Draft lesson.

Postnikov towers and Grothendieck spectral sequences

A spectral sequence records successive obstructions and the associated pieces of a filtered answer. This lesson constructs those obstructions from canonical cohomological truncations. It proves the exact-couple construction and convergence, then obtains the Grothendieck spectral sequence from the derived-composition theorem. The two examples include both a collapsing sequence and a sequence with an explicitly nonzero higher differential.

We use Chain homotopies, mapping cones and triangles for cochain and shift signs, and Derived morphisms through quasi-isomorphism roofs, Section 5, for the K-injective and K-projective tests. Triangles after quasi-isomorphism localization proves the derived triangulation and natural exact-sequence triangles. Injective replacements and right derived functors constructs bounded-below replacements, exact derived functors and derived Hom. Universal cohomology and acyclic resolutions, Section 5, proves the canonical composition comparison and its exact acyclicity criterion.

Let \(\mathcal B\) be an abelian category with enough injectives, and let \(G:\mathcal B\to\mathcal C\) be additive and left exact, with \(\mathcal C\) any abelian category. We work in fixed ambient universes. All complexes to which a total derived functor is applied represent objects of \(D^+\), but no upper cohomological bound is imposed.

1. Functorial truncations and their triangles

For a cochain complex \(X\), write \(Z^qX=\ker d_X^q\) and \(B^qX=\operatorname{im}d_X^{q-1}\). Define the two smart truncations by

The incoming differential of the first complex factors through its cycle object; the outgoing differential of the second factors through its quotient. Their natural maps are \(\tau_{\leq q}X\to X\to\tau_{\geq q+1}X\). Direct calculation identifies their cohomology with \(H^nX\) in the retained ranges and with zero in the other ranges.

A chain map preserves cycles and boundaries and thus induces maps on both truncations. Homotopies descend too. At the upper cutoff restrict the last source component to cycles: the discarded outgoing-homotopy term is then zero. At the lower cutoff pass the target component to its quotient: the discarded incoming-homotopy term is a boundary and becomes zero. In all other degrees the original homotopy equation remains unchanged. Hence these are functors on the homotopy category. Their cohomology calculation shows that they preserve quasi-isomorphisms, so the roof universal property descends them to functors on \(D^+\), with the same natural maps for every derived arrow.

There is a natural distinguished triangle

To construct it, take the kernel \(K\) of the termwise epimorphism \(X\to\tau_{\geq q+1}X\). It has terms \(X^n\) below and through degree \(q\), term \(B^{q+1}X\) in degree \(q+1\), and zero higher terms. Its differential onto that last term is epic. The natural inclusion \(\tau_{\leq q}X\to K\) is a quasi-isomorphism: it identifies all cohomology through \(q\), and the cohomology in degree \(q+1\) of \(K\) is zero. Transport the natural short-exact-sequence triangle for \(0\to K\to X\to\tau_{\geq q+1}X\to0\) along this isomorphism. This gives (1.2). Every construction is natural for chain maps, and its naturality extends to inverted quasi-isomorphisms and their composites, hence to all derived arrows.

The consecutive truncations give another natural triangle:

Indeed their termwise quotient has only two possibly nonzero terms, \(X^{q-1}/Z^{q-1}X\simeq B^qX\) in degree \(q-1\) and \(Z^qX\) in degree \(q\), with the inclusion differential. Its projection to \(H^qX[-q]\) is a natural quasi-isomorphism. Transport the quotient short-exact-sequence triangle along that projection. This also proves the naturality of \(\partial_q\), including its sign, with the fixed cone conventions.

Two useful consequences specify these truncations intrinsically. If \(H^nY=0\) for \(n>q\) and \(H^nZ=0\) for \(n<q+1\), then

For the proof choose a bounded-below representative of \(Y\), then truncate it above at \(q\); it has no terms above \(q\). Replace a lower smart truncation of \(Z\) by an injective complex with zero terms below \(q+1\). Every chain map between those representatives is zero because their nonzero degree ranges do not overlap. The K-injective Hom test identifies its homotopy Hom with derived Hom, proving (1.4).

Apply exact Hom sequences to (1.2). Both Hom terms involving the tail and its shift by \(-1\) vanish for a source supported in cohomological degrees at most \(q\), by (1.4). Thus \(\tau_{\leq q}X\to X\) is universal for maps from such sources. Dually, \(X\to\tau_{\geq q+1}X\) is universal for maps to targets supported in degrees at least \(q+1\); the head and its shift by \(1\) give the two vanishing terms. These bijections include uniqueness and naturality. They will identify a specified two-stage extension with its canonical truncation tower in Example 6.2.

Finally \(RG\) preserves lower cohomological bounds. If \(H^nX=0\) below \(a\), first represent \(X\) by its lower smart truncation starting in degree \(a\), then use the injective replacement starting in that same degree. Applying \(G\) term by term leaves the terms below \(a\) zero, so

All comparisons to other representatives are the canonical derived-functor comparisons already proved.

2. Derive an exact couple without omitted verifications

An exact couple has objects \(D,E\) and arrows \(i:D\to D\), \(j:D\to E\), \(k:E\to D\), with

The objects can also be graded families, with the appropriate degree changes attached to these arrows. Each assertion below is then read componentwise at the indicated source and target. This formulation requires no coproduct of all graded components in the ambient category.

Set \(d=jk\). Since \(kj=0\), we have \(d^2=0\). Put

These expressions define morphisms in any abelian category, not merely for groups. To verify expressions using representatives, take a map from a test object into the object in question. A lift through an image or quotient is available after pulling back its defining epimorphism and precomposing the test object with that epimorphic projection. Finite successive lifts give an epimorphic composite. Equalities proved after those changes descend by cancellation of the epimorphism. Pullbacks of epimorphisms remain epimorphisms by the finite abelian proof in Extending right exact functors from generators, Section 1. Thus the following checks are categorical checks of the kernel, image and quotient assertions.

The arrow \(i'\) exists because \(i\) sends its image into its image. For \(j'\), the arrow \(jx\) is a cycle since \(kj=0\). If two lifts have the same \(ix\), their difference is in \(\ker i=\operatorname{im}k\); the difference of their images under \(j\) is therefore a boundary \(jk(e)\). This proves the well-defined quotient arrow \(j'\). For \(k'\), a cycle satisfies \(jke=0\), so \(ke\in\ker j=\operatorname{im}i=D'\). Boundaries are killed by \(k\), since \(kjk=0\). Thus \(k'\) factors through \(E'\).

The three compositions \(j'i'\), \(k'j'\) and \(i'k'\) are zero: their representative expressions are respectively \([jix]=0\), \(kjx=0\), and \(ike=0\). Here the last vanishing uses \(\operatorname{im}k=\ker i\).

We check exactness at each place. If \(j'(ix)=0\), then \(jx\) is a boundary, so after a covering lift it equals \(jke\). Hence \(x-ke\in\ker j=\operatorname{im}i\), say \(x-ke=iy\). Applying \(i\) gives \(ix=i^2y\), proving \(\ker j'=\operatorname{im}i'\). If \(k'([e])=0\), then \(ke=0\), so \(e=jx\) after a covering lift. Therefore \([e]=j'(ix)\), proving \(\ker k'=\operatorname{im}j'\). Finally suppose \(i'(ix)=0\). Then \(ix\in\ker i=\operatorname{im}k\), so choose \(e\) with \(ke=ix\). It is a cycle because \(de=jke=jix=0\), and \(k'([e])=ix\). This proves \(\ker i'=\operatorname{im}k'\). All reverse inclusions follow from the zero compositions. Thus (2.2) is again an exact couple.

A commuting morphism of original couples induces maps on images, kernels and quotient cohomology. Substitution into the representative formulas proves that it commutes with \(i',j',k'\). The construction consequently preserves identities and composition and is functorial. Iterating it produces cohomology pages and all their induced differentials; exactness at every iteration has been proved.

3. The pages of the Postnikov tower

Fix \(X\in D^+(\mathcal B)\), with \(H^qX=0\) for \(q<m\). Apply exact \(RG\) to the tower (1.3). Write \(K_q=RG(\tau_{\leq q}X)\). The cohomology sequences of its triangles form the graded exact couple

Its arrows have degrees

The first map comes from the tower inclusion, the second from its quotient in (1.3), and the third from the connecting arrow. The latter raises total cohomological degree by one. Reading the long cohomology sequence shows exactly the three kernel-image equalities (2.1), at every bidegree.

Start this couple at page two. Its initial differential \(d_2=jk\) has degree \((2,-1)\). Apply Section 2 repeatedly, calling the resulting objects \(D_r,E_r\). After each derivation the degree of \(i\) stays \((-1,1)\) and that of \(k\) stays \((2,-1)\). The new \(j'\) sends \(ix\) to \([jx]\), so its degree increases by \((1,-1)\). Induction therefore gives

Every differential squares to zero because it arises from an exact couple. The retained part of that couple is, with its displayed embedding,

At \(r=2\) this is the identity image. Taking one further image of the tower map proves the next case, so induction proves (3.4).

Left exactness gives \(R^pG=0\) for \(p<0\); the cohomological lower bound on \(X\) gives vanishing for \(q<m\). Thus every page is supported in \(p\geq0,q\geq m\), since taking kernels and quotient images preserves those zero components. For fixed \((p,q)\), the incoming arrow is zero once \(r>p\), and the outgoing arrow is zero once \(r>q-m+1\). The page consequently stabilizes for

Denote that stable object by \(E_\infty^{p,q}\). This is an eventual finite page at each bidegree; its definition does not require an infinite intersection or union in \(\mathcal C\).

All tower arrows are natural for every derived map of \(X\), by Section 1. Section 2 then gives a morphism on every page that commutes with every differential, and these morphisms preserve identities and composition. This proves functoriality of the entire spectral sequence, not just its second page.

4. Prove convergence to a finite filtration

For \(n\in\mathbb Z\), set \(V^n=H^n(RG(X))\) and define an increasing filtration by actual images:

It is increasing because the maps factor through the tower. If \(q<m\), then \(\tau_{\leq q}X\) has zero cohomology, is a zero object of the derived category, and \(F_qV^n=0\).

For \(q\geq n\), apply \(RG\) to (1.2). The tail has cohomology only in degrees at least \(q+1\), and (1.5) gives the same lower bound for its derived image. Its degree-\(n-1\) and degree-\(n\) cohomology vanish. The exact cohomology sequence therefore makes \(H^n(K_q)\to V^n\) invertible. Hence \(F_qV^n=V^n\) for \(q\geq n\). For \(n<m\) the whole object is zero by (1.5). In every other degree the filtration is finite, with stages from \(m\) through \(n\).

Theorem 4.1. The stable pages are naturally the associated pieces of this finite filtration:

Proof. Fix \(n=p+q\) with \(p\geq0,q\geq m\). Choose \(r\) large enough that \(q-r+1<m\) and \(q+r-3\geq n\), and also large enough for (3.5). By (3.4), the source of \(j_r\) mapping into \(E_r^{p,q}\) is

The last identification uses the already proved isomorphism from that high tower stage to \(V^n\). The immediately preceding \(i_r\) has image \(F_{q-1}V^n\): its source is the image from \(K_{q-1}\) to \(K_{q+r-3}\), and this latter stage also maps isomorphically to \(V^n\).

The target of \(k_r\) on \(E_r^{p,q}\) is \(D_r^{p+2,q-1}\). Formula (3.4) describes it as an image from \(H^{n+1}(K_{q-r+1})\). That source is zero because its tower index is below \(m\). Thus \(k_r=0\). Exactness of the derived couple makes \(j_r\) epic and its kernel the stated image of \(i_r\). This identifies \(E_r^{p,q}\) with the quotient in (4.2). At our chosen stage \(E_r\) is already stable. The identifications come from the original tower maps and the quotient universal property, so commute with every morphism of towers. Choosing a later sufficiently large \(r\) gives the same quotient identification, by the derived-couple formulas. This proves (4.2) and its naturality. \(\square\)

Consequently we have constructed, with all differential bidegrees and a finite filtration in each total degree,

The arrow means exactly the proved statement (4.2). It does not identify the abutment with a direct sum of its pieces. Additional extension information may be needed to recover that object.

The upper edge map \(V^n\to G(H^nX)\) is induced by the quotient of the stage \(K_n\), after identifying \(H^n(K_n)\) with \(V^n\). Its image is the stable first-column object \(E_\infty^{0,n}\), embedded into \(E_2^{0,n}\) through the successive kernels: there are no incoming differentials in that column. At the bottom row the map \(R^pG(H^mX)\to V^{p+m}\) is induced by \(K_m\to RG(X)\); its image is \(F_mV^{p+m}\). Its kernel is precisely the successive incoming boundaries, since there are no outgoing differentials from that row. These descriptions follow from the same couple and quotient maps and are natural.

5. Derive the Grothendieck spectral sequence

Let \(F:\mathcal A\to\mathcal B\) be additive and left exact, where \(\mathcal A\) also has enough injectives. Suppose

The composition theorem in the prerequisite identifies \(R(GF)\) with \(RG\,RF\) by its canonical transformation, naturally and with its augmentation and choice comparisons. Apply (4.4) to \(X=RF(Y)\), for \(Y\in D^+(\mathcal A)\). Its lower cohomological bound exists by the replacement construction. Transport the abutment along the canonical composition isomorphism. The result is

Every total degree has the finite filtration already constructed. Every derived arrow of \(Y\) gives a map of the tower through \(RF\), and hence a map of every page and of that filtration. Different replacement choices give the canonical isomorphisms of derived functors; applying functorial truncation and the functorial exact-couple construction transports all pages compatibly with their cocycle comparisons. This proves choice coherence as well as existence.

For an object \(A\) in degree zero, \(H^qRF(A[0])=R^qF(A)\) and the lower bound is zero. Thus (5.2) becomes the first-quadrant Grothendieck spectral sequence

This proof constructs the sequence through truncation towers. Its exact-couple and convergence arguments were supplied in Sections 2–4, so neither a double resolution nor a separate convergence theorem is an unproved input here.

6. Two computed examples

Example 6.1 (invariants in two stages). For a group \(\Gamma\) and normal subgroup \(N\), take \(\mathcal A=\mathbb Z[\Gamma]\text{-Mod}\), \(\mathcal B=\mathbb Z[\Gamma/N]\text{-Mod}\), \(F(M)=M^N\), and \(G(V)=V^{\Gamma/N}\). Normality makes the quotient action on \(M^N\) well-defined: conjugating an element of \(N\) by \(\gamma\) stays in \(N\). Both invariants functors are additive and left exact, because kernels and maps into a kernel can be tested on the fixed elements. Their composite is the \(\Gamma\)-invariants functor.

Here are all the required injective assertions. For any unital ring \(R\) and injective abelian group \(D\), the left module \(C_R(D)=\operatorname{Hom}_{\mathbb Z}(R,D)\), with \((a\phi)(r)=\phi(ra)\), is injective. Evaluation at one gives the natural adjunction \(\operatorname{Hom}_R(V,C_R(D))\simeq\operatorname{Hom}_{\mathbb Z}(V,D)\), whose inverse takes \(\ell\) to \(v\mapsto(r\mapsto\ell(rv))\). Under this adjunction a map extends across a module monomorphism because its underlying group map extends into \(D\). If \(M\hookrightarrow D\) is an injective embedding of the underlying group, the map \(m\mapsto(r\mapsto\iota(rm))\) embeds \(M\) into \(C_R(D)\), as evaluation at one recovers the given monomorphism. The prerequisite proves that abelian groups have enough injectives, so this proves enough injectives for every module category used here.

Inflation of quotient-group modules is exact: it leaves underlying groups and maps unchanged. Maps from an inflated module to \(M\) have image in \(M^N\), giving the adjunction with \(F\). As the right adjoint of an exact functor, \(F\) preserves injectives by the extension proof in the prerequisite. Therefore (5.1) holds.

Write \(H^q(\Gamma,M)\) for the derived functors of invariants. The underlying groups of \(R^qF(M)\) are \(H^q(N,M)\). To verify this, restriction of injective \(\Gamma\)-modules to \(N\) preserves injectives: its left adjoint is induction \(\mathbb Z[\Gamma]\otimes_{\mathbb Z[N]}-\), and the group ring is free as a right subgroup ring on coset representatives. Induction is consequently a direct sum of copies of the underlying module and is exact. Restricting an injective resolution therefore computes \(N\)-cohomology. Its \(N\)-invariant complex carries the quotient action, giving the asserted identification and that action. Formula (5.3) now reads

For \(\Gamma=\mathbb Z^2\), \(N=\mathbb Z\) the first factor and \(M=\mathbb Z\) trivial, the cyclic-group resolution is \(0\to\mathbb Z[t,t^{-1}]\xrightarrow{t-1}\mathbb Z[t,t^{-1}]\to\mathbb Z\to0\). The augmentation kernel is generated by \(t-1\), since every Laurent monomial minus its value at one is divisible by \(t-1\). That multiplier is injective: the highest nonzero exponent of a finite Laurent polynomial prevents its product with \(t-1\) from vanishing. Thus this is a two-term free resolution. The K-projective Hom test, or comparison with an injective target resolution, makes its Hom complex a derived Hom model. With trivial coefficients its differential is zero. It computes cyclic cohomology \(\mathbb Z\) in degrees zero and one and zero otherwise.

The second central generator acts trivially on both resulting groups. Its action on a \(\Gamma\)-injective resolution extends its identity action on \(M\), so the homotopy comparison theorem gives identity on derived cohomology. Applying the same cyclic calculation in the second factor makes the second page a square of four copies of \(\mathbb Z\), at \(p,q\in\{0,1\}\). Every higher differential has zero target or source by its bidegree. Hence \(H^0(\mathbb Z^2,\mathbb Z)=\mathbb Z\), \(H^2(\mathbb Z^2,\mathbb Z)=\mathbb Z\), and higher degrees are zero. Degree one has a short exact filtration with two pieces \(\mathbb Z\); it splits because a lift of the generator of its quotient defines a section. Therefore \(H^1(\mathbb Z^2,\mathbb Z)\simeq\mathbb Z^2\), with no canonical splitting asserted by the spectral sequence alone.

Example 6.2 (a nonzero differential removes infinitely many entries). Let \(R=k[\varepsilon]/(\varepsilon^2)\), and let \(k=R/(\varepsilon)\). Enough injectives for this module category was proved in the prerequisite; it also follows from the coinduction proof just given. Take \(G=\operatorname{Hom}_R(k,-)\). A projective resolution \(P\to k[0]\) has \(P^n=R\) for \(n\leq0\), zero above zero, differential multiplication by \(\varepsilon\) in every negative degree, and augmentation reduction. Its kernels and images in negative degrees are both \(k\varepsilon\), and the degree-zero augmentation has that kernel. All terms are free, so the bounded-above K-projective test applies.

The complex \(\operatorname{Hom}(P,k[0])\) has one copy of \(k\) in every nonnegative degree and zero differential. Thus \(\operatorname{Ext}^p_R(k,k)=k\) for \(p\geq0\). For a concrete generator \(e_p\), use the chain map

The chain equation below the cutoff holds since the consecutive coefficients differ by \((-1)^p\), precisely the shift differential sign. At the cutoff both relevant outgoing maps are zero. The component in degree \(-p\) is identity, so composition with the shifted augmentation represents the nonzero Hom class and hence the generator.

Let \(\xi=e_2:k[0]\to k[2]\). Complete \(\xi[-1]:k[-1]\to k[1]\) to a distinguished triangle and rotate twice backwards. This gives an object \(X\) and a triangle

Its cohomology sequence gives \(H^0X=k\), \(H^1X=k\), and zero in every other degree. By the truncation universal property, the first arrow factors through \(\tau_{\leq0}X\); that factor is a cohomology isomorphism and hence invertible. The triangle comparison axiom then compares (6.3) to the canonical truncation triangle with isomorphisms on all vertices. Choose the identification of \(H^1X\) with \(k\) from this comparison. Its tower connecting arrow is now exactly the displayed \(\xi[-1]\), with its sign fixed by that triangle.

The second page of (4.4) has \(k\) at every \((p,0)\) and \((p,1)\), for \(p\geq0\), and zero elsewhere. Its \(d_2\) is induced by that connecting arrow. On the generators it is postcomposition with \(\xi[p]\):

To check the product, represent it on \(P\) by \(q[p+2]t_2[p]t_p\). In degree \(-p-2\), the coefficient of \(t_p\) is \((-1)^{-2p}=1\), and that of \(t_2[p]\) is identity. The composite therefore represents the nonzero normalized generator in degree \(p+2\); the Hom differential there is zero. This proves (6.4) for every \(p\), not just \(p=0\).

Every such differential is an isomorphism. Page three has only \(E_3^{0,0}=k\) and \(E_3^{1,0}=k\). Thus the abutment is \(H^0RG(X)=k\), \(H^1RG(X)=k\), and zero in all other degrees. As an independent check, apply derived Hom to (6.3). In its long sequence the connecting maps are exactly multiplication by \(\xi\), just proved invertible in every nonnegative Ext degree. The same kernel-cokernel calculation gives those two surviving cohomology objects. The initial two infinite rows therefore do not by themselves determine the answer: their nonzero differential is essential.

7. Four graded exercises with complete solutions

Exercise 1 (foundation: keep the total degree). Derive the bidegree of \(d_r\) from the maps of the exact couple. Show that every differential raises total degree by one and that the stable-page bound (3.5) follows from the supported region.

Solution. The tower map raises \(q\) by one and keeps \(p+q\) fixed, so \(i\) has degree \((-1,1)\). The quotient map initially keeps both indices. The connecting map lowers the tower index by one and raises cohomological degree by one, so \(k\) has degree \((2,-1)\). Therefore \(jk\) starts with degree \((2,-1)\). In the derived couple, \(j'\) applies \(j\) to a preimage under \(i\), so its degree changes by the negative of the degree of \(i\), namely \((1,-1)\); the degree of \(k'\) remains unchanged. After \(r-2\) derivations, \(d_r\) has degree \((r,1-r)\), whose coordinate sum is one. Its possible incoming source has first coordinate \(p-r\), and its possible outgoing target has second coordinate \(q-r+1\). Those components are zero respectively for \(r>p\) and \(r>q-m+1\). When both inequalities hold, the cohomology quotient giving the next page changes nothing. This proves eventual stabilization with the exact bound in (3.5).

Exercise 2 (intermediate: derive the five-term sequence). In (5.3), prove the natural exact sequence

Solution. Write \(V^n=R^n(GF)(A)\). At \((1,0)\) neither an incoming nor an outgoing differential can be nonzero, so \(F_0V^1=E_\infty^{1,0}=E_2^{1,0}\). At \((0,1)\) there are no incoming differentials, and the only possible outgoing one is \(d_2:E_2^{0,1}\to E_2^{2,0}\). Thus \(V^1/F_0V^1=E_\infty^{0,1}=\ker d_2\). Inclusion of that kernel into \(E_2^{0,1}\), following the filtration quotient, gives the middle arrow of (7.1) and exactness through its first four positions. At \((2,0)\) the only possible incoming arrow is this same \(d_2\); no outgoing arrow exists. Hence \(F_0V^2=E_\infty^{2,0}=\operatorname{coker}d_2\). Compose the cokernel projection from \(E_2^{2,0}\) with the inclusion of this lowest filtration piece into \(V^2\). Its kernel is exactly \(\operatorname{im}d_2\), which proves exactness at the fifth term. The identifications with the terms in (7.1) use its second-page formula. Every map was a filtration, kernel or quotient map from the natural couple, so the sequence is natural in \(A\).

Exercise 3 (advanced: comparison on page two). Suppose a morphism between two sequences constructed in this lesson is invertible on every second-page component. Prove it is invertible on the abutment in every degree, assuming the indicated finite filtrations.

Solution. An isomorphism commuting with a differential identifies its kernel and image, so induces an isomorphism on its quotient cohomology. Induction makes the morphism invertible on every page. For each fixed bidegree choose a stage at which both sequences have stabilized; their stable components are therefore isomorphic. Use (4.2) to identify them with the associated pieces of the two filtrations. Choose a common lower bound for the filtrations. In a fixed total degree there are finitely many stages, both start with zero and both end with the respective abutments. Induct through those stages using the short exact sequences of a filtration stage, its preceding stage and its quotient. If the maps on the preceding stage and quotient are isomorphisms, the map on the middle stage is an isomorphism: injectivity follows by sending a kernel element into the quotient then into the preceding stage; surjectivity follows by lifting first its quotient image and then its residual in the preceding stage, using epimorphic covering lifts in an arbitrary abelian category. This proves the induction and the asserted isomorphism on the final objects. Finite-filtration convergence supplies the needed start and end; page isomorphisms alone would not supply those hypotheses for a different convergence problem.

Exercise 4 (synthesis: degeneration does not specify extensions). Explain why a stable page with two pieces \(\mathbb Z/2\) in one total degree cannot distinguish \(\mathbb Z/4\) from \(\mathbb Z/2\oplus\mathbb Z/2\). Contrast this with the splitting conclusion for the two free pieces in Example 6.1.

Solution. Filter \(\mathbb Z/4\) by \(0\subset2\mathbb Z/4\subset\mathbb Z/4\). Its subobject and quotient are both \(\mathbb Z/2\). Filter \(\mathbb Z/2\oplus\mathbb Z/2\) by its first summand. Again both pieces are \(\mathbb Z/2\), but every element of this second group is killed by two, whereas the first group has an element of order four. They are not isomorphic despite their identical associated graded objects. The first extension does not split, since every homomorphism from \(\mathbb Z/2\) to \(\mathbb Z/4\) lands in the subgroup killed by two and consequently has zero composite to the quotient. The second splits by the second summand. In Example 6.1 the quotient is \(\mathbb Z\); choose a lift of its generator and extend by integer multiples to obtain a section. That separate projectivity argument supplies the splitting. The stable page by itself supplies only the short exact filtration and its two pieces.

References and further reading

• The Stacks Project, spectral sequences from exact couples, for the exact-couple terminology. Section 2 above verifies its representative, quotient, exactness and functoriality steps.
• The Stacks Project, the Grothendieck spectral sequence, for the derived-composition statement. Sections 1–5 above construct it by canonical truncation towers and prove its finite-filtration convergence.