Algebraic Geometry: Limits and Coits Limits Definition.. Let I be a small category, C be any category, and F : I C be a functor. If for each object i I and morphism m ij Mor I (i, j) there is an associated C i = F(i) C and commuting morphisms h ij = F(m ij ) Mor(C ), then the collection (F, C i, h ij ) is called a diagram indexed by I. Definition.2. The it of a diagram is an object C i C together with a collection of morphisms f s : C i C s, which satisfies the following equivalent properties: C i!g C i C j h jk =F(m jk ) C k g k C k The first diagram commutes for every morphism m jk Mor I (k, j) and the collection ( C i, f s ) is universal(final) with respect to this property. In the second diagram, for any obj(c ) and maps such that g k = h jk g j for every m jk Mor I (k, j), there exists a unique morphism g such that the diagram commutes for every k I. By universality, if the it of a diagram exists then it is unique. Exercise.3. Construct fibred products, usual products, and equalizers as its of diagrams. Proof. Take I to be any set such that Mor I (i, j) = if i j, so the only morphisms are the identity maps. Thus, using the (second) universal property we have that C i = i I C i!g g k C k where each is the usual projection map and g k is arbitrary. Take I to be any set {i, i 2, i 3 } such that Mor I (i, i 3 ) = {m 3 }, Mor I (i 2, i 3 ) = {m 23 }, Mor I (i, i 2 ) = and Mor I (i j, i k ) = i > k, and the other morphisms are the identity maps. For any category C, using the (first) universal property we have that i i 2 m 23 i 3 m 3 C i f i2 f i C i C i2 C i3 F(m 23 ) F(m 3 ) where f, f 2 are the appropriate projections onto C and C 2 ; thus, by universality C i is the same as the fibred product C i Ci3 C i2. Take I to be any set {i, i 2 } such that Mor I (i, i 2 ) = {m, n}, Mor I (i 2, i ) =, and the other morphisms are the identity maps. For any category C, using the (first) universal property
we have the commutative diagram on the right n i i 2 m C i f i C i F(m) F(n) f i2 C i2 where f i and f i2 are the maps satisfying f i2 = F(m) f i = F(n) f i so by uniqueness, C i is the same as the equalizer of F(m) and F(n). Note that in the case where Mor I (i, i 2 ) = then the it is just ( C i, f i ) for any isomorphism f i. Exercise.4. Prove that in the category Sets, the object A together with the obvious projection maps to each A i constitute the it A i { } A = (a i ) i I i I A i : F(m jk )(a j ) = a k, m jk Mor I (j, k) Mor(I ) Proof. e recall that the product i I A i would itself be the it of the diagram (F, A i, {}); that is if there are no non-identity morphisms in Mor(I ). Otherwise, consider the diagram (F, A i, h jk ) where h jk = F(m jk ), then place the it in the following composite diagram g j A A i g k A j h jk A k in which we wish to show that the map exists and is unique. By construction, the outer triangle commutes by setting the projection g j ((a i ) i I ) = a j and analogously for g k, so it follows that g k = h jk g j. Since we are working in the category of sets, it makes sense to consider the elementwise description of the map. For a given element u A i let (u) = b A j and (u) = c A k, so h jk (b) = c. Defining (u) = (f i (u)) i I it is easy to that exists, since the image belongs to A; explicitly, we have h jk ( (u)) = ( (u)) for all m jk Mor I (j, k). Furthermore, g i (u) = f i (u) for every i I, so the upper triangles of the diagram commute, that is, the maps f i factor through ; thus, by universality, A is the it of the diagram (F, A i, h jk ) given that is unique. To establish uniqueness, suppose that there exists a map φ satisfying g i φ(u) = f i (u), i I, then by the definition of g i which necessitates that φ =. g i φ(u) = g i ((a i ) i I ) = a i = f i (u) i I, u A i Limits need not exist for every diagram of an index set, particularly if I is not a small category. However, we have the following existence theorem. Theorem.5. Let I be an index category with D F = (F, F(i), F(m jk )) being a diagram indexed by I, and C F(i) a category. The category C possesses its for I, if C has equalizers and has all products which are indexed by obj(i ) and Mor(I ). 2
Proof. Let F : I C be a functor and suppose products exist as stated in the theorem, and for any morphism m Mor(I ) denote by Ω m the codomain of m; that is, the object j such that m Mor I ( i, j). P = F(i) Q = F(Ω m ) i obj(i ) m Mor(I ) Now consider the parallel morphisms p : P Q and p 2 : P Q, where Θ m denotes the domain of m; that is, the object i such that m Mor I (i, j). p = π Ωm p 2 = F(m) π Θm m Mor(I ) m Mor(I ) Here we have denoted by π Ωm the projection from P onto the subproduct of objects of the form F(Ω m ), and π Θm is defined analogously. Under the assumption of existence of equalizers we know that the equalizer of p, p 2 exists; we now prove that this equalizer E is the it of D F. E e φ j!φ E φ k e 2 F(j) g j g k F(m jk ) F(k) P p p 2 Q By definition, Ω mjk = k and Θ mjk = j so if π i : P F(i) is the obvious projection, we set = π j e = π k e By commutativity of the bottom triangle in the left-hand diagram, the maps φ j and φ k in the right-hand diagram exist and are unique. In particular, from the universal property of products the bottom triangles commute in the following diagram P π Θm F(Θ m ) g j!h p p 2!g Q π Ωm g k F(Ω m ) F(m) and by definition of p, p 2 the outer square commutes. Thus we can take φ j = h and φ k = g, then by universality of the equalizer we know φ exists and is unique, so it suffices to define = φ. f i = π i e = π i e φ = π i φ i = g i That = F(m jk ) under this choice of follows from the fact that e p = e p 2 = e 2. Theorem.6. The covariant functor H A = Hom C (A, ) preserves its in any locally small category C. Proof. Let I be an index category with D F = (F, C i, h ij ) being a diagram indexed by I, and C a locally small category. Assuming that its exist, this gives rise to the commutative 3
diagram on the left, C i H A ( ) H A ( C i ) H A ( ) C j h jk C k H A (C j ) H A (h jk ) H A (C k ) which for some A obj(c ) is mapped to the right diagram under the H A functor. Now, consider the diagram D H A = (H A, H A (C i ), g ij ) indexed by obj(f(i )) C i. It is immediate that g ij = H A (h jk ), and so the associated it of this diagram can be recognized as the inner commutative triangle of the diagram H A ( ) H A (C j ) H A ( ) H A ( C i )!φ HA (C i ) H A (h jk ) H A ( ) H A ( ) H A (C k ) It is seen if we write out the explicit set descriptions that φ is the unique inclusion morphism on H A ( C i ), so is the inclusion morphism on H A (C i ); it follows that φ = φ =. 2 Coits Definition 2.. The coit of a diagram is an object C i C together with a collection of morphisms f s : C s C i, such that by reversing the arrows on the diagrams in Definition.2 we retain commutativity. By universality the coit of a diagram is unique if it exists. Exercise 2.2. Cofibred products, usual coproducts, and coequivalences as coits of diagrams. Analogous to the exercise above, the coproduct i I exists over any discrete index category, and is obtained by reversing the arrows in the definition of product. Take I to be any set {i, i 2, i 3 } such that Mor I (i 3, i ) = {m 3 }, Mor I (i 3, i 2 ) = {m 32 }, Mor I (i 2, i ) = and Mor I (i j, i k ) = i < k, and the other morphisms are the identity maps. For any category C, using the (first) universal property we have that i i 2 m 32 i 3 m 3 C i f i2 C i2 f i F(m 32 ) C i C i3 F(m 3 ) where f, f 2 are the appropriate inclusions into C i ; thus, by universality C i is the same as the cofibred product C i C i 3 C i2. Taking I to be any set {i, i 2 }, and Mor I (i, i 2 ) = {m, n}, Mor I (i 2, i ) =, the coit of the diagram is the coequalizer. n i i 2 m C i f i C i F(m) F(n) f i2 C i2 4
where f i and f i2 are the maps satisfying f i = f i2 F(m) = f i2 F(n). Unlike for equalizers, in the case where Mor I (i, i 2 ) = then the coit is just ( C i, f i2 ) for any isomorphism f i2. Exercise 2.3. (a) Provide an explicit construction showing that Q = Z. N n (b) Let S be a set and C be the category with objects C obj(c ) if C S. Provide a construction showing that the union of certain of these subsets is a coit. Proof. Take C to be the obvious category, and N as a partially ordered index set but with the relation Mor N (a, b) = { } if and only if a divides b; otherwise Mor N (a, b) =. Now, we define the functor according to: F(a) = Z, F( ) = h a ab : Z Z defined by h a b ab( z ) = z ab = w; a a ab b the indexed diagram is thus given by D F = (F, Z, h n ab). e consider the following diagram g b f b Q f a g a Z b h ab Z a where obj(c ) is any object satisfying g a = g b h ab. The inner triangle is commutative, with inclusion maps f n : Z Q defined by f n n(x) = [x], where [x] denotes the equivalence class of x. Hence, denoting gcd(z, n) = α and p = n/α, it suffices to define ([z/n]) = g p ( z/α ) = g n/α p( z ), p providing a well-defined map which makes the upper triangles commute. ( z ) ([ ( ) ( ) z z z ( z ) f n = = g p = g n h pn = g n n n]) z p p n n n Z where h pn always exists, since gcd(z, n) always exists. For uniqueness of it suffices to note that the choice of representative for [z/n] does not matter; if w/m [z/n] and gcd(w, m) = β then p = m/β and w/β = z. It now follows from universality that Q along with the maps f n form the coit of the diagram D F, as intended. Suppose that C i obj(c ) and A = i I C i, where I is an arbitrary set which has the form of a discrete category. Thus, Mor I (j, k) = {id I } if and only i = k, and we obtain the indexed diagram D F = (F, C i, h jj ), where h jj : C j C j is the identity map. Furthermore, it is quickly verified that C forms a poset category under the stipulation Mor C (A, B) = { } if and only if A B. The diagram to check for commutativity is thus simply A g j C j where obj(c ) is any object, and : C j A and g j : C j are set inclusion. It is clear that = id C is the unique map establishing commutativity, thus A along with the inclusion maps form the coit of the diagram D F. Theorem 2.4. Let I be an index category with D F = (F, C i, h ij ) being a diagram indexed by I, and C C i a category. The category C possesses coits for I, if C has coequalizers and has all coproducts which are indexed by obj(i ) and Mor(I ). Proof. Dualize the proof for existence of its. 5
Restricting attention to the category Sets, we recall that a directed set is a non-empty set with associated preorder (Λ, ) satisfying: α, β Λ, c Λ such that α c and β c. Coits exist for such index categories, and eschewing some of the categorical perspective, we can prove some interesting results concretely. Proposition 2.5. The coit of exact sequences over a directed set Λ, is exact. More generally, in the case that D F = (F, C α, f αβ ) is a directed system of chain complexes, then the coit commutes with homology.. H n(c α ) = H n ( C α ) = H n (C) Proof. Let Λ be a directed set and A = (F, A α, f ij ) be a directed system of unitary R- modules. Similarly consider a class of given directed systems {A, B, } where for each α i the set {A αi, B αi, } forms an exact sequence. This gives rise to an associated commutative diagram where the exact sequences form the rows and directed systems form the columns. A αi a αi f ij A αj a αj k A αk a αk b αi c αi B αi C αi g ij h ij b αj c αj B αj C αj g jk h jk b αk c αk B αk C αk Taking the coits of each column results in the chain with maps induced by {a α, b α, } A α a B α b C α c Let [b] Ker(b) B α, then there exists a representative b i B αi such that ψ i (b i ) = [b], and so given φ i : C αi C α we have φ i (b αi (b i )) = [0]. This means that there exists α m α i such that h im (b αi (b i )) = 0 m C αm, and since h im = h ij h jm we have the relation h im b αi = b αm g im. Thus, b αm (g im (b i )) = 0 m C αm which shows that g im (b i ) = b m Ker(b αm ), and by exactness there exists a m A αm such that b m = a αm (a m ). If i : A αi A α, by commutativity we have the relation a( m (a m )) = ψ m (a αm (a m )) = ψ m (b m ) = [b], but then m (a m ) = [a] is the element whose image under a belongs to Ker(b). It follows that Ker(b) Im(a) which gives one side of the of proof exactness between these two maps. Now, let [b] Im(a) B α, then a[a] = [b] and there exists representatives a i A αi, b m B αm such that i (a i ) = [a] and ψ m (b m ) = [b]. From commutativity and exactness we have b m Im(a αm ) so b m = a αm (f im (a i )) b[b] = ba( i (a i )) = b(ψ m (b m )) = φ m (b αm (b m )) = φ m (b αm (a αm (a m ))) = φ m (0 m ) = [0] C α Thus, Im(a) Ker(b) which proves exactness. Repeating this argument for every adjacent pair of coit modules proves that they form an exact sequence. 6
Set f α : C α C and let h αβ : H n (C α ) H n (C β ), also φ α : H n (C α ) H n (C α ) and ψ α : H n (C α ) H n (C). e are interested in proving injectivity and surjectivity of the map h : H n (C α ) H n (C) where hφ α = ψ α. Firstly, given a homology class [c] H n (C) then we can find a representative c α C α such that f α (c α ) = [c]. e will assume that all boundary operators have been suitably chosen within the respective complexes and so will denote them all by. Since [c] = [0], then [0] = (f α (c α )) = f α ( (c α )) which means that (c α ) is a representative of [0] and so c α belongs to a homology class c α H n (C α ). Thus, given any [c] we have a corresponding A = φ α (c α ) H n (C α ) such that h(a) = ψ α (c α ) = [c] which proves surjectivity. Injectivity depends only on the kernel of h being trivial, so suppose there exists A H n (C α ) such that h(a) = [0]. There exists a representative c α of [0] such that A = φ α (c α ) for c α H n (C α ) and so ψ α (c α ) = [0] = f α (0 β ), where 0 β C α for β α. Thus ψ α (c α ) = ψ β (h αβ (c α )) = ψ β (f αβ (c α )) = ψ β (0 β ) and since f αβ preserves identities we have c α = 0 α and A = 0 as desired. Proposition 2.6. Let Λ be a directed set, then any coit G α of torsion free Abelian groups G α is torsion free. More generally, any finitely generated subgroup of G α is realized as a subgroup of some G α. Proof. Let Λ be a directed set and D F = (F, G α, f ij ) be a directed system of torsion free Abelian groups. For m N, suppose that there exists an m-torsion element [0] [b] G α, then we have m[b] = [mb] = [0] where mb G αi and 0 G αj. By equality of the equivalence classes, α k α i, α j such that f ik (mb) = k (0) = b k G αk, and since each f ij is a homomorphism it must map identities to identities. Thus, b k = 0 k G αk which implies that f ik (mb) = f ik (b) + + f ik (b) = 0 k but we know that f ik (b) k (0) for any k. It follows that f ik (b) 0 k, but mf ij (b) = 0 k so f ik (b) is a torsion element of G αk. This contradicts the hypothesis that all groups of the directed system are torsion free, and so G α is torsion free. Since G α is torsion free and Abelian it is free-abelian and hence every subgroup is also free-abelian; this means that for every subgroup, finitely generated is equivalent to finitely presented. Let H G α be finitely generated(presented) defined by relations r i H = [h ], [h 2 ],..., [h m ] R(H) = {r, r 2,..., r n } where each r i is a word on the m generators. For each h i we have the map ψ i : G αi G such that [h i ] ψ i (G αi ). By directedness, we can choose a representative g si G αs of [h i ] for α s larger than or equal to all other α i for i m, so this gives [h i ] = ψ s (g si ). For some α t α s all relations hold in G αt, since there are only finitely many, and so the map [h] i g ti defines a homomorphism from H to G αt. Another class of index categories for which coits always exist, is obtained when the index category is filtered; the following definition generalizes the notion of a directed poset. Definition 2.7. A nonempty category I is filtered if (a) For every pair i, j obj(i ) there exist k obj(i ) and morphisms m ik Mor I (i, k) and m jk Mor I (j, k). (b) For every pair φ, Mor I (i, j) there exists ψ Mor I (j, k) satisfying ψ = ψ φ. 7
Exercise 2.8. Prove that in the category Sets, the object A/ together with the obvious injection maps from each A i constitute the coit A i, of any diagram indexed by a filtered category. { A = (a i, i) } A i and A/ is defined under the equivalence relation i I (a i, i) (a j, j) f Mor C (A i, A k ) and g Mor C (A j, A k ) : f(a i ) = g(a j ) Proof. Consider the diagram D F = (F, A i, h jk ) indexed by some filtered category I, where we may assume that a map h ij (a i ) = a j exists. Otherwise, since I is filtered there exists a morphism h ik, and we can replace A j with A k in the following diagram A/ g j A i f i g i A j h ij A i By the filtered property of I we know that for each i, j there exist morphisms h ik and h jk in D F, which we take to be the desired morphisms f and g respectively. In particular, we can show that h ik (a i ) = h jk (a j ) A k, since by elementary arrow chasing it can be seen from the diagram below A k h jk A i f i h ik f i = h ij = h jk h ij = h ik A j h ij A i h jk (a j ) = h jk h ij (a i ) = h ik (a i ) The outer triangle of the first diagram can thus be made to commute by defining the maps g i (a i ) = [(a i, i)], i I. Explicitly, the morphism properties proven before guarantee that (a i, i) (a j, j), hence g i (a i ) = [(a i, i)] = [(a j, j)] = g j (a j ) whenever h ij exists. It only remains to prove that a unique map exists which makes the diagram commute for any i, j. e define elementwise which is possible in Sets by ([(a s, s)]) = f s (a s ) for all s I. It follows that g s (a s ) = ([(a s, s)]) = f s (a s ), s I, and that is unique in this regard; by universality, A/ and the maps g s form the it of D F. Exercise 2.9. Let I be a filtered category and C to belong to the category Mod R, then for any diagram D F = (F, M i, h jk ) indexed by I, describe the coit M i as an R-module. Proof. By Exercise 2.8, we take as the building block, the underlying set structure M/. (i) Before addition in M/ can be defined we need to establish what [0] is. Given that all morphisms in Mor(C ) map zero elements to zero elements, (a i, i) [0] if and only if there exists h ik such that h ik (a i ) = 0 M k. 8
(ii) Addition of elements belong to the same M i is just regular addition in that module. If a i M i and a j M j then since I is filtered, there exists M k and morphisms h jk, h ik. e define addition in M by a i a j = h ik (a i ) + h jk (a j ) = b k M k [(a i, i)] [(a j, j)] = [(h ik (a i ) + h jk (a j ), k)] = [(b k, k)] M/ To show that this addition is well defined in M/, consider another possible value a s a t = h sn (a s ) + h tn (a t ) = b n M n where we wish to show that (b k, k) (b n, n) if (a i, i) (a s, s) and (a j, j) (a t, t). By definition, the stated equivalences imply existence of morphisms φ and φ 2 such that φ (a i ) = φ 2 (a s ) M l ; likewise φ 3 (a j ) = φ 4 (a t ) M l. For the pair l, l there exists M α and morphisms satisfying h lα φ (a i ) = h lα φ 2 (a s ) M α h l α φ 3 (a j ) = h l α φ 4 (a t ) M α h lα φ (a i ) + h l α φ 3 (a j ) = h lα φ 2 (a s ) + h l α φ 4 (a t ) M α Now define the canonical sum [(a i, i)] [(a j, j)] to be the equivalence class [(h lα φ (a i ) + h l α φ 3 (a j ), α)] and analogously for [(a s, s)] [(a t, t)]; it is obvious that these two classes are the same. However, since [(a i, i)] [(a j, j)] = [(b k, k)] and [(a s, s)] [(a t, t)] = [(b n, n)] it folows from transitivity of the equivalence relation that (b k, k) (b n, n). e also note that by the definition of the canonical sum and b, associativity of addition is preserved, since for any i, j, k ([(a i, i)] [(a j, j)]) [(a k, k)] = [(h iz (a i ) + h jz (a j ), z)] [(a k, k)] = [(b z, z)] [(a k, k)] [(a i, i)] ([(a j, j)] [(a k, k)]) = [(a i, i)] [(h jw (a j ) + h kw (a k ), w)] = [(a i, i)] [(b w, w)] for some z, w which exists by filteredness. Now, there exist z and w since all morphisms h ij are R-module homomorphisms, we have [(b z, z)] [(a k, k)] = [(h zz h iz (a i ) + h zz h jz (a j ) + h kz (a k ), z )] = [(c z, z )] [(a i, i)] [(b w, w)] = [(h iw (a i ) + h ww h jw (a j ) + h ww h kw (a k ), w )] = [(c w, w )] For some M α we want to choose certain morphisms h z α and h w α such that h z α(c z ) = h w α(c w ), which would imply (c w, w ) (c z, z ). m α = h z α h zz h iz (a i ) + h z α h zz h jz (a j ) + h z α h kz (a k ) M α n α = h w α h iw (a i ) + h w α h ww h jw (a j ) + h w α h ww h kw (a k ) M α There exist morphisms (through diagram chasing) h iα, h jα, h kα such that which provides the necessary condition. [m α = n α = h iα (a i ) + h jα (a j ) + h kα (a k ) (iii) Multiplication be elements of R is defined by r a i = [(ra i, i)] M/, where it is clear that r a i = r a j if and only if (a i, i) (a j, j). Moreover, distributivity follows by setting r (a j a k ) = r a j r a k. 9
(iv) Finally, that M/ is the coit of D F follows from taking g i and to be as defined in Exercise 2.8, and noting that all maps are now endowed with the structure of R-module homomorphisms. Exercise 2.0. Let A be an integral domain and S a multiplicative subset containing. Show that S A = A in the category Mod S s A. Proof. Take C to belong to the category Mod A, and note that S is a filtered set since for any s, s 2 S there exists t S such that t = s s 2. Formally, we have the relation Mor S (s, t) = { } if and only if r S such that rs = t; otherwise Mor S (s, t) =. Now, we define the functor according to: F(s) = A, F( ) = h s st : A A defined by h s t st(x) = r x. Such a definition r makes sense due to the fact that A is an integral domain, so /s s 2 is always well defined and the indexed diagram is given by D F = (F, A, h s st). If S A has the underlying set structure A/, where A = { ( a, s) s s S A}, then by the previous exercise it can be realized as the s coit module. However, by Exercise 2.8 it suffices to prove that S A is the coit in Sets. g s2 f s2 S A g s f s s 2 A h s s 2 s A As usual, obj(c ) is any object satisfying g s = g s2 h s s 2, and the inner triangle is commutative if f s : A s S A is defined by f s (x) = [x]. For an integral domain, [a /s ] = [a 2 /s 2 ] if and only if s a 2 = s 2 a, hence it suffices to define ([a/s]) = g p ( b ) where p is chosen such that p a i s i [a/s], r i such that r i p = s i and r i b = a i. Such a p exists if we set b and p to share no divisors, since if rb = b and rp = p just choose b, p. a i s i b p = a ip = bs i = p bs i, b a i p = p s i, b a i As in Exercise 2.3 this provides a unique map which makes the upper triangles commute; explicitly, for any a s s A f s ( a s ) ([ a ]) ( ) ( ) b b ( a ) = = g p = g s h ps = g s s p p s where existence of h ps is guaranteed by the definition of p. It follows from universality, that S A along with the maps f s form the module coit of the diagram D F, as intended. Exercise 2.. If I is a non-filtered category and C belongs to the category Mod R, then for any diagram D F = (F, M i, h jk ) indexed by I,show that the coit M i is M/N M = i I M i N = a i h ij (a i ) hij Proof. Note that N is the module generated by elements a i h ij (a i ) and for each pair i, j such that Mor I (i, j), ranges over all morphisms h ij Mor I (M i, M j ). As usual, we consider the 0
following diagram g l f l M/N g k M l h kl M k where we define f i (a i ) = a i + N = [a i ], and thus (a k ) = f l h kl (a k ) since for any h kl (a k ) f l h kl (a k ) = a k + N h kl (a k ) + N = (a k h kl (a k )) + N = N Thus, for any finite sum m M/N we describe as the map defined by ( ) ([m]) = (a i + N) = (a i + N) = g i (a i ) i i i and uniqueness is proven as in Exercise 2.9. It only remains to show that M/N and the maps described constitute a module structure with associated R-module homomorphisms. This follows from the algebraic properties of quotient modules.