{i} X_{i} together with the projections _{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 _{i} X_{i} such that the projections _{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 case is similar. The projections p_{1}, p_{2} of X Y determines two projections q_{1}: Z[X Y] --> X and q_{2}: Z[X Y] --> Y extending p_{1}, p_{2} , which then induces a function t: Z[X Y] --> X Y extending the identity function of X Y such that q_{1} = p_{1}t and q_{2} = p_{2}t. 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 = p_{1} and vs = p_{2}. Also u and v induces a function s': Z --> X Y such that p_{1}s' = u and p_{2}s' = v. Thus we have p_{1}s's = p_{1} and p_{2}s's = p_{2}, which implies that s is injective, therefore bijective. By (1.2) X Y carries an algebraic structure such that s is an isomorphism. Then p_{1} = us and p_{2} = 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 = p_{1}r and v = p_{2}r. Let t': Z[W] --> W be the canonical morphism and r': Z[W] --> Z[X Y] be the morphism induced by r. Then p_{1}rt' = q_{1}t' = p_{1}tr' and p_{2}rt' = q_{2}t' =p_{2}tr' 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 p_{1}: X _{S} Y --> X and p_{2}: X _{S} Y --> Y such that fp_{1} = gp_{2}, 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 = p_{1}t and v = p_{2}t. 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 p_{1}: X _{S} Y --> X sending (a, b) to a, and the projection p_{2}: 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 p_{1}, p_{2}: 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 p_{1}, p_{2} be the projections of Y _{X} Y. Then p_{1}h = p_{2}h, which implies that h(Z[Y _{X} Y]) 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 _{i} X_{i} be the products of X = X_{i} with itself indexed by i. Let f: Y --> X_{i} be the morphism determined by f_{i}: Y --> X = X_{i}. Let g: Y --> 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}: X_{1} --> 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} = sp_{1} for each i. Proposition 3.10. Suppose {X_{i}} is a set of objects. Let r_{i}: X_{i} --> Z[_{i} X_{i}] be the inclusions. Then there is a unique surjective morphism t: Z[_{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 Y] and s: Y --> Z[X Y] be the inclusions. Consider the two morphisms u: Z[X] --> Z[X Y] and v: Z[Y] --> Z[X Y] determined by the inclusion X --> X Y and Y --> X Y. Let a: Z[X] --> X and b: Z[Y] --> Y be the canonical morphisms. Let t: Z[X 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 Y] --> B' be the morphism induced by the function (f, g): X 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. |