Draft lesson.

Hereditary categories and cohomology splittings

Original text public domain (CC0).

In a semisimple category every complex contracts onto its cohomology, even when it is unbounded. In a hereditary category every bounded derived object also splits into its cohomology objects, but the splitting can depend on choices. Principal ideal domains give examples of the second phenomenon for all modules, including modules with arbitrarily large generating sets.

Let \(\mathcal A\) be an abelian category in the ambient universe of Universes and small categories. We use the axiom of choice for sets in that universe. Complexes are cochain complexes: \(X[r]^n=X^{n+r}\), with differential \((-1)^r d_X\). Write \(D(\mathcal A)\) for localization of the homotopy category at quasi-isomorphisms, and

Cohomological bounds, denominators and truncation adjunctions proves the localization Hom formulas, full bounded embeddings, negative Ext vanishing, degree-zero full faithfulness and the natural truncation triangles. Split triangles, Section 2 proves that a triangle with zero last arrow is a split triangle with its specified inclusion and projection. Its Proposition 2.1 identifies an invertible first arrow by a zero third vertex. Chain homotopies, mapping cones and triangles, Lemma 4.1 proves the exact Hom sequences in the homotopy category. We use their derived versions from Triangles after quasi-isomorphism localization.

1. Semisimplicity gives an actual contraction

An abelian category is semisimple if every short exact sequence splits. This definition imposes no finite-length condition. The equivalent section, retraction and biproduct criteria are proved in Splittings, Section 1. We first keep the maps that the splitting produces.

Put \(Z^n=\ker d_X^n\), \(B^n=\operatorname{im}d_X^{n-1}\), and \(H^n=Z^n/B^n\). In a semisimple category choose splittings of

for each integer \(n\). There are only set-many choices. Together they identify \(X\), as a complex with its actual differential, with

To check the differential, its restriction to the cycle summand is zero. On the last summand its composite with the chosen section of \(X^n\twoheadrightarrow B^{n+1}\) is precisely the prescribed inclusion of \(B^{n+1}\) into the first summand of \(X^{n+1}\). Thus (1.2) uses the same image objects in adjacent degrees.

Let \(G(X)\) be the complex with term \(H^n\) in degree \(n\) and all differentials zero. Define \(i:G(X)\to X\) by middle-summand inclusions and \(p:X\to G(X)\) by middle-summand projections. These are chain maps and \(pi=1\). Define the degree-minus-one map

Direct calculation gives

Hence \(i,p\) are mutually inverse in the homotopy category. If \(X\) is acyclic then \(G(X)=0\), and (1.4) contracts its identity to zero. This proof covers unbounded complexes: each degree uses only a three-term biproduct. It never takes an infinite coproduct in \(\mathcal A\).

The cone of any quasi-isomorphism is acyclic, and is therefore zero in the homotopy category by (1.4). The zero-cone criterion makes that quasi-isomorphism invertible already in the homotopy category. Consequently its localization functor \(K(\mathcal A)\to D(\mathcal A)\) is an equivalence. Indeed the identity functor on \(K\) already inverts every denominator and factors through the localization; both resulting composites are the identity by the localization property. In particular it preserves and reflects every Hom group.

For \(A,B\in\mathcal A\), the complexes \(A[0]\) and \(B[r]\) have nonzero terms only in degrees zero and \(-r\). If \(r\ne0\), every chain map between them is zero. Thus semisimplicity implies

Conversely suppose \(\operatorname{Ext}^1(B,A)=0\) for all objects \(A,B\). A short exact row \(0\to A\to E\xrightarrow q B\to0\) gives its natural derived triangle, whose last arrow \(\delta:B\to A[1]\) lies in that zero group. Exactness of \(\operatorname{Hom}_D(B,-)\) then lifts \(1_B\) to an arrow \(s:B\to E\) in \(D\). Degree-zero full faithfulness turns it into an actual arrow of \(\mathcal A\) satisfying \(qs=1_B\). Thus every row splits. Conversely, a section of a particular row makes its boundary zero: consecutive triangle arrows satisfy \(\delta q=0\), so \(\delta=\delta qs=0\).

We have proved that the following three conditions are equivalent: semisimplicity; vanishing of all degree-one Ext groups; vanishing of all nonzero-degree Ext groups. No enough-injectives or enough-projectives hypothesis was used for this equivalence. For one particular exact row, vanishing of its particular group \(\operatorname{Ext}^1(B,A)\) is already sufficient to split that row.

Example 1.1. Over a field \(k\), take the complex \(k^2\to k^3\to k^2\) in degrees \(0,1,2\), with

The composite is zero. Its cohomology is one copy of \(k\) in each degree, represented respectively by the second coordinate, the third coordinate and the second coordinate. Include these coordinates to obtain \(i\), and project to them to obtain \(p\). Set \(h^1(a,b,c)=(a,0)\), \(h^2(s,t)=(0,s,0)\), and set the other homotopy components to zero. In degree zero, \(h^1d^0(u,v)=(u,0)\). In degree one, \(d^0h^1+h^2d^1\) sends \((a,b,c)\) to \((a,b,0)\). In degree two, \(d^1h^2(s,t)=(s,0)\). These are exactly the three components of \(1-ip\).

2. A submodule of an arbitrary-rank free PID module is free

A principal ideal domain is a nonzero commutative integral domain \(R\) in which every ideal is principal. Let

be a free module. Choice well-orders its basis, giving the ordinal \(\lambda\). For each \(\alpha\le\lambda\), put \(F_\alpha=\bigoplus_{\beta<\alpha}Re_\beta\). For a submodule \(N\subseteq F\), put \(N_\alpha=N\cap F_\alpha\).

At a successor stage, the \(e_\alpha\)-coordinate map on \(N_{\alpha+1}\) has kernel \(N_\alpha\) and image an ideal \(I_\alpha\subseteq R\). If \(I_\alpha=0\), then \(N_{\alpha+1}=N_\alpha\). If \(I_\alpha\ne0\), choose a nonzero generator \(a_\alpha\) and choose \(v_\alpha\in N_{\alpha+1}\) with \(e_\alpha\)-coordinate \(a_\alpha\). Every \(x\in N_{\alpha+1}\) has that coordinate \(t a_\alpha\) for some \(t\in R\); hence \(x-tv_\alpha\in N_\alpha\). Moreover, \(tv_\alpha\in N_\alpha\) implies \(t a_\alpha=0\), and the domain hypothesis gives \(t=0\). Therefore

At a nonzero limit ordinal \(\mu\le\lambda\),

Indeed an element of \(F_\mu\) has finite support. If its support is nonempty, the successor of its largest index is still strictly less than the limit \(\mu\); this element therefore belongs to an earlier \(F_\alpha\). The zero element belongs to every stage. Intersecting with \(N\) gives (2.3).

Transfinite induction now proves that the chosen \(v_\beta\), for \(\beta<\alpha\) with \(I_\beta\ne0\), form a basis of \(N_\alpha\). The zero stage has the empty basis. Equation (2.2) proves the successor step. At a limit, (2.3) gives spanning by finite combinations from an earlier stage. Every finite proposed linear relation also lies in one earlier stage, where independence has already been proved. At \(\alpha=\lambda\) this is a basis of all of \(N\).

All choices here are indexed by the set of stages of a fixed basis. There is no assumption that \(N\) is finitely generated. We used principality of the whole ideal \(I_\alpha\), which itself can initially be presented by arbitrarily many elements. The domain hypothesis was used to cancel its nonzero generator. The argument proves freeness of \(N\); it does not claim that \(N\) is a direct summand of \(F\).

Example 2.1. The inclusion \(2\mathbb Z\subset\mathbb Z\) has a free source of rank one. It has no retraction: if \(r(1)=2t\), then \(r(2)=4t\) cannot equal two. Thus its quotient \(\mathbb Z/2\) is not a free module, and a free submodule need not be a summand.

3. Projective complexes compute the actual derived Hom

An object \(P\) is projective when every arrow from \(P\) to the target of an epimorphism lifts through that epimorphism. A free \(R\)-module on any set is projective: choose a preimage for the image of each basis vector, and extend by finite linear combinations. This works for arbitrary basis cardinalities. Every module \(M\) has the surjection from the free module on its underlying set, sending \(e_m\) to \(m\).

We need to connect such resolutions to definition (0.1).

Lemma 3.1. Let \(P\) be a bounded-above complex of projective objects in any abelian category. Every chain map \(f:P\to A\) to an acyclic complex is null-homotopic. For every complex \(Y\), the natural map

is an isomorphism.

Proof. Choose \(b\) with \(P^n=0\) for \(n>b\). We construct \(h^n:P^n\to A^{n-1}\) downwards, with all components above \(b\) zero. Suppose the equations

have been obtained for \(j>n\). Set \(e^n=f^n-h^{n+1}d_P^n\). The chain-map equation and the already obtained equation in degree \(n+1\) give

Thus \(e^n\) factors through \(Z^nA\). Acyclicity makes \(A^{n-1}\to Z^nA\) an epimorphism. Projectivity supplies a lift \(h^n\), establishing (3.2). At the first stage the same calculation uses \(P^{b+1}=0\). Downward recursion supplies every integer degree; no termwise stabilization or infinite limit in \(\mathcal A\) is required. Equation (3.2) is the desired null-homotopy.

If \(t:Y\to V\) is a quasi-isomorphism, its cone and every shift of that cone are acyclic. The first part makes all their Hom groups from \(P\) zero in \(K\). Applying the exact Hom sequence of the cone triangle shows that

Use the outgoing denominator formula from the cohomological-bounds prerequisite. For each denominator \(t:Y\to V\), its numerator group \(\operatorname{Hom}_K(P,V)\) is identified with \(\operatorname{Hom}_K(P,Y)\) by \(t_*^{-1}\). For a transition \(u:V\to V'\) with \(ut=t'\), these identifications agree because \(u_*t_*=t'_*\). The outgoing index is nonempty and filtered; its colimit of this constant group is the same group. This proves (3.1), including injectivity and naturality. Equivalently every outgoing roof has the unique homotopy class of a chain-map representative supplied by (3.4). \(\square\)

Now let \(R\) be a PID and \(M\) any module. Take the free cover \(F_0\twoheadrightarrow M\). Its kernel is free by Section 2; call it \(F_1\). We have the exact row

Its two-term complex \(P=(F_1\xrightarrow j F_0)\), in degrees \(-1,0\), is bounded above and projective, and its augmentation \(P\to M[0]\) is a quasi-isomorphism. Lemma 3.1 gives \(\operatorname{Ext}^r(M,N)=\operatorname{Hom}_K(P,N[r])\).

For \(r=0\), a chain map is an arrow \(F_0\to N\) killing \(j(F_1)\), hence an arrow \(M\to N\); there are no nonzero homotopy components. For \(r=1\), its only component is an arbitrary arrow \(F_1\to N\). Two such arrows are homotopic precisely when their difference is \(t j\) for an arrow \(t:F_0\to N\): that is exactly the degree-minus-one homotopy equation in degree \(-1\). For \(r\ge2\) or \(r<0\), the nonzero degrees of \(P\) and \(N[r]\) do not overlap, so every chain map is zero. We have proved

This computes the derived-category Ext, rather than defining another group by a chosen resolution. It proves that the category of all \(R\)-modules is hereditary: by definition every Ext group in degree at least two vanishes.

A field is a PID: an ideal containing a nonzero element contains one. The integers are a PID: a nonzero ideal has a least positive element \(a\), and division with remainder makes every element divisible by \(a\). For a field \(k\), the domain \(k[x]\) is a PID. Choose a nonzero polynomial of least degree in a nonzero ideal. Polynomial division leaves a remainder in that ideal of smaller degree, so the remainder is zero. Division works by canceling the leading term with a multiple of the chosen polynomial; its leading coefficient is invertible in \(k\), and the degree strictly decreases until the process stops. Products of nonzero polynomials have nonzero leading coefficient, which verifies the domain hypothesis. Thus all modules over \(k\), \(\mathbb Z\), and \(k[x]\) for a field \(k\) satisfy (3.6).

Vector spaces are semisimple: choose a basis of the quotient in a short exact row, lift its basis vectors, and extend linearly to a section. For basis existence, well-order the underlying set of the vector space, and retain a vector exactly when it is outside the span of all previously retained vectors. Every vector is either retained or already in that earlier span, giving spanning. A nontrivial finite relation among retained vectors would express its last vector as a linear combination of earlier ones, contradicting its selection. This proves independence and supplies the required basis with choice. The integer example is not semisimple by Example 2.1. Nor is the module category of \(k[x]\): the quotient row for \((x)\subset k[x]\) has nonzero quotient \(k[x]/(x)\), but every arrow from that quotient to \(k[x]\) is zero, since its image would be killed by the nonzero element \(x\) of a domain. Hence that row has no section.

4. Bounded cohomology splits in a hereditary category

Let now \(\mathcal A\) be any hereditary abelian category, defined using (0.1). It need not have enough injectives or projectives. Let \(X\in D(\mathcal A)\) have cohomology only in the finite interval \([a,b]\). The bounded-representative theorem identifies this with an object of \(D^b(\mathcal A)\). Write \(U_nX=\tau_{\le n}X\).

Theorem 4.1. There exists an isomorphism

whose cohomology maps are the identities under the displayed identifications. The sum is finite. When \(X\) is acyclic, the assertion instead says that \(X\) is zero.

Proof. The natural layer triangle is

Below \(a\), \(U_{a-1}X\) is acyclic and hence zero. Suppose by induction that \(U_{n-1}X\) has been identified with \(\bigoplus_{j=a}^{n-1}H^j(X)[-j]\), compatibly with its cohomology. Under its shifted identification, the \(j\)-th component of \(\delta_n\) lies in

The equality of Hom groups applies the translation equivalence to both arguments; the exponent is \(n-j+1\ge2\). A map into a finite biproduct is determined by its components, so \(\delta_n=0\).

Choose a section \(s_n:H^n(X)[-n]\to U_nX\) of the second arrow of (4.2). The split-triangle theorem makes \([u_n,s_n]\) an isomorphism from \(U_{n-1}X\oplus H^n(X)[-n]\) to \(U_nX\), where \(u_n\) is its actual first arrow. For degrees below \(n\), the map \(u_n\) gives the existing cohomology identifications. In degree \(n\), the projection in (4.2) induces the canonical identification with \(H^nX\); its section therefore induces the identity. This extends the induction with the required cohomology maps. At \(n=b\), the natural map \(U_bX\to X\) is a quasi-isomorphism. Composing gives (4.1). \(\square\)

The choices of sections are meaningful. In a split triangle \(A\xrightarrow u E\xrightarrow v C\to A[1]\), fix one section \(s\). Every other section is uniquely \(s+u\alpha\), where \(\alpha\in\operatorname{Hom}_D(C,A)\). Indeed the difference is killed by \(v\), and the split coordinates identify the kernel of \(v_*\) with the image of the injective map \(u_*\). In stage (4.2), the possible components of this difference belong to \(\operatorname{Ext}^{n-j}(H^nX,H^jX)\). For \(j=n-1\), this is degree one, which a hereditary category does not require to vanish.

Example 4.2. Abelian groups are hereditary by Section 3. Let \(A=\mathbb Z[0]\), \(C=(\mathbb Z/2)[-1]\), and \(X=A\oplus C\). Reduction gives the exact row \(0\to\mathbb Z\xrightarrow2\mathbb Z\to\mathbb Z/2\to0\). Its boundary is a nonzero arrow \(\mathbb Z/2\to\mathbb Z[1]\): a zero boundary would give a group section of reduction by the proof in Section 1, but every homomorphism \(\mathbb Z/2\to\mathbb Z\) is zero. Shifting that boundary gives a nonzero \(\alpha:C\to A\).

Let \(i_A,i_C,p_A,p_C\) be the biproduct maps of \(X\), and set \(N=i_A\alpha p_C\). Then \(N^2=0\), since \(p_Ci_A=0\). Therefore \(1_X+N\) is invertible, with inverse \(1_X-N\). It differs from the identity because \(p_A N i_C=\alpha\ne0\). Nevertheless it induces the identity on every cohomology object: \(C\) has cohomology only in degree one, while \(A\) has cohomology only in degree zero, so all cohomology maps of \(\alpha\) vanish.

There can consequently be no choice of (4.1) natural for all derived arrows and compatible with these cohomology identifications. Naturality for \(1_X+N\), whose cohomology is the identity, would make it act as the identity under an isomorphism (4.1). Canceling that isomorphism would imply \(1_X+N=1_X\), a contradiction.

5. The finite Hom calculation after a splitting

Choose splittings (4.1) for \(X,Y\in D^b(\mathcal A)\). Finite biproduct coordinates and translation give

Both sums have only finitely many nonzero object positions. To verify the formula, the component from \(H^iX[-i]\) to \(H^jY[-j]\) is \(\operatorname{Ext}^{i-j}(H^iX,H^jY)\). Negative degrees vanish by the cohomological-bounds prerequisite. Degrees at least two vanish by heredity. Degree zero is ordinary Hom and degree one is the second summand in (5.1). The diagonal components are exactly the cohomology maps, because the chosen splittings induce the identity on cohomology. The off-diagonal components and the displayed group decomposition depend on those choices.

Call an arrow a ghost if all its cohomology maps are zero. Formula (5.1) shows that a ghost has only degree-one components. If \(f:X\to Y\) and \(g:Y\to Z\) are ghosts between bounded derived objects, every possible nonzero component of their composite passes from degree \(i\) to degree \(i-1\) and then to degree \(i-2\). Such a composite lies in \(\operatorname{Ext}^2(H^iX,H^{i-2}Z)=0\). Matrix composition is a finite sum of these zero arrows. Thus

This conclusion is independent of the splittings used to prove it. Ghosts can themselves be nonzero, as Example 4.2 shows. Cohomology therefore detects isomorphisms while retaining a possible nonzero, square-zero collection of invisible arrows.

6. Four graded exercises with complete solutions

Exercise 1 (foundation: a vanishing Ext group splits one row). Let \(0\to A\xrightarrow i E\xrightarrow q B\to0\) be exact in an arbitrary abelian category. Assume only \(\operatorname{Ext}^1(B,A)=0\). Prove that the specified row splits. Give an inverse to the resulting map \(A\oplus B\to E\).

Solution. The row triangle has boundary \(\delta\in\operatorname{Ext}^1(B,A)\), so \(\delta=0\). The exact sequence \(\operatorname{Hom}_D(B,E)\to\operatorname{Hom}_D(B,B)\to\operatorname{Hom}_D(B,A[1])\) lifts \(1_B\) to \(s\) with \(qs=1_B\). By degree-zero full faithfulness, \(s\) is an actual arrow of the abelian category. The map \(1_E-sq\) is killed by \(q\), and therefore factors uniquely as \(it\) through its kernel \(i\). Monicity gives \(ti=1_A\), by composing \(it=1_E-sq\) with \(i\). Composing with \(s\) gives \(its=0\), and monicity gives \(ts=0\). Hence

Thus \([i,s]\) is the required split isomorphism with inverse \((t,q)\). This does not impose any condition on other Ext groups, or assume a supply of injective objects.

Exercise 2 (intermediate: infinitely many basis vectors). Let \(F=\bigoplus_{n\ge0}\mathbb Ze_n\). Let \(N\) be generated by \(v_0=2e_0\) and \(v_n=e_n+e_0\) for \(n\ge1\). Prove that these vectors are a basis of \(N\). Compute \(F/N\) and show that \(N\) is not a summand of \(F\).

Solution. Spanning is the definition of \(N\). In a finite relation \(\sum_n a_nv_n=0\), the coordinate of \(e_n\), for every \(n\ge1\), is \(a_n\), so all those coefficients vanish. The coordinate of \(e_0\) then gives \(2a_0=0\), hence \(a_0=0\). This proves independence; a relation in a free module is finite, even though the basis is infinite.

Send a vector \(\sum b_ne_n\) to \(\sum b_n\bmod2\). The finite-support condition defines this homomorphism \(F\to\mathbb Z/2\). It is surjective and kills every \(v_n\). If a vector lies in its kernel, subtract \(\sum_{n\ge1}b_nv_n\); the remainder is \((b_0-\sum_{n\ge1}b_n)e_0\). Its coefficient is even because it differs from the even number \(\sum b_n\) by \(-2\sum_{n\ge1}b_n\). It is therefore a multiple of \(v_0\), proving that the kernel is exactly \(N\). Thus \(F/N\simeq\mathbb Z/2\).

If the quotient map had a section, it would embed \(\mathbb Z/2\) into \(F\). But \(F\) is torsion-free: \(2x=0\) forces each of its finitely many integer coordinates to be zero. Every map from \(\mathbb Z/2\) to \(F\) is consequently zero, so no section exists. This free submodule is not a direct summand.

Exercise 3 (advanced: compute all degrees for cyclic groups). For positive integers \(m,n\), compute \(\operatorname{Ext}^r_{\mathbb Z}(\mathbb Z/m,\mathbb Z/n)\) for every integer \(r\). Also compute \(\operatorname{Ext}^1_{\mathbb Z}(\mathbb Z/m,\mathbb Z)\).

Solution. Use the projective complex \(\mathbb Z\xrightarrow m\mathbb Z\) in degrees \(-1,0\). Multiplication by the positive integer \(m\) is injective, so its augmentation resolves \(\mathbb Z/m\). With target \(N\), the degree-zero maps are exactly elements killed by \(m\), and degree-one homotopy classes are \(N/mN\), by (3.6). All degrees other than zero and one vanish.

For \(N=\mathbb Z/n\), put \(g=\gcd(m,n)\), \(m=gm'\), and \(n=gn'\). The condition \(mx\equiv0\pmod n\) is equivalent to \(n'\mid m'x\), hence to \(n'\mid x\), because \(m',n'\) are relatively prime: integers \(u,v\) with \(um'+vn'=1\) prove the implication. The killed subgroup is generated by the class of \(n'\) and has order \(g\), so Hom is \(\mathbb Z/g\). The quotient \((\mathbb Z/n)/m(\mathbb Z/n)\) is \(\mathbb Z/(m\mathbb Z+n\mathbb Z)=\mathbb Z/g\). The equality of ideals follows from integer division, or the same Bézout identity after multiplying by \(g\). Therefore the Ext groups in degrees zero and one are both isomorphic to \(\mathbb Z/g\), and all other degrees are zero. If \(m=1\) or \(n=1\), then \(g=1\) and these groups are zero as required. With \(N=\mathbb Z\), degree one is \(\mathbb Z/m\), while degree zero is zero since an integer killed by \(m>0\) is zero.

Exercise 4 (synthesis: a splitting cannot be natural). In a hereditary category, show that composing any two ghosts between bounded derived objects gives zero. Then, for \(X=\mathbb Z[0]\oplus(\mathbb Z/m)[-1]\) with \(m\ge2\), exhibit a nonidentity automorphism inducing the identity on all cohomology. Deduce the failure of a natural cohomology splitting.

Solution. Choose cohomology splittings for the three objects of a composable pair. By (5.1), a ghost only has components from cohomological degree \(i\) to degree \(i-1\), represented by degree-one Ext. Every term in the finite matrix product of two such arrows goes from degree \(i\) to degree \(i-2\), a degree-two Ext group. Heredity makes every such group zero, proving that the composite is zero.

The exact row \(0\to\mathbb Z\xrightarrow m\mathbb Z\to\mathbb Z/m\to0\) does not split: an integer annihilated by \(m\) is zero, so every potential section is zero. Its derived boundary is nonzero by the exact Hom argument of Exercise 1, applied as the equivalence between a zero boundary and a section. Shifting gives a nonzero \(\alpha:(\mathbb Z/m)[-1]\to\mathbb Z[0]\). Let \(N=i_{\mathbb Z}\alpha p_{\mathbb Z/m}\) on the stated biproduct. Then \(N^2=0\), \((1+N)(1-N)=(1-N)(1+N)=1\), and \(p_{\mathbb Z}Ni_{\mathbb Z/m}=\alpha\ne0\). Hence \(1+N\) is an automorphism different from the identity. Every cohomology map of \(N\) is zero because its nonzero component joins objects concentrated in different degrees.

If an isomorphism between every bounded derived object and its shifted cohomology sum were natural in all arrows, applying naturality to this automorphism would compare \(1+N\) with the identity automorphism of that sum, since its cohomology maps are identities. The isomorphism could be canceled, forcing \(N=0\), which contradicts the calculation. Thus the existence theorem does not admit such a natural choice.

Freely accessible references

• The Stacks Project, Projective resolutions, tag 0643, for the bounded-above projective Hom statements.
• The Stacks Project, Ext groups, tag 06XP, for the derived-category definition and its resolution interfaces.
• The Stacks Project, Structure of modules over a PID, tag 0ASL, for further finite-module results.