3. Products and Coproducts 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 {Xi} is a set of objects in a category. A product of {Xi} is an object X, together with a collection of morphisms pi: X --> Xi, with the universal property that, if Y is another object together with a collection of morphisms ti: Y --> Xi, there is a unique morphism s: Y --> X such that ti = p1s for each i.  Example 3.3.1. If {Xi} is a set of sets, the product set i Xi together with the projections i Xi --> Xi is the product of {Xi} in the category of sets.  Proposition 3.4. Suppose {Xi} is a set of objects. Then there is unique algebra structure on the product set i Xi such that the projections i Xi --> Xi are morphisms, which is the product of objects {Xi} 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 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]) 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 fi, gi: Y --> X is a set of pairs of morphisms. A common coequalizer of {fi, gi} is a morphism k: X --> K of objects such that kfi = kgi for all i, with the universal property that, if m: X --> M is another morphism of object with mfi = mgi for all i, then there is a unique morphism n: K --> M such that m = nk.  Proposition 3.8. Any set fi, gi: Y --> X of pairs of morphisms has a common coequalizer.  Proof. Let i Xi be the products of X = Xi with itself indexed by i. Let f: Y --> Xi be the morphism determined by fi: Y --> X = Xi. Let g: Y --> Xi be the morphism determined by gi: Y --> X = Xi. Then the equalizer of f, g is the common equalizer of fi, gi.  Definition 3.9. Suppose {Xi} is a set of objects. A coproduct of {Xi} is an object X, together with a collection of morphisms qi: X1 --> X, with the universal property that, if Y is another object together with a collection of morphisms ti: Xi --> Y, there is a unique morphism s: X --> Y such that ti = sp1 for each i.  Proposition 3.10. Suppose {Xi} is a set of objects. Let ri: Xi --> Z[ i Xi] be the inclusions. Then there is a unique surjective morphism t: Z[ i X] --> B such that each tri is a morphism and (B, {tri}) is the coproduct of {Xi}.  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.  [Next Section][References][Notations][Home] 