Homological Algebra in Abelian Categories Rylee Lyman March 3, 2019 1 Preliminaries Recall that a category is a collection of objects with arrows between them. We use the word \collection" advisedly, making no claim that we are dealing with sets. We require each object A to have an associated arrow A 1A A called the identity. We also require composition of arrows to be defined whenever it makes sense, and we require this composition to be associative. Furthermore, given f : A ! B, f1A = f = 1Bf We follow the treatment in [Mac71] to which the curious reader is referred. In category theory, a dual statement is one in which all the arrows have been reversed. For example, an arrow m is monic if it is \right-cancellative," i.e. mf = mg =) f = g The dual notion is an epi. An isomorphism is an arrow that has a two-sided inverse. Exercise 1. Isomorphisms are both monic and epi The converse does not always hold. A standard example is the inclusion Z ! Q in the category of rings. An object z is initial if, given another object t, there is a unique arrow z A: Dually, an object is terminal if there is a unique arrow to the object from any given object. A zero object is one that is both initial and terminal. Exercise 2. If z and w are both initial, terminal, or zero objects, there exists a unique isomorphism z ! w. We will denote our zero objects as 0, as well as any arrow that factors through the zero object. 1 A kernel for a map f : A ! B is an object ker f together with an arrow m: ker f ! A (sometimes also denoted ker f) such that the following diagram commutes. ker f m A 0 f B We say a diagram commutes when following any directed path in the diagram yields equal arrows, so the content of the statement here is just that fm = 0 as arrows ker f ! B. Actually that's not all that's required. A kernel must also be universal for this property. That is, for any other choice of object and arrow that makes the diagram commute, there is a unique arrow (indicated as dashed in the figure below) that makes the diagram commute. K ker f m A 0 f 0 B The dual notion is a cokernel. The following are good practice, and facts we need. Exercise 3. Kernels and cokernels, when they exist, are unique up to unique isomorphism. Moreover, kernels are monic and cokernels are epi. Exercise 4. If m: A ! B is monic in a category with a zero object, 0 ! A is a kernel for m. Dually, if e: A ! B is epi in a category with a zero object, B ! 0 is a cokernel for e. Exercise 5. In a category with a zero object, 1B : B ! B is a cokernel for 0: A ! B. Therefore any cokernel of a zero arrow is an isomorphism. Finally, a (binary) product of two objects A and B is a third object A × B and morphisms π1 : A × B ! A, π2 : A × B ! B, that satisfies the universal property that given maps f : C ! A and g : C ! B, there exists a unique map f × g : C ! A × B that makes the following diagram commute. C f g f×g A A × B B π1 π2 The dual notion is a coproduct, A q B, for which the diagram is A ι1 A q B ι2 B fqg f g C 2 Exercise 6. Products and coproducts, when they exist, are unique up to unique isomorphism. 2 Abelian Categories Definition 1. An abelian category is a category which 1. has a zero object 2. every pair of objects has a product and a coproduct and the unique map [(1A × 0) q (0 × 1B)]: A q B ! A × B is an isomorphism. 3. every arrow has a kernel and a cokernel 4. every monic arrow is a kernel and every epi arrow is a cokernel. Because the dual of each statement is itself, proving a statement about abelian categories proves the dual statement as well. Rather than binary co- products and products, we will think instead of an object A ⊕ B that is both a product and a coproduct. Exercise 7. Given A⊕B in an abelian category, use [(1A×0)q(0×1B)] = 1A⊕B to show that π1ι1 = 1A, π2ι2 = 1B, π1ι2 = 0 and π2ι1 = 0. Conclude that πi and ιi are epi and monic, respectively. Show also that [(1A q 0) × (0 q 1A)] = 1A⊕B. Exercise 8. The unique map ∆ = (1A × 1A) described below is a monomor- phism. A 1 1 A ∆ A A A ⊕ A A π1 π2 We would like to prove the following Theorem 1. Given arrows f : A ! B, g : A ! B in an abelian category, there is a third arrow f +g : A ! B. This new arrow is uniquely determined by f and g as (f + g) = (f q g)∆, and gives the collection of arrows A ! B the structure of an abelian group. Before we can, we need some preliminaries. Exercise 9. In an abelian category, a monic arrow is the kernel of its cokernel. Dually, an epi is the cokernel of its kernel. Proposition 1. In an abelian category, an arrow that is both monic and epi is an isomorphism. (Proposition 1.5.1 in [Bor94].) Proof. In view of the previous exercises, note that if f : A ! B is both monic and epi, then 0 ! A is a kernel of f, and since f is epi, f = coker(ker f) = coker 0, which is an isomorphism. 3 Exercise 10. Show that ι1π1 + ι2π2 = 1A⊕B, and that addition of arrows distributes over function composition, i.e. that (f + g)h = (fh) + (gh), similarly k(f + g) = (kf) + (kg). Use the above to show that f q g = fπ1 + gπ2, f × g = ι1f + ι2g and thus in particular ∆ = ι1 + ι2. Use symmetry of the diagram for ∆ to show addition of arrows is commutative. Exercise 11. Show that (A ⊕ B) ⊕ C and A ⊕ (B ⊕ C) are isomorphic, and thus that addition of arrows is associative. Proposition 2. 0: A ! B is the identity for addition of arrows. Proof. We note that the following diagram commutes because π1ι2 = 0 and π1ι1 = 1 A ι1 A ⊕ A ι2 A fπ1 f 0 B and conclude that fπ1 = f q 0. Now observe that fπ1∆ = fπ1(1A × 1A) = f1A = f. So far we have constructed a commutative monoid structure on arrows A ! B, and to do it we have only needed biproducts and zero objects. It is the existence of inverses for which we will need the extra hypotheses. Proposition 3. In an abelian category with a biproduct A ⊕ B, ι1 = ker π2, π2 = coker ι1, and similarly for pairs with indices swapped. Proof. Suppose we have f : C ! A⊕B with π2f = 0. Now, f = (ι1π1+ι2π2)f = ι1π1f +ι20 = ι1π1f. since ι1 is monic, the factorization (π1f) is unique, proving that ι1 is a kernel for π2. Now if g : A⊕B ! C is such that fι1 = 0, f = f(ι1π1+ι2π2) = 0π1+fι2π2 = fι2π2, and since π2 is epic, this factorization is unique. Proposition 4. In an abelian category, every pair of arrows A ! B has an equalizer and a coequalizer. Moreover, the following are equivalent: 1. m is monic 2. ker m = 0 3. mf = 0 =) f = 0. Indeed, dually if fe = 0 =) f = 0, then e is epi. (Propositions 1.5.3 and 1.5.4 in [Bor94].) Proof. by an equalizer for f; g : A ! B we mean an arrow m: M ! A such that fm = gm, such that for any other arrow c: C ! A with fc = gc, there is a unique arrow (dashed), which makes the diagram commute. f M m A B g c C 4 As usual, a coequalizer is the dual notion, and when equalizers or coequalizers exist they are unique up to unique isomorphism. So given f; g : A ! B, consider the map f × g : A ! B ⊕ B together with ∆: B ! B ⊕ B, and write F = coker ∆(f × g). Since ∆ is monic, ∆ = ker(coker ∆), so we get a map e: ker F ! B such that the diagram commutes. ker F e B ker F ∆ f×g A B ⊕ B F coker ∆ coker ∆ Indeed this is a universal filling of the square (a pullback), for if we have another object and pair of arrows A c C d B such (f × g)c = ∆d, then F c = coker ∆(f × g)c = (coker ∆)∆d = 0, so c factors through ker f and one checks that the remaining upper triangle commutes. d C ker F e B c ker F ∆ f×g A B ⊕ B F coker ∆ coker ∆ We claim that the map ker F is the desired equalizer. Indeed, f ker F = π1(f ×g) ker F = π1∆e = e, and similarly g ker F = π2(f ×g) ker F = π2∆e = e. If c is another map with fc = gc, then (fc) × (gc) = (f × g)c factors as ∆fc, so F c = coker ∆(f × g)c = (coker ∆)∆fc = 0, so c factors through ker F , as desired. By duality, we have also constructed coequalizers. Note that our proof actually shows that equalizers are kernels, and thus monic, so coequalizers will be cokernels, and thus epi. Certainly if m is monic, then ker m = 0, and mf = 0 =) f = 0. Moreover, if mf = 0 =) f = 0, m ker m = 0 tells us that ker m = 0.