Draft lesson.

Relative replacements and finite resolution bounds

Original text public domain (CC0).

A useful resolution class need not consist of injective objects. We construct replacements in a full additive class, first for bounded-below complexes and then, under a uniform finite condition, for arbitrary complexes. Every infinite construction becomes constant in each degree. We also compare the derived category of a thick abelian subcategory with the complexes whose cohomology belongs to that subcategory.

Let \(\mathcal A\) be an abelian category, and let \(\mathcal J\subset\mathcal A\) be a full additive subcategory. Finite biproducts in \(\mathcal J\) are taken with their ambient biproduct structure. When a chosen universal object is only isomorphic to an object of \(\mathcal J\), transport its maps through that isomorphism. Work in a fixed ambient universe containing the category and the choices used in each construction.

We use cochain complexes. Put \(Z^nX=\ker d_X^n\), \(B^nX=\operatorname{im}d_X^{n-1}\), and \(H^nX=Z^nX/B^nX\). For a chain map \(f:X\to Y\), our cone is

Cohomological bounds, denominators and truncation adjunctions, Sections 1–3, proves the derived localizations, their fraction equality tests, the full embeddings of all three bounded variants, and their representatives with prescribed cohomological bounds. Derived morphisms through quasi-isomorphism roofs, Section 5, proves the K-injective Hom test. The kernel, cokernel and coimage calculations below use the abelian axiom proved in Kernels, cokernels and the abelian comparison, Sections 1–4. No enough-injectives assumption is imposed until Section 4.

1. One pushout stage and its exact cohomology maps

We will use two elementary pushout facts. A pushout of a monomorphism is monic, and a pushout identifies the cokernels of its parallel arrows. Here is a direct verification with the maps retained.

For a monomorphism \(i:D\hookrightarrow E\) and any \(v:D\to Q\), form

The arrow \((v,-i)\) is monic because its second component is monic. The cokernel map \(Q\oplus E\to P\) therefore has kernel exactly this copy of \(D\). If \((t,0):T\to Q\oplus E\) maps to zero, it factors uniquely through \((v,-i)\); the second component says \(i\) kills that factor, so the factor and then \(t\) are zero. Thus \(Q\to P\) is monic. A map from \(P\) killing \(Q\) is precisely a map from \(E\) killing \(i\), by the cokernel universal property. Hence

is exact, with the indicated maps. This proves both facts in the case we need, without treating arbitrary abelian objects as sets of elements. They agree with the free references Stacks, Tag 08N4 and Tag 08N3.

Suppose \(X^j=0\) for \(j<m\). At stage \(p\geq m-1\), suppose a complex \(Y\) is zero above \(p\), has its nonzero terms in \(\mathcal J\), and a chain map \(f:X\to Y\) induces isomorphisms in degrees below \(p\) and a monomorphism in degree \(p\). Set

The map \(\bar f|_K:K\to Q=H^pY\) is monic. Define two pushouts

Lemma 1.1. There is an exact sequence

and the induced map \(S\to W\) is monic.

Proof. Write \(Q_0=Q/\bar f(K)\) and \(I=\operatorname{im}\bar d\). The abelian comparison identifies \(D/K\) with \(I\). Thus \(\bar f\), after quotienting its target by \(\bar f(K)\), gives a map \(I\to Q_0\). The universal properties identify

Indeed, compatible maps out of \(Q\) and \(Z\) kill \(\bar f(K)\), and their induced maps out of \(Q_0\) and \(Z\) agree on \(I\); the reverse construction gives the original compatible maps. Since \(I\hookrightarrow Z\) is monic, (1.2) gives \(0\to Q_0\to S\to Z/I\to0\). Here \(Z/I=C\), and the kernel of \(Q\to Q_0\) is the displayed copy of \(K\). This proves (1.5). Finally, associativity of these pushouts identifies \(W\) with \(S\amalg_Z E\). Its map from \(S\) is a pushout of the monomorphism \(Z\hookrightarrow E\), so is monic. \(\square\)

Choose an object \(J\in\mathcal J\) and a map \(g:W\to J\) whose restriction to \(S\) is monic. Append \(Y^{p+1}=J\), and define

The pushout equation makes these maps commute with \(d_X^p\) and \(f^p\). The new differential kills \(\operatorname{im}d_Y^{p-1}\), so its composite with the preceding differential is zero. All other complex equations were unchanged, and the new last differential is zero.

Since \(S\to J\) is monic, the kernel of \(Q\to J\) is exactly \(K\). Therefore \(H^pf\) becomes an isomorphism. Moreover \(\operatorname{im}d_Y^p\simeq Q_0\), and (1.5) gives the monomorphism

The map in (1.8) is the induced cohomology map: the map on cycles \(Z\to J\) is precisely the map used in (1.4)–(1.7). Thus both of the induction's cohomology assertions retain their specified maps.

2. Two bounded-below replacement criteria

Call \(\mathcal J\) cogenerating if every object of \(\mathcal A\) admits a monomorphism into an object of \(\mathcal J\).

Theorem 2.1. If \(\mathcal J\) is cogenerating, every complex \(X\) with \(X^j=0\) for \(j<m\) admits a quasi-isomorphism \(X\to Y\), where \(Y^j=0\) for \(j<m\) and \(Y^j\in\mathcal J\) for every \(j\).

There is also an alternative sufficient criterion for this same conclusion for a given \(X\):

1. \(\mathcal J\) is closed under extensions and under cokernels of monomorphisms between its objects;
2. for each monomorphism \(S\hookrightarrow W\) with \(S\in\mathcal J\), there is \(W\to J\), \(J\in\mathcal J\), whose composite \(S\to J\) is monic;
3. every \(H^jX\) belongs to \(\mathcal J\).

Proof. Start with the zero target at \(p=m-1\). Its cohomology in and below that degree is zero, as is the source's. Under cogeneration, embed each \(W\) of (1.4) into an object of \(\mathcal J\). Its restriction to \(S\) is monic, so (1.7) supplies the next stage.

Under the alternative criterion we maintain also \(Q\in\mathcal J\). It holds initially with \(Q=0\). At a stage, \(K=H^pX\in\mathcal J\) and \(K\to Q\) is monic. Thus \(Q_0\in\mathcal J\). Since \(C=H^{p+1}X\in\mathcal J\), the short exact sequence \(0\to Q_0\to S\to C\to0\) puts \(S\) in \(\mathcal J\). Apply condition 2 to \(S\hookrightarrow W\). It gives exactly the map \(g\) needed for (1.7). The new image is \(Q_0\in\mathcal J\); its monomorphism into \(J\) has cokernel \(J/Q_0\in\mathcal J\). This is the new top \(Q\), so the extra invariant is preserved.

Append one term at each successive stage. The old terms, maps and differentials are retained. Every fixed term and outgoing differential is final after finitely many stages, so these final values define a complex \(Y\) and a chain map \(X\to Y\). Every complex or chain-map equation holds at a finite stage. For a fixed \(k\), the cohomology map in degree \(k\) is final after the next term has been appended, and is then the isomorphism proved in Section 1. Hence the final map is a quasi-isomorphism. No exactness of an infinite colimit has been assumed. \(\square\)

The alternative criterion does not imply that the replacement map is monic in every degree. Only the map from \(S\) in (1.4) is required to stay monic. The distinction matters in Exercise 1.

Let \(K^*(\mathcal J)\) denote the full homotopy subcategory of complexes with terms in \(\mathcal J\), with \(*=+,b\), or no bound. Fullness holds because chain maps and chain homotopies between such complexes have all their components in the full subcategory \(\mathcal J\). Cones and shifts stay there, since each cone term is a finite biproduct. The complexes acyclic in \(\mathcal A\) form a triangulated null system. Let \(S_{\mathcal J}\) be the maps whose cones are acyclic.

Lemma 2.2 (replacement density). Suppose every object of an ambient bounded variant \(K^*(\mathcal A)\) admits an outgoing quasi-isomorphism into \(K^*(\mathcal J)\). Then the natural exact functor

is an equivalence. The same argument applies to a full triangulated subcategory of \(D^*(\mathcal A)\) if just its representatives and their quasi-isomorphic comparison objects admit these replacements.

Proof. Essential surjectivity follows by replacing a representative. An arrow between class complexes has an outgoing roof \(I\xrightarrow aV\xleftarrow sJ\), with \(s\) a quasi-isomorphism. The bounded denominator comparison gives a target in the relevant ambient bounded variant. Replace it by \(t:V\to L\), with \(L\) a class complex. The roof \((ta,ts)\) lies inside the full homotopy subcategory and represents the same arrow. This proves fullness.

For faithfulness, the ambient fraction equality test compares two roofs after maps to a common target \(W\), with the compared denominators quasi-isomorphisms. Replace \(W\) by a class complex \(L\). Postcomposition makes all target maps lie between class complexes; their equalities are equalities of homotopy classes there, by fullness of the homotopy inclusion. The same denominator and equality conditions therefore hold in the class localization. This proves faithfulness, including the equality to zero case. All quasi-isomorphism assertions use ambient cohomology, so the inverted class is exactly \(S_{\mathcal J}\).

The functor sends the chosen cone triangles to their ambient distinguished triangles and commutes with shifts, hence is exact. Its fully faithful, essentially surjective inverse is exact: represent the first arrow of any distinguished triangle in the source localization, take its cone triangle, and compare its image with the given triangle. The isomorphisms on the first two vertices make the third comparison an isomorphism by the two-vertex criterion. Thus the given triangle is an image cone triangle up to isomorphism. \(\square\)

Theorem 2.1 and this lemma give \(S_{\mathcal J}^{-1}K^+(\mathcal J)\simeq D^+(\mathcal A)\) under cogeneration. This conclusion is about a localization of class complexes; it does not assert that their acyclic complexes are contractible.

3. The precise uniform finite condition

Fix an integer \(d\geq0\). Consider the following condition:

For \(d\geq1\), exactness is required at every term after the first displayed term; no initial zero is imposed. For \(d=0\), (3.1) means that every object of \(\mathcal A\) belongs to \(\mathcal J\). This condition in particular makes \(\mathcal J\) closed under isomorphism: for \(d\geq1\), place an isomorphism \(J\to V\) after \(d-1\) zero objects; for \(d=0\) there is nothing to prove.

Proposition 3.1. Suppose \(\mathcal J\) is cogenerating and satisfies (3.1). If \(X\) is zero outside \([a,b]\), there is a quasi-isomorphism \(X\to T\) with terms in \(\mathcal J\), zero outside \([a,b+d]\). Consequently

Proof. Take the bounded-below replacement \(q:X\to Y\) from Theorem 2.1. Then \(H^jY=0\) for \(j>b\). For \(d\geq1\) and \(r\geq b+d\), the sequence

is exact in the sense of (3.1). There are exactly \(d\) terms from \(\mathcal J\). At its interior terms the cohomology degrees are at least \(r-d+1\geq b+1\); the final surjectivity is \(H^rY=0\). Thus \(Z^rY\in\mathcal J\).

Put \(r=b+d\), and truncate \(Y\) above \(r\), using \(Z^rY\) as its term in degree \(r\). The resulting complex \(T\) has the claimed bounds and its inclusion into \(Y\) is a quasi-isomorphism. The map \(q\) factors as a chain map \(X\to T\): the source is zero above \(b\leq r\), and its degree-\(r\) image consists of cycles when that source term can be nonzero. Its composite with \(T\to Y\) is \(q\), so it too is a quasi-isomorphism. For \(d=0\), every object lies in the class and one can simply take \(T=X\). Applying Lemma 2.2 to bounded complexes proves (3.2). \(\square\)

The cutoff \(b+d\) is a proved sufficient bound. Cogeneration alone does not guarantee bounded class representatives; Exercise 2 gives an explicit obstruction.

4. Injectives, actual homotopies and small Hom sets

Let \(\mathcal I\) be the full subcategory of injective objects. An object is injective if any map to it from a subobject extends to the containing object.

Lemma 4.1. Every map from an acyclic complex \(N\) to a bounded-below complex \(I\) of injective objects is null-homotopic. The rest of \(\mathcal A\) need not have enough injectives.

Proof. Suppose \(I^n=0\) for \(n<m\), and set \(h^n=0:N^n\to I^{n-1}\) for \(n\leq m\). Suppose the homotopy equations are proved below \(n\). The residual \(r^n=f^n-d_I^{n-1}h^n\) kills \(d_N^{n-1}\): the chain-map equation and the preceding homotopy equation give

At the starting degree the target's lower zero term gives the same equality. Since \(N\) is acyclic, \(\operatorname{coker}d_N^{n-1}\) identifies with \(\operatorname{im}d_N^n=Z^{n+1}N\). Thus \(r^n\) factors through this subobject of \(N^{n+1}\). Extend its factor to \(h^{n+1}:N^{n+1}\to I^n\) by injectivity. This gives \(f^n=d_I^{n-1}h^n+h^{n+1}d_N^n\). Upward induction proves every equation. \(\square\)

This is the bounded test of the earlier roof lesson, with its exact subobject and indices displayed. An acyclic bounded-below complex of injectives is therefore contractible, by applying the lemma to its identity map. In \(K^+(\mathcal I)\) each quasi-isomorphism has a zero cone, hence is already invertible. If \(\mathcal A\) has enough injectives, Theorem 2.1 and Lemma 2.2 give

If \(\mathcal I\) also satisfies (3.1), Proposition 3.1 gives \(K^b(\mathcal I)\simeq D^b(\mathcal A)\): its bounded acyclic complexes are also contractible. Finite biproducts of injectives are injective by extending each component separately, so the asserted homotopy categories and their cone triangles are well defined.

Corollary 4.2. If \(\mathcal A\) is locally \(\mathcal U\)-small for a Grothendieck universe \(\mathcal U\) containing \(\mathbb Z\), and has enough injectives, all Hom sets of \(D^+(\mathcal A)\) are \(\mathcal U\)-small.

Proof. Choose a bounded-below representative \(X\) and a bounded-below injective replacement \(Y\to I\). The K-injective Hom test gives

Choose \(\mathcal U\)-set encodings of the sets \(\operatorname{Hom}_{\mathcal A}(X^n,I^n)\). Their \(\mathbb Z\)-indexed product is \(\mathcal U\)-small. Chain maps form a subset of that product, defined by the differential equations. Quotienting this set by chain homotopy remains \(\mathcal U\)-small. This proves the conclusion for every chosen pair. No smallness of the collection of all objects is asserted. \(\square\)

5. Arbitrary complexes under the finite condition

Assume throughout this section that \(\mathcal J\) is cogenerating and satisfies (3.1).

Lemma 5.1 (initial tail). For any complex \(X\) and integer \(n\), there is a quasi-isomorphism \(X\to Y\) whose target terms lie in \(\mathcal J\) in degrees at least \(n\).

Proof. Replace the stupid upper tail \(X_{\geq n}\), which is \(X\) in degrees at least \(n\) and zero below, by a bounded-below class complex \(I\). Form \(Y\) by retaining \(X^j\) for \(j<n\) and using \(I^j\) for \(j\geq n\). The boundary differential is

The two adjacent composites are zero by the chain-map equation and \(d_X^2=0\). Identity below \(n\) and \(f\) above define a chain map. Compare the short exact sequences of complexes with upper tails as their subcomplexes. Their tail map is a quasi-isomorphism; the quotient complexes are the identical stupid lower tail, with identity comparison. The long exact cohomology sequences then give an isomorphism on the middle complex in every degree: in the five consecutive terms around that degree, the four tail/quotient maps are isomorphisms, and exactness gives injectivity and surjectivity of the middle map. Hence \(X\to Y\) is a quasi-isomorphism. \(\square\)

Lemma 5.2 (downward splice). Suppose \(X^j\in\mathcal J\) for \(j\geq n\). For any \(n'<n\), there is a quasi-isomorphism \(q:X\to Z\) with \(Z^j\in\mathcal J\) for \(j\geq n'\), and \(q^j=1_{X^j}\) for \(j\geq n+d\).

Proof. By Lemma 5.1 choose a quasi-isomorphism \(f:X\to Y\) with \(Y^j\in\mathcal J\) for \(j\geq n'\). Its cone \(M=C(f)\) is acyclic, and \(M^j\in\mathcal J\) for \(j\geq n-1\). Set

If \(d\geq1\), acyclicity gives an exact sequence of exactly \(d\) class terms

Its first degree is \(a-d=n-1\), so (3.1) puts \(L\) in \(\mathcal J\). For \(d=0\), every object already belongs to the class. Define

Its lower differentials are those of \(Y\), its upper ones those of \(X\), and the boundary maps are

where \(L\hookrightarrow M^a=Y^a\oplus X^{a+1}\). These pair formulas mean morphisms with the given projections, not an assumption that \(\mathcal A\) is a module category. A member of the kernel satisfies \(d_Y^ay+f^{a+1}x=0\) and \(d_X^{a+1}x=0\). These equalities make (5.5) land in the stated kernel and make both adjacent composites zero.

Define \(q:X\to Z\) and \(p:Z\to Y\) by

The degree-\(a\) value of \(q\) lies in \(L\), since \(f\) is a chain map and \(d_X^2=0\). At the lower boundary, \(q^ad_X^{a-1}=(d_Y^{a-1}f^{a-1},0)=d_Z^{a-1}q^{a-1}\). At the upper boundary, \(d_Z^aq^a=d_X^a=q^{a+1}d_X^a\). For \(p\), the lower equation is immediate, and the upper equation is \(d_Y^ay=f^{a+1}(-x)\), precisely the kernel equation. All other equations are those of \(f\) or identity. Thus both maps are chain maps, and \(pq=f\).

We check the three affected cohomology degrees. At \(a-1\), the kernel of \(y\mapsto(d_Yy,0)\) is \(Z^{a-1}Y\); its boundaries are unchanged. Thus \(H^{a-1}Z=H^{a-1}Y\), and \(H^{a-1}q=H^{a-1}f\). At \(a\), the cycles in \(L\) with \(-x=0\) identify with \(Z^aY\) by \(y\mapsto(y,0)\). Its boundaries identify with \(B^aY\), so again \(H^aZ=H^aY\) and the induced map is \(H^af\).

At \(a+1\), acyclicity identifies \(L=\operatorname{im}d_M^{a-1}\). In particular the map

is epic. Its composite with \(d_Z^a\) is \((y,x)\mapsto d_X^ax\), which has image \(B^{a+1}X\). Epicity proves \(\operatorname{im}d_Z^a=B^{a+1}X\). The next differential is \(d_X^{a+1}\), so \(H^{a+1}Z=H^{a+1}X\), and \(q\) induces identity there. At every lower degree cohomology is that of \(Y\) with map \(Hf\); at every higher degree it is that of \(X\) with identity. Therefore \(q\) is a quasi-isomorphism.

All the terms of (5.4) in degrees at least \(n'\) lie in \(\mathcal J\): lower ones by the choice of \(Y\), the cut term by (5.3), and upper ones by the hypothesis on \(X\), since \(a+1=n+d\geq n\). Formula (5.6) is identity in precisely that upper range. The cut must be one degree below the promised range; it is \(n+d-1\), not \(n+d\). \(\square\)

Theorem 5.3. Every complex \(X\) admits a quasi-isomorphism into a complex of \(\mathcal J\), and

Proof. First choose \(X\to Y_0\) by Lemma 5.1, with class terms in degrees at least some integer \(n_0\). Put \(n_k=n_0-k\). Repeatedly apply Lemma 5.2 to obtain quasi-isomorphisms

where \(Y_k\) has class terms in degrees at least \(n_k\), and the transition from stage \(k\) is identity in degrees at least \(n_k+d\).

For a fixed degree \(j\), all transitions are literal identities in that degree once \(j\geq n_k+d\). Both this term and its outgoing differential are then constant, since degree \(j+1\) is also in the fixed range. Take those eventual values to define \(Y^j\) and \(d_Y^j\). They belong to \(\mathcal J\). A complex equation uses three adjacent terms and thus holds at a common finite stage. The finite composites from \(X\) give eventually constant components, hence a chain map \(X\to Y\). In degree \(j\), choose a common stage at which the terms in degrees \(j-1,j,j+1\) and the corresponding components have stabilized. The induced cohomology map is then that of a finite composite of quasi-isomorphisms, so is invertible.

This construction even gives the colimit of (5.9) in complexes, without assuming arbitrary colimits exist in \(\mathcal A\). The map from each stage to its stable value is the finite transition composite up to a stable stage in each degree. These components satisfy the chain equations at common stages and form a compatible cocone. Any other compatible cocone determines a unique map from each stable term to its target. Compatibility again proves the chain equations and uniqueness. This is the colimit universal property for this particular eventually constant diagram. It uses neither AB5 nor exactness of filtered colimits. Lemma 2.2 now proves (5.8). \(\square\)

6. Thick abelian subcategories and their derived comparisons

Let \(\mathcal B\subset\mathcal A\) be a strictly full additive subcategory closed under kernels and cokernels of maps between its objects, and under extensions. We call this a thick abelian subcategory. Its kernels and cokernels are the ambient ones, so it is abelian and its inclusion is exact. Closure under all ambient subobjects and quotients is the stronger Serre condition; it is not needed here. Write

as a full subcategory. If the original full additive class is not replete, use its isomorphism closure and interpret membership as being isomorphic to an object of the class. The inclusion into that closure is fully faithful and essentially surjective; an inverse chooses class representatives and conjugates every arrow by the chosen isomorphisms. Both functors are additive and exact because they transport the ambient kernel and cokernel maps. Applying them termwise gives inverse equivalences on complexes and homotopies, preserves cohomology and quasi-isomorphisms, and hence gives inverse equivalences on each derived localization. This transports the conclusions back to the original class with its chosen representatives.

Proposition 6.1. Each category (6.1) is triangulated. Exact inclusion induces a natural exact comparison

Proof. Cohomology of complexes from \(\mathcal B\) is computed by the same kernels, images and quotients in either category. Its finite biproducts and shifts therefore have cohomology in \(\mathcal B\). For a triangle \(X\xrightarrow fY\to Z\to X[1]\), its long exact cohomology sequence gives

If \(X,Y\) belong to (6.1), the first and last objects belong to \(\mathcal B\) by its cokernel and kernel closures for the indicated maps between its objects, and the middle one belongs by extension closure. Rotation and shifts give the other two-vertex cases. Cohomological lower or upper bounds remain bounded under the same exact sequence; the bounded case has both bounds. Repleteness follows from invariance of cohomology under isomorphism. This proves all the required inherited triangle operations. If a retract is present, its cohomology is the image of an idempotent endomorphism \(e\) of the original cohomology. This image is \(\ker(1-e)\), a kernel of a map between objects of \(\mathcal B\), and hence belongs to \(\mathcal B\). Thus the category is also closed under retracts.

Termwise inclusion gives a functor on complexes and homotopies, retains shifts and cone matrices, and carries quasi-isomorphisms to quasi-isomorphisms. Its localization is therefore exact and lands in (6.1), proving (6.2). \(\square\)

Theorem 6.2. Suppose that for every monomorphism \(S\hookrightarrow W\), with \(S\in\mathcal B\), there is a map \(W\to B\), with \(B\in\mathcal B\), whose composite with \(S\) is monic. Then

Dually, if for every epimorphism \(W\twoheadrightarrow S\), with \(S\in\mathcal B\), there is \(B\to W\), \(B\in\mathcal B\), whose composite with \(S\) is epic, then

Proof of (6.4). A bounded-below representative of an object of \(D^+_{\mathcal B}(\mathcal A)\) has cohomology in \(\mathcal B\). The alternative criterion in Theorem 2.1 applies: thickness gives its extension and monic-cokernel closure, and the hypothesis gives its compatible enlargement. No subobject or quotient closure for arbitrary ambient arrows is used. Thus this representative has an outgoing quasi-isomorphism to a bounded-below \(\mathcal B\)-complex.

For any two such class complexes, the ambient denominator targets and equality witnesses have the same cohomology as a class complex whenever they occur with a quasi-isomorphic denominator. They too admit these replacements. The restricted density argument of Lemma 2.2 consequently proves full faithfulness as well as essential surjectivity of \(\delta^+\). Its source localization is exactly \(D^+(\mathcal B)\), because its cohomology and quasi-isomorphisms are computed intrinsically or ambiently with the same result.

For the bounded case, the full embeddings \(D^b\hookrightarrow D^+\) for both abelian categories imply full faithfulness of \(\delta^b\). To prove essential surjectivity, replace a bounded ambient representative by a bounded-below \(\mathcal B\)-complex \(Y\). Choose \(b\) above its nonzero cohomology. Its upper smart truncation has top term \(Z^bY\), which belongs to \(\mathcal B\) as the kernel of \(d_Y^b:Y^b\to Y^{b+1}\), a map between two objects of \(\mathcal B\); all its other terms also belong. This is a bounded \(\mathcal B\)-complex with a quasi-isomorphism to \(Y\). No uniform finite condition is needed here. Exactness was already proved in Proposition 6.1, so both are exact equivalences.

Proof of (6.5). Pass to the opposite abelian category. An epimorphism \(W\to S\) becomes a monomorphism \(S^{\mathrm{op}}\to W^{\mathrm{op}}\), and the stipulated composite becomes the monic composite of (6.4). The thick subcategory is \(\mathcal B^{\mathrm{op}}\). To retain the complex indices explicitly, put

This reverses a chain map \(X\to Y\) into \(RY\to RX\). If its homotopy has components \(h^j:X^j\to Y^{j-1}\), the reversed homotopy has components \((h^{-j+1})^{\mathrm{op}}\); reversing the homotopy equation gives exactly the homotopy equation for (6.6). Kernels and cokernels exchange, and the abelian image comparison gives \(H^j(RX)=(H^{-j}X)^{\mathrm{op}}\). Thus this construction preserves quasi-isomorphisms, exchanges bounded-above and bounded-below complexes, preserves bounded complexes, and is an involution on the localized categories up to their chosen opposite identifications. Apply (6.4) in \(\mathcal A^{\mathrm{op}}\) and reverse it to get full faithfulness and essential surjectivity in (6.5). Their exactness follows independently from Proposition 6.1, so no unverified assertion about opposite cone signs is needed. \(\square\)

The general comparisons (6.2) exist without the enlargement hypotheses. Those hypotheses prove the bounded equivalences just stated; no equivalence for arbitrary unbounded complexes is inferred from them alone.

7. Four graded exercises with complete solutions

Exercise 1 — A class that is not cogenerating (introductory)

Let \(\mathcal A\) be all vector spaces over a field \(k\), and \(\mathcal J\) its finite-dimensional subcategory. Verify the alternative criterion of Theorem 2.1. Explain what it says for a bounded-below complex with finite-dimensional cohomology, why \(\mathcal J\) is not cogenerating, and why its replacement maps cannot in general be termwise monic.

Solution. Kernels and quotients of finite-dimensional spaces are finite-dimensional. An extension of spaces of dimensions \(r,s\) has dimension \(r+s\), by choosing a basis in the subspace and lifting a quotient basis. Thus extension and monic-cokernel closure hold. For finite-dimensional \(S\hookrightarrow W\), extend a basis of \(S\) to one of \(W\), and project onto the span of the first basis. The resulting \(W\to S\) restricts to identity on \(S\); this proves compatible enlargement. Theorem 2.1 therefore replaces every bounded-below complex with finite-dimensional cohomology by one whose terms are finite-dimensional.

An infinite-dimensional space cannot embed in a finite-dimensional one: the images of any finite independent family under an embedding remain independent, contradicting the target's finite dimension when the family is large enough. Hence the class is not cogenerating. For the bounded-below acyclic complex \(W\xrightarrow1W\) in degrees 0 and 1, with \(W\) infinite-dimensional, all cohomology is zero. Its class replacement exists (the zero complex is one), but no termwise monomorphism from it into finite-dimensional terms can exist. Finally, if a complex of finite-dimensional terms is quasi-isomorphic to \(X\), every \(H^nX\) is finite-dimensional, since it is a subquotient of its degree-\(n\) term. The finite-cohomology hypothesis is therefore necessary in this example.

Exercise 2 — Cogeneration without bounded representatives (intermediate)

Let \(\mathcal A\) be finite-dimensional \(k\)-vector spaces and \(\mathcal J\) its full subcategory of even-dimensional spaces. Prove that it is additive and cogenerating but fails (3.1) for every \(d\geq0\). Construct a bounded-below class replacement of \(k[0]\), and prove that no bounded class complex represents \(k[0]\).

Solution. Zero and finite sums of even-dimensional spaces have even dimension, and the full subcategory inherits addition of maps. Every \(V\) embeds in \(V\oplus V\) by its first inclusion, so it is cogenerating.

Put \(N:k^2\to k^2\), \(N(u,v)=(0,u)\), and \(p:k^2\to k\), \(p(u,v)=u\). Then \(N^2=0\) and \(\operatorname{im}N=\ker N=\ker p=k(0,1)\). For any \(d\geq1\), the sequence of \(d\) copies of \(k^2\), joined by \(N\), and ending in \(p:k^2\to k\to0\), is exact at every term after the first. Its final object \(k\) has odd dimension. This disproves (3.1) for that \(d\), including \(d=1\) where only the epimorphism is required. For \(d=0\), the claim that every space has even dimension is already false.

Let \(Y^j=k^2\) for \(j\geq0\), zero otherwise, and \(d_Y^j=N\) for \(j\geq0\). Its degree-zero cohomology is \(\ker N=k(0,1)\), and its higher cohomology is \(\ker N/\operatorname{im}N=0\). The map \(k[0]\to Y\) sending \(t\) to \((0,t)\) in degree zero is a chain map and an isomorphism on every cohomology. It is the required bounded-below replacement.

For a finite complex \(T\), write \(z_j=\dim Z^jT\), \(b_j=\dim B^jT\). The exact sequences defining cycles and cohomology give \(\dim T^j=z_j+b_{j+1}\) and \(z_j=\dim H^jT+b_j\). Taking the finite alternating sum cancels the two boundary sums, so

If all the terms are even-dimensional, the left side is even. A representative of \(k[0]\) would have right side 1. This contradiction proves that no bounded class representative exists.

Exercise 3 — The splice boundary and its signs (advanced)

Take \(\mathcal J=\mathcal A\), all \(k\)-vector spaces, so \(d=0\). Let \(n=0\), \(n'=-1\), \(X=k[0]\), and let \(Y\) have \(Y^{-1}=k^2\), \(Y^0=k^3\), differential \((u,v)\mapsto(u,v,0)\), and zero other terms. Let \(f^0(t)=(0,0,t)\). Compute its cone and the splice of Lemma 5.2. Show why cutting at \(n+d\) would not make the map identity in degree \(n+d\).

Solution. The map is a quasi-isomorphism: \(H^{-1}Y=0\), and \(H^0Y=k^3/(k^2\oplus0)\) identifies with \(k\) by its last coordinate, under which \(H^0f\) is identity. Formula (0.1) gives \(M^{-1}=k^2\oplus k\), \(M^0=k^3\), with differential \((u,v,t)\mapsto(u,v,t)\). The other terms vanish. Thus the cone is acyclic, \(\ker d_M^{-1}=0\), and \(\ker d_M^0=k^3\).

The correct cut is \(a=n+d-1=-1\). Formula (5.4) gives \(Z^{-1}=0\), \(Z^0=k\), zero other terms, so \(Z=X\). At degree \(-1\), the source of \(q^{-1}\) is zero; at degree 0, (5.6) gives \(q^0=1_k\). The map \(p^0\) is the inclusion of the third coordinate, and \(pq=f\). All differentials of \(X\) and \(Z\) are zero, so the chain equations and cohomology maps are precisely these identities.

If instead one cut at \(a=0=n+d\), the cut term would be \(\ker d_M^0=k^3\). The prescribed cut component would be \(q^0(t)=(0,0,t)\), because \(d_X^0=0\). It is not an isomorphism, much less an identity, even though it induces the right cohomology after the retained lower differential. Such a cut fixes only the strictly higher degrees. Moving the cut to \(n+d-1\) proves the stronger fixed-degree range in the lemma.

Exercise 4 — Two intrinsic comparisons (advanced)

For all \(k\)-vector spaces \(\mathcal A\) and finite-dimensional spaces \(\mathcal B\), prove both enlargement hypotheses of Theorem 6.2 and all three bounded comparison equivalences. For a bounded complex \(X\) of arbitrary vector spaces with finite-dimensional cohomology, exhibit a bounded \(\mathcal B\)-complex representing it. Explain why a claim about its terms' dimensions alone would not prove full faithfulness of the derived comparison.

Solution. Subobjects, quotients and extensions of finite-dimensional spaces are finite-dimensional by the basis argument of Exercise 1; the full subcategory is therefore thick. Its monomorphism enlargement is the basis projection constructed there. Given an epimorphism \(W\to S\) with \(S\) finite-dimensional, lift a finite basis of \(S\) to \(W\). Those lifts are independent, since applying the epimorphism to a linear relation gives a relation among the basis vectors. Their finite span \(B\subset W\) maps isomorphically onto \(S\). Thus \(B\hookrightarrow W\) is the required epimorphism enlargement.

Theorem 6.2 gives \(D^+(\mathcal B)\simeq D^+_{\mathcal B}(\mathcal A)\), \(D^-(\mathcal B)\simeq D^-_{\mathcal B}(\mathcal A)\), and \(D^b(\mathcal B)\simeq D^b_{\mathcal B}(\mathcal A)\). To exhibit a bounded representative, choose vector-space splittings \(Z^jX=B^jX\oplus H_j\) and \(X^j=Z^jX\oplus L_j\). The restriction \(d_X^j:L_j\to B^{j+1}X\) is an isomorphism: its kernel is \(L_j\cap Z^jX=0\), and its image is all of \(B^{j+1}X\). Thus \(X\) is the direct sum, degreewise, of the zero-differential complex with terms \(H_j\simeq H^jX\) and the two-term identity complexes supplied by these isomorphisms. Explicitly, define \(h^j:X^j\to X^{j-1}\) by projecting onto \(B^jX\), applying \((d_X^{j-1}|_{L_{j-1}})^{-1}\), and including \(L_{j-1}\). The composite \(d_X^{j-1}h^j\) is identity on \(B^jX\) and zero on the other summands; \(h^{j+1}d_X^j\) is identity on \(L_j\) and zero on the other summands. If \(i,p\) are inclusion and projection of the \(H_j\)-complex, this proves \(1_X-ip=d_Xh+hd_X\), while \(pi=1\). Both \(i,p\) are chain maps. Hence the finite complex \(\bigoplus_jH^jX[-j]\), with zero differential, is a bounded finite-dimensional representative.

This construction gives essential surjectivity for the example. Full faithfulness additionally concerns every derived arrow and every equality between arrows. A mere count of term dimensions supplies neither. In the proof of Lemma 2.2 and Theorem 6.2, quasi-isomorphic denominator targets and common equality witnesses are replaced inside \(\mathcal B\), preserving the actual homotopy classes. That argument proves surjectivity and injectivity of the Hom comparison, completing the equivalences.

Freely accessible references

Stacks, Tag 013R gives the bounded-below injective null-homotopy test. The complete component calculation is also in the earlier roof lesson and Section 4 here. Stacks, Section 13.17, Tag 06UP treats cohomology-defined triangulated subcategories and gives a bounded-above comparison under a subobject epimorphism condition. Its condition implies the epimorphism enlargement of Theorem 6.2 by using the subobject inclusion. Our proof states both variances and the bounded restriction with their maps. The general additive-class replacements, finite cutoff and explicit unbounded splice are proved in Sections 1–5.