Draft lesson.

Filtered complexes and Cartan–Eilenberg resolutions

A filtered complex turns successive lifting problems into a spectral sequence. A double complex supplies two such filtrations. We construct every page directly, prove convergence under a finite condition in each degree, and build Cartan–Eilenberg resolutions with their comparison maps. The resulting two spectral sequences compute a bounded-below derived functor. One example has a nonzero second differential; another computes the derived Hom of a two-term complex.

We use Kernels, cokernels and the abelian comparison for the abelian image calculus. Chain homotopies, mapping cones and triangles fixes cochain, Hom and shift signs. Derived morphisms through quasi-isomorphism roofs, Section 5, proves the K-injective and K-projective tests. Injective replacements and right derived functors constructs bounded-below replacements, their comparisons, right derived functors and enough injectives in abelian groups. All categories and indexing sets are taken in fixed ambient universes.

1. Quotients that give all spectral pages

Let \(C\) be a cochain complex in an abelian category \(\mathcal A\). A decreasing filtration consists of subcomplexes \(F^pC\), for all integers \(p\), with \(F^{p+1}C\subset F^pC\). No infinite sum or limit is used in constructing the pages. Put \(n=p+q\), and define

Here the inverse image and intersection are pullbacks of subobjects. Since the differential preserves the filtration, \(Z_0^{p,n}=F^pC^n\). The induced differential on the quotient \(E_0\) has bidegree \((0,1)\).

We will use a precise convention for computations with subobjects. If an arrow factors through the sum \(U+V\), pull back the epimorphism \(U\oplus V\to U+V\); after that epimorphic change of source it has a decomposition into a part in each subobject. An arrow into a quotient can likewise be lifted after pulling back its quotient epimorphism. Pullbacks of epimorphisms are epimorphisms by the finite abelian proof in Extending right exact functors from generators, Section 1. A finite number of these changes can be made on one common epimorphic cover. Equations and vanishing of arrows can be checked after such a cover because epimorphisms cancel on the right. Thus the representative calculations below are valid in any abelian category. They do not assume that objects have ordinary elements.

For \(r\geq1\), set

Both denominator terms are subobjects of the numerator. For the first, its differential lies in \(F^{p+r}\); for the second, its image lies in \(F^p\) and its next differential is zero. When \(r=1\), the formula gives exactly the kernel of the quotient differential on \(E_0\), divided by its image. Indeed a class in \(F^p/F^{p+1}\) is a cycle precisely when its differential lies in \(F^{p+1}\); its boundaries are images of \(dF^p\). Hence \(E_1=H(E_0,d_0)\).

The differential of \(C\) defines

The image belongs to the target numerator since \(dx\in F^{p+r}\) and \(d^2x=0\). Changing \(x\) by a term in \(Z_{r-1}^{p+1,n}\) changes its image by a term in the target's boundary denominator; changing it by a boundary has no effect. This proves well-definedness. The equality \(d_r^2=0\) follows from \(d^2=0\).

Theorem 1.1. The formulas give canonical, natural identifications \(E_{r+1}=H(E_r,d_r)\) for every \(r\geq1\).

Proof. A representative \(x\in Z_r^{p,n}\) is in the kernel of \(d_r\) exactly when, after an epimorphic cover,

Then \(x-v\in Z_{r+1}^{p,n}\), and it represents the same class in \(E_r\). Conversely every member of \(Z_{r+1}^{p,n}\) maps to the kernel. Thus this subobject surjects onto that kernel.

To identify its quotient by incoming images, suppose \(x\in Z_{r+1}^{p,n}\) is zero in \(E_r\). Write \(x=a+db\), with \(a\in Z_{r-1}^{p+1,n}\) and \(b\in Z_{r-1}^{p-r+1,n-1}\). Since \(dx=da\in F^{p+r+1}\), we have \(a\in Z_r^{p+1,n}\). Also \(b\in F^{p-r}\) and \(db=x-a\in F^p\), so \(b\in Z_r^{p-r,n-1}\). Its boundary is therefore an incoming \(d_r\)-image. More generally, any incoming image has a representative \(dy\) with \(y\in Z_r^{p-r,n-1}\). Subtracting it reduces to the preceding zero-class calculation. Consequently the quotient of the kernel by those images is

which is (1.2) on page \(r+1\). The reverse inclusions hold too: the first term is a zero class on page \(r\), and the second is an incoming image. All maps used are restrictions, images or quotient maps of the original differential, so the identifications are independent of the covering lifts and natural. \(\square\)

A chain map preserving each filtration induces maps on all these subobjects and quotients, and commutes with every differential and page identification. Identities and compositions are inherited from the underlying chain maps. If two such maps have a filtration-preserving chain homotopy, their maps on \(E_0\) are chain homotopic and therefore agree on \(E_1\). They then agree on every later page by Theorem 1.1.

2. Finite convergence and comparison

Assume now that for each degree \(n\) there are integers \(a_n<b_n\) with

The bounds may depend on the degree. In particular, the filtration need not have a single finite length for the entire complex. Define the induced cohomology filtration by

Writing \(Z^n=\ker d^n\) and \(B^n=\operatorname{im}d^{n-1}\), this image is the image of \(F^pC^n\cap Z^n\) in \(Z^n/B^n\). It is all of \(H^n(C)\) for \(p\leq a_n\), and zero for \(p\geq b_n\), so it is finite.

Theorem 2.1. Under (2.1), every position stabilizes after finitely many pages, and its stable value is naturally

Proof. Fix \(p,n\). Choose \(r\) so large that \(F^{p+r}C^{n+1}=0\) and \(F^{p-r+1}C^{n-1}=C^{n-1}\); these conditions hold for every larger \(r\) too. Then the numerator in (1.2) is \(F^pC^n\cap Z^n\), its first denominator term is \(F^{p+1}C^n\cap Z^n\), and the second is \(F^pC^n\cap B^n\). For the last equality, a boundary in \(F^p\) can be lifted through \(C^{n-1}\to B^n\); its lift has differential in \(F^p\), which is exactly the defining condition on the indicated \(Z_{r-1}\). Thus the page is

This is the quotient in (2.3). To see the kernel explicitly, the inverse image of \(F^{p+1}H^n\) in \(F^pC^n\cap Z^n\) consists of a cycle in \(F^{p+1}\) plus a boundary that lies in \(F^p\). Such decompositions are justified by the epimorphic-cover convention. The formula remains identical on subsequent pages, and all identifications commute with filtered chain maps. No interchange of infinite limits, images or intersections occurs. \(\square\)

Corollary 2.2. If a filtered chain map between two complexes satisfying (2.1) is an isomorphism on some whole page, it is a quasi-isomorphism. Its cohomology maps are isomorphisms of filtered objects, including the inverse maps.

Proof. The page identifications propagate the isomorphism to every later page, and then to the stable pieces. In each total degree only finitely many pieces are nonzero. For the two finite filtrations, choose a common range by extending with zero and the whole object. Induct through \(0\to F^{p+1}H\to F^pH\to\operatorname{gr}^pH\to0\), starting at the zero end. A map of short exact sequences that is an isomorphism on the subobject and quotient is an isomorphism in the middle: its kernel is zero by projecting first to the quotient and then to the subobject; to lift a target arrow, lift its quotient after an epimorphic cover and correct the remaining subobject error. This also proves surjectivity. The induction gives isomorphisms on every filtration step and on the whole cohomology. Their inverses preserve each step. \(\square\)

3. The two filtrations of a double complex

A double complex \(L^{a,b}\) has commuting differentials \(d_h\) and \(d_v\) of bidegrees \((1,0)\) and \((0,1)\). Suppose \(L^{a,b}=0\) unless \(a\geq m\) and \(b\geq0\). Its total complex uses finite sums:

The two mixed terms in \(D^2\) cancel because the horizontal differential changes \(a\) by one; the two pure terms are zero. Each total degree has only finitely many summands, even if the complex has no upper bound.

The column filtration and row filtration are

Both satisfy (2.1): \(P^pT^n=T^n\) for \(p\leq m\) and is zero for \(p>n\); \(Q^pT^n=T^n\) for \(p\leq0\) and is zero for \(p>n-m\). Theorems 1.1 and 2.1 therefore give two convergent spectral sequences with finite filtrations on \(H^nT\).

For the column filtration, the quotient differential is \((-1)^p d_v\). Multiplication in degree \(q\) by \((-1)^{pq}\) identifies that column complex with its ordinary vertical complex. Alternatively, cycles and boundaries are identical, so their quotient cohomology has its direct identity identification. Using this latter identification, the next differential is the map induced by \(d_h\). Thus

These conventions specify the signs of higher maps through (1.3), rather than by an unspecified choice of page isomorphisms. A representative of a page-\(r\) class can be assembled from terms \(x_i\in L^{p+i,q-i}\), for \(0\leq i<r\). Its successive equations are

To justify this description, reduce a representative in \(P^pT^{p+q}\) modulo \(P^{p+r}\). Its only surviving components are the displayed \(x_i\), and the condition that its differential lie in \(P^{p+r}\) is exactly these equations. Components farther to the right belong to the denominator \(Z_{r-1}^{p+1}\) and change the final obstruction only by a target denominator. When the objects are not modules, all choices of representatives and decompositions are made after epimorphic covers as in Section 1.

For the row filtration the complementary degree is the horizontal degree. Its quotient differential is \(d_h\); on the first cohomology page the differential is \((-1)^q d_v\). Hence

For the second line, identify the signed vertical complex with the ordinary one by multiplication in degree \(p\) by \((-1)^{pq}\). Both spectral sequences have \(d_r\) of bidegree \((r,1-r)\), using their own displayed indexing. Their abutment is the same total cohomology; their induced filtrations need not agree.

Proposition 3.1. A morphism of such double complexes that is a quasi-isomorphism in every column is a quasi-isomorphism on total complexes. The same is true with rows in place of columns.

Proof. A common lower bound works for source and target. For columns the map is an isomorphism on (3.3)'s first page. For rows it is an isomorphism on (3.5)'s first page. Apply Corollary 2.2 to the corresponding degree-wise finite filtration. \(\square\)

4. Constructing Cartan–Eilenberg resolutions and their comparisons

Assume \(\mathcal A\) has enough injectives. For a complex \(X\), retain the notation \(B^aX\), \(Z^aX\), \(H^aX\). Call a short exact sequence of complexes cyclewise exact when the induced sequences on all boundary, cycle and cohomology objects are short exact too. Call a monomorphism cyclewise exact when it belongs to such a sequence with its quotient complex.

Choose monomorphisms \(B^aX\to J_B^a\) and \(H^aX\to J_H^a\) into injectives. Extend the first to a map \(\alpha^a: X^a\to J_B^a\). Since \(H^aX=Z^aX/B^aX\) is a subobject of \(X^a/B^aX\), extend the second to a map \(\beta^a: X^a/B^aX\to J_H^a\). Define the complex

The tuple notation describes morphisms into finite biproducts; \(\iota_B^{a+1}\) denotes the chosen embedding of the boundary object. Since \(\alpha^{a+1}\) extends that embedding, \(e\) is a chain map. Its last coordinate kills a kernel arrow only when that arrow factors through \(Z^aX\); then its middle coordinate forces it into \(B^aX\), and its first coordinate kills it only when it is zero. Thus every \(e^a\) is monic. Moreover

The maps induced by \(e\) on boundaries and cohomology are the chosen embeddings, and its map on cycles is monic.

Lemma 4.1. The sequence \(0\to X\to J(X)\to X_1\to0\), where \(X_1\) is the quotient complex, is cyclewise exact.

Proof. Its ordinary cohomology long exact sequence has zero connecting maps because \(H^aX\to H^aJ(X)\) is monic in every degree. It gives short exact cohomology sequences. Boundaries of the quotient lift from boundaries of \(J(X)\): lift a source of their differential through the degree-wise quotient epimorphism and then take its differential. The kernel of \(B^aJ(X)\to B^aX_1\) is the intersection of \(B^aJ(X)\) with \(X^a\). An arrow in this intersection is a cycle of \(X\), and its class is killed in \(H^aJ(X)\). Cohomology injectivity makes it a boundary of \(X\). This proves the short exact boundary sequences. Cycles of the quotient lift to cycles of \(J(X)\) because their cohomology connecting map is zero: explicitly, the differential of a degree-wise lift lies in \(X^{a+1}\); it is a boundary there, so subtract a lift from \(X^a\). The kernel is \(Z^aX\). This proves the cycle sequences. All lifts are along epimorphisms, so the argument applies in \(\mathcal A\). \(\square\)

Lemma 4.2. If \(A\to B\) is a cyclewise exact monomorphism, every chain map \(A\to J(X)\) extends to \(B\to J(X)\).

Proof. A chain map \(f\) into (4.1)'s split complex is specified by two families: maps \(f_B^a: A^a\to J_B^a\), and maps \(f_H^a: A^a/B^aA\to J_H^a\). The third coordinate in degree \(a\) is necessarily \(f_B^{a+1}d_A^a\). The middle coordinate kills boundaries by the chain-map equation. Conversely these families give a chain map, since \(d_A^2=0\).

The arrows \(A^a\to B^a\) are monic. So are \(A^a/B^aA\to B^a/B^aB\): the intersection of \(A^a\) with the boundary object of \(B\) is the boundary object of \(A\), by injectivity of \(H^aA\to H^aB\). Extend each family through these monomorphisms using the corresponding injective object. The resulting maps have the same third-coordinate relation and therefore form the required extension. Each degree involves only three coordinates; no infinite biproduct is needed. \(\square\)

Theorem 4.3. Every bounded-below complex \(K\) has a Cartan–Eilenberg injective resolution, with comparisons of complex maps unique up to a vertical homotopy consisting of horizontal chain maps.

Proof. Suppose \(K^a=0\) for \(a<m\). In (4.1) choose the zero injective for every zero boundary or cohomology object below the bound, including \(B^mK=0\). Then \(J(K)^a=0\) for \(a<m\). Lemma 4.1 makes its quotient \(K_1\) bounded below by the same \(m\). Repeat the construction on that quotient and on all subsequent quotients. Put

Each arrow is a horizontal chain map, so the horizontal and vertical differentials commute; consecutive vertical maps compose to zero. The augmented vertical sequences on each of \(K^a\), \(B^aK\), \(Z^aK\) and \(H^aK\) are exact because every successive short exact sequence is cyclewise exact. Their terms are injective by (4.2) and finite biproduct closure of injectives. Thus they are injective resolutions of those four objects. There are no terms with \(a<m\) or \(j<0\). These are all the required Cartan–Eilenberg properties.

For a chain map \(f: K\to L\), let \(I,J\) be two resolutions constructed in this way. Extend \(K\to L\to J^{\bullet,0}\) across \(K\to I^{\bullet,0}\) using Lemma 4.2. The resulting horizontal chain map induces a quotient map \(K_1\to L_1\). Extend it across \(K_1\to I^{\bullet,1}\), and continue. This gives a double-complex morphism extending \(f\); the quotient construction makes every vertical square commute.

For two such comparisons \(u,v\) extending the same \(f\), their difference in degree zero vanishes on \(K\) and therefore factors through \(K_1\). Extend that factor to a horizontal chain map \(h^1: I^{\bullet,1}\to J^{\bullet,0}\). It gives \((u-v)^0=h^1d_I^0\). Inductively, if \(h^1,\ldots,h^j\) have been chosen, the map \((u-v)^j-d_J^{j-1}h^j\) kills the image of \(d_I^{j-1}\): substitute the preceding homotopy equation and use \(d_J^2=0\). It factors through \(K_{j+1}\), and Lemma 4.2 extends the factor to \(h^{j+1}\). Hence

Every \(h^j\) is a horizontal chain map. The same extension argument also works when the target is any Cartan–Eilenberg resolution whose horizontal rows split as (4.1): its injective boundary and cohomology objects supply the same extension property. \(\square\)

The total homotopy is \(H|_{I^{a,j}}=(-1)^a h^j\). The two vertical terms in \(DH+HD\) give (4.4), and the horizontal terms cancel because \(h^j\) commutes with \(d_h\) while \((-1)^a\) changes sign in the next horizontal degree. Thus comparison maps are also unique up to ordinary total-complex homotopy. Comparisons lifting identities in both directions compose to maps homotopic to identities. Comparisons of compositions are homotopic to composites by the same uniqueness proof.

5. Both derived spectral sequences and their naturality

Let \(F: \mathcal A\to\mathcal B\) be additive and left exact, with \(\mathcal B\) abelian and \(\mathcal A\) having enough injectives. Let \(K\to I\) be a Cartan–Eilenberg resolution as in Theorem 4.3.

View \(K\) as a double complex concentrated in vertical degree zero. Its map to \(I\) is a quasi-isomorphism in every column, so Proposition 3.1 proves

The total complex is bounded below and every term is a finite sum of injectives. The earlier K-injective test therefore makes it a K-injective replacement. Applying \(F\) to it computes \(RF(K)\). Finite sums commute with an additive functor, including all inclusions, projections and signs, so \(F\operatorname{Tot}(I)=\operatorname{Tot}(FI)\).

In the column spectral sequence of \(FI\), each column is \(F\) of an injective resolution of \(K^p\). Consequently

Its first differential is \(R^qF(d_K^p)\): the horizontal map of columns extends \(d_K^p\), and injective-resolution comparison identifies its induced vertical cohomology map with that derived map. This also shows that the middle expression is a complex.

For the row spectral sequence, each row of \(I\) has split boundary and cycle sequences. Any additive functor preserves those split sequences, so horizontal cohomology after \(F\) is exactly \(F\) of the horizontal cohomology object. Those cohomology objects, as a vertical complex, resolve \(H^qK\). The signed vertical differential on the first page is \((-1)^q\) times its resolution differential; the degree-\(p\) scaling \((-1)^{pq}\) identifies it with the ordinary resolution. We obtain

Both abutments have finite filtrations, proved by Theorem 2.1. No exactness of arbitrary products or sums in \(\mathcal A\) or \(\mathcal B\) is required.

Here is the complete choice and map comparison. A comparison of Cartan–Eilenberg resolutions is a filtered total map for both filtrations. Two comparisons of the same chain map are related by (4.4). For the column filtration their signed total homotopy preserves the horizontal degree, so they give identical maps from page one. For the row filtration, take horizontal cohomology first. Since each \(h^j\) is a horizontal chain map, it induces a homotopy on the first-page vertical complexes: the signed total homotopy is \((-1)^q h\) on horizontal degree \(q\), and (4.4) remains precisely its homotopy equation for the signed vertical differential. Thus the comparisons agree from page two. All later maps then agree by Theorem 1.1. Identity comparisons and composition comparisons have the same property. Together with their total homotopies this proves coherent choice-independence of the spectral sequences and abutments, from the specified pages.

The second sequence, from page two together with its filtered abutment, is a functor of the derived object \(K\in D^+(\mathcal A)\). Indeed a quasi-isomorphism induces isomorphisms on every \(H^qK\), hence on (5.3)'s entire second page and every later page. Its abutment is the isomorphism \(RF(K)\to RF(L)\); Corollary 2.2 makes the cohomology map an isomorphism on each filtration step. Chain-homotopic maps give the same second-page map because they give the same \(H^q\)-map, and the same derived abutment map; they therefore give the same later-page maps too. The roof universal property now inverts quasi-isomorphisms and extends the construction to \(D^+\). This argument also proves that the resulting maps have the displayed \(R^pF(H^q-)\) second-page description for every derived arrow.

The first sequence is natural in the complex \(K\), and its abutment is its derived value. It need not depend only on the derived isomorphism class on its first or second page: those pages use the individual terms \(K^p\). No such stronger independence is asserted.

6. Two computations

Example 6.1: a genuine second differential. Over a field \(k\), let the only nonzero objects of a double complex be

Set \(d_h(a)=c\), \(d_h(b)=e\), \(d_v(b)=c\), and all other differentials zero. The commuting-square condition holds: each possibly nonzero mixed composite ends at a zero differential. The vertical complex of column one is \(kb\to kc\), an isomorphism. Thus the column first page has only \(k[a]\) in position \((0,1)\) and \(k[e]\) in position \((2,0)\). Its first differential is zero.

To represent the former class on page two, use \(a+b\). With (3.1)'s sign, \(D(a)=c\) and \(D(b)=e-c\), so \(D(a+b)=e\). Formula (3.4) gives \(d_2[a]=[e]\), with coefficient \(+1\). The second differential is an isomorphism, and every third-page term is zero. Direct totalization confirms this: the map from degree one, with basis \((a,b)\), to degree two, with basis \((c,e)\), is

It is invertible over every field, so the total complex is acyclic. For the row filtration, both nonzero rows are already exact: \(kb\to ke\) and \(ka\to kc\). Its first page is zero. The different collapsing pages are compatible with the common zero abutment.

Example 6.2: a two-term derived Hom. Work in abelian groups, fix \(m\geq2\), and put \(F(A)=\operatorname{Hom}(\mathbf Z/m,A)\). Let \(K\) be \(\mathbf Z\xrightarrow{m}\mathbf Z\) in degrees zero and one. Its cohomology is \(H^0K=0\), \(H^1K=\mathbf Z/m\).

The free resolution \(P^{-1}=\mathbf Z\xrightarrow{-m}P^0=\mathbf Z\to\mathbf Z/m\) and the Hom differential compute \(R^0F(A)=A[m]\), \(R^1F(A)=A/mA\), and zero above degree one. To apply it to \(K\), the same bounded-projective test shows that \(\operatorname{Hom}^\bullet(P,K)\) computes \(RF(K)\): compare with a bounded-below injective replacement of \(K\). Hom out of \(P\) preserves that quasi-isomorphism, and Hom into the K-injective replacement inverts the quasi-isomorphism \(P\to(\mathbf Z/m)[0]\). These two comparisons identify its cohomology with the required derived Hom.

Ordering the degree-one terms as \((\operatorname{Hom}(P^0,K^1),\operatorname{Hom}(P^{-1},K^0))\), the resulting total Hom complex is

Its cohomology is zero in degree zero, \(\mathbf Z/m\) in degree one, and \(\mathbf Z/m\) in degree two, with no other degrees. The first sequence (5.2) has only row \(q=1\): its entries in horizontal degrees zero and one are \(\mathbf Z/m\), and its differential is multiplication by \(m\), hence zero. No higher differential can remain in that row. The second sequence (5.3) has only \(q=1\), with \(R^0F(\mathbf Z/m)=\mathbf Z/m\) at \(p=0\) and \(R^1F(\mathbf Z/m)=\mathbf Z/m\) at \(p=1\). It likewise has no higher differential. Both sequences give exactly the directly computed groups in total degrees one and two. Enough injectives in this example was proved in the linked injective-replacement lesson; no existence theorem is being assumed just from a citation.

7. Graded exercises with solutions

Exercise 1 (routine). Let a double complex be supported in horizontal degrees \(0,1\) and vertical degrees \(0,1\). Show that its column spectral sequence stabilizes on page two, and identify the filtration of each total cohomology group. Explain why the stable terms do not always determine that group up to isomorphism.

Solution. The first page is supported in those four positions, and every later page is a subquotient there. For \(r\geq2\), a differential raises horizontal degree by \(r\); its source or target is outside the two allowed columns. All those differentials are zero, so \(E_2=E_\infty\). The column filtration is \(P^0T=T\), \(P^1T\) the column-one subcomplex, and \(P^2T=0\). Consequently, for each \(n\), it gives a short exact sequence

Terms outside the stated support are zero; in particular only total degrees zero, one and two can occur. The stable page specifies the subobject and quotient but does not specify the extension class. Already in abelian groups, \(\mathbf Z/4\) and \((\mathbf Z/2)^2\) both have a subobject \(\mathbf Z/2\) and quotient \(\mathbf Z/2\). They are not isomorphic: the first has an element of order four and the second does not.

Both possibilities occur in double complexes with exactly the requested support. Given \(0\to A\to M\xrightarrow{\pi}B\to0\), embed \(B\) into an injective abelian group \(J\), with embedding \(\eta\). Put \(L^{0,0}=0\), \(L^{0,1}=J\), \(L^{1,0}=M\), \(L^{1,1}=J\). Take \(d_h: L^{0,1}\to L^{1,1}\) to be the identity and \(d_v: L^{1,0}\to L^{1,1}\) to be \(\eta\pi\). The mixed square has zero source and thus commutes. Its sole total differential is \(J\oplus M\to J\), \((x,y)\mapsto x-\eta\pi y\). It is epic and its kernel is the graph of \(\eta\pi\), isomorphic to \(M\), so \(H^1T=M\). The column-one subcomplex \(M\to J\) has \(H^1=A\), whose inclusion into that graph is the original \(A\to M\). On the first page the remaining \(J\to J/\eta B\) differential is the quotient map, leaving \(E_2^{0,1}=B\) and \(E_2^{1,0}=A\), and all other terms zero. Take \(A=B=\mathbf Z/2\), \(J=\mathbf Q/\mathbf Z\), and either of the two middle groups above. Their stable pages are identical while their total cohomology groups differ.

Exercise 2 (intermediate). In complexes of \(k\)-vector spaces, filter \(C=(k[t]\hookrightarrow k[[t]])\), in degrees zero and one, by powers of \(t\). For \(p\leq0\) take the whole complex. Show that both term filtrations are exhaustive and separated, that every page from page one is zero, and that total cohomology is not zero. Identify exactly how finite convergence fails.

Solution. For \(p\geq0\), take \(F^pC^0=t^pk[t]\) and \(F^pC^1=t^pk[[t]]\). A polynomial or power series lying in every \(t^p\)-subspace has every coefficient zero, so both filtrations are separated. Taking \(p=0\) shows exhaustiveness. For every \(p\geq0\), the associated graded complex is \(k\xrightarrow{1}k\), because the inclusion preserves the coefficient of \(t^p\). For negative \(p\) its graded complex is zero. Thus \(E_1=0\), and Theorem 1.1 makes every later page zero.

The inclusion of polynomials in power series is monic, so \(H^0C=0\), while \(H^1C=k[[t]]/k[t]\). This quotient is nonzero; for example \(\sum_{i\geq0}t^i\) is not a polynomial. Any power series has a polynomial truncation through degree \(p-1\). Subtracting it does not change its class in the quotient and leaves a series in \(t^pk[[t]]\). Hence \(F^pH^1C=H^1C\) for every \(p\). The stable page is the zero associated graded of this constant nonzero filtration. The induced cohomology filtration is not separated, and condition (2.1) fails for both nonzero terms. Thus the finite-filtration convergence theorem does not apply. Separation on the terms alone did not imply separation on their cohomology.

Exercise 3 (advanced). Show why termwise injectivity does not replace the Cartan–Eilenberg row conditions. Use \(J=(\mathbf Q\to\mathbf Q/\mathbf Z)\), in degrees zero and one, with the quotient differential, and \(F=\operatorname{Hom}(\mathbf Z/2,-)\).

Solution. Both terms are divisible abelian groups and therefore injective by the proof in the injective-replacement lesson. Its boundary in degree one is \(\mathbf Q/\mathbf Z\), its cycle in degree zero is \(\mathbf Z\), and its only nonzero cohomology is \(H^0J=\mathbf Z\). The group \(\mathbf Z\) is not injective: the identity on \(\mathbf Z\subset\mathbf Q\) cannot extend to a map \(\mathbf Q\to\mathbf Z\), since twice the image of \(1/2\) would have to be one. Thus this row has noninjective cycles and cohomology and cannot be a split row of the kind (4.1).

After applying \(F\), its two terms are zero and \(\mathbf Z/2\): there is no nonzero two-torsion in \(\mathbf Q\), and the two-torsion subgroup of \(\mathbf Q/\mathbf Z\) has exactly the classes of zero and \(1/2\). Consequently \(H^1(FJ)=\mathbf Z/2\), whereas \(F(H^1J)=0\). This is the precise failure of commuting horizontal cohomology with \(F\). The complex \(J\) still is a valid bounded-below K-injective replacement of \(\mathbf Z[0]\); indeed its sole cohomology is \(\mathbf Z\). Accordingly that nonzero group is \(R^1F(\mathbf Z)\). Termwise injectivity is enough for that derived computation, but not for the split-row identification used to obtain (5.3).

Exercise 4 (challenging). For two independently chosen Cartan–Eilenberg resolutions of each complex, construct coherent identifications of the row spectral sequences from page two. Prove that they commute with derived arrows and preserve the finite filtrations on the abutment. Explain the different claim that is available for the column sequence.

Solution. For a fixed complex \(K\), use Theorem 4.3 to choose a comparison lifting \(1_K\) in each direction. Their composites lift \(1_K\) as well, so (4.4) gives vertical homotopies, consisting of horizontal chain maps, to the identities. Taking horizontal cohomology turns those into homotopies of the first-page signed vertical complexes. Therefore they are inverse on page two and on every subsequent page. Any two identity comparisons give the same page-two map by the same argument. For three resolutions the direct comparison and composite comparison also lift the same identity, so their maps agree; this proves the cocycle and all composition compatibilities.

For a chain map \(f: K\to L\), compare the two possible composite maps between the chosen resolutions. Both lift \(f\), so their page-two maps agree. These comparisons are consequently natural. Their page-two descriptions are \(R^pF(H^qf)\), as follows from the induced comparison on the vertical resolution of \(H^qK\). A quasi-isomorphism gives an isomorphism in every second-page position. The comparisons propagate this to all later pages. The total maps represent \(RF(f)\), and their homotopies make this representation independent of choices. For a quasi-isomorphism \(f\), Corollary 2.2 makes the cohomology map an isomorphism on every finite filtration step, and its inverse preserves those steps. For a general chain map the induced map preserves the steps. Chain-homotopic maps have the same cohomology maps on \(K\), hence the same second-page maps, and the same derived abutment map. Finally invert a quasi-isomorphism in a roof; the construction inverts its compatible page and filtered-abutment maps, proving naturality for every derived arrow and independence of the roof representation.

For the column sequence, the total comparison homotopies preserve column filtration already before taking a page. They give canonical choice-independent identifications from page one, natural in chain maps. However a quasi-isomorphism of complexes need not give an isomorphism of the term-derived first page. For example \(0\to(\mathbf Z\xrightarrow{1}\mathbf Z)\) is a quasi-isomorphism of total complexes, but for \(F\) the identity functor its target column first page has nonzero entries \(\mathbf Z\) in positions \((0,0)\) and \((1,0)\), while its source page is zero. The next differential cancels these entries. The earlier page is thus not a functor of the derived isomorphism class, in agreement with Section 5.

Even its second page can fail to be derived-invariant. Take the exact complex \(K=(\mathbf Z\hookrightarrow\mathbf Q\twoheadrightarrow\mathbf Q/\mathbf Z)\) in degrees zero, one and two, and \(F=\operatorname{Hom}(\mathbf Z/2,-)\). The last two terms are injective. Their positive derived groups vanish, while their degree-zero images are zero and \(\mathbf Z/2\), respectively. For the first term, \(R^0F(\mathbf Z)=0\) and \(R^1F(\mathbf Z)=\mathbf Z/2\), with zero higher groups, by Example 6.2's resolution calculation. Thus the column second page has exactly \(\mathbf Z/2\) in positions \((0,1)\) and \((2,0)\). The first differential has no possible nonzero target, so these entries have survived from page one. Finite convergence to \(RF(K)=0\) forces the only possible second differential between them to be an isomorphism. The same zero derived object has the zero-complex representative, with zero second page. This proves the second-page failure asserted in Section 5.

8. Further reading

• The Stacks Project, spectral sequences of filtered complexes, including the distinction between stabilization and convergence.
• The Stacks Project, spectral sequences of double complexes, for the two filtrations and totalization.
• The Stacks Project, Cartan–Eilenberg resolutions and the two derived spectral sequences.

The finite condition in each degree and the common lower bound are hypotheses of the proved convergence and totalization results. Arbitrary unbounded replacements or convergence require additional arguments.