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A finite filtration of a cochain complex produces a convergent spectral sequence. The induced abutment filtration is the image of $H^n(F^{p}C)\to H^n(C)$. Finiteness in each degree suffices; a bounded range of cochain degrees is not required.

A cochain complex with a decreasing filtration by cochain subcomplexes. A filtration finite in each degree is bounded in the degreewise sense; uniform bounds $F^{a}C=C$ and $F^{b}C=0$ are sufficient. The associated spectral sequence starts with the graded quotient complexes.

For a bounded complex $P$ of projective right modules and a left module $M$ of finite projective dimension, a finite projective resolution of $M$ yields a bounded double cochain complex. Its two spectral sequences give the displayed abutment. Over a noncommutative ring, the handedness of the two tensor factors is essential.

Use integral coefficients throughout, allowing a nontrivial orientation character. A dimension-$d$ Poincare duality group has a finite resolution by finitely generated projective group-ring modules andas abelian groups. The action on the top copy of $\mathbb{Z}$ may be by signs. This group-ring criterion is particularly suited to the Lyndon–Hochschild–Serre spectral sequence.

Apply that sequence to the extension with coefficient module $\mathbb{Z}G$:As an $N$-module, $\mathbb{Z}G$ is a direct sum of copies of $\mathbb{Z}N$ indexed by the cosets in $Q$. Cohomology of $N$ commutes with this direct sum because its projective resolution is finitely generated. Hence the inner cohomology vanishes except when $b=n$, where it is a direct sum of rank-one groups indexed by $Q$. The $Q$-action shifts these summands freely and transitively, with possible signs from the orientation module. Choosing a generator in one summand and transporting it by $Q$ identifies the resulting left $Q$-module with its regular module $\mathbb{Z}Q$. The residual right group action records the orientation twist; it need not be trivial.

Now the duality of $Q$ says that the only nonzero $E_2$ term is $(a,b)=(q,n)$ and that this term is $\mathbb{Z}$ as an abelian group. No differential can enter or leave it. Therefore $H^{\ast }(G;\mathbb{Z}G)$ has exactly the duality pattern in degree $n+q$. The usual extension-resolution construction also supplies a resolution by finitely generated projective modules: combine the finite kernel and quotient resolutions, lift the quotient differentials and add the necessary homotopy corrections. Its total length is at most $n+q$. Alternatively, the same spectral sequence for arbitrary coefficient modules gives the dimension bound, while the extension finiteness lemma supplies finite generation of the resolution. The surviving top term makes the bound sharp. This proves the extension rule for Poincare duality groups:Requiring a trivial top orientation action would need an additional orientation condition; the untwisted conclusion is not implicit in the hypotheses.

We next need two precise auxiliary dimension results. The Strebel infinite-index subgroup theorem says that an infinite-index subgroup of an integral $P{D}^d$ group has integral cohomological dimension at most $d-1$. The Stallings-Swan theorem says that a group of integral cohomological dimension at most one is free, including the trivial group. Since $G/N$ contains an element of infinite order, $N$ has infinite index in $G$, so $\mathrm{cd}N\le 2$. If its dimension is at most one, the required free-group conclusion follows, with finite rank because $N$ is finitely presented.

Suppose instead that $\mathrm{cd}N=2$. Finite presentation gives a partial free resolution with finitely generated modules through degree two. The kernel in degree one is finitely generated and, because the integral cohomological dimension is two, projective. Truncating at this kernel gives a finite resolution by finitely generated projective modules. Thus $N$ has the finiteness needed for group-ring cohomology to commute with direct sums.

Choose an infinite-order element $\bar{t}\in G/N$, lift it to $t\in G$, and let $H$ be the preimage of $⟨\bar{t}⟩$. Then $H=N⋊⟨t⟩$. Restriction of $\mathbb{Z}H$ to $N$ yields copies of $\mathbb{Z}N$ indexed by all powers of $t$. For every $j$, the quotient generator acts on $H^j(N;\mathbb{Z}H)$ by shifting this direct sum, possibly with an automorphism on each copy. Its invariants vanish, since an invariant finite-support sequence must be zero. Its coinvariants are one copy of $H^j(N;\mathbb{Z}N)$: the shift relations identify successive copies, and any twists can be removed by transporting the identifications along the integer index.

The infinite cyclic group has integral cohomological dimension one, so only columns zero and one occur in the spectral sequence. Column zero vanishes and column one consists of these coinvariants. We obtain the cyclic-extension shift of group-ring cohomologyas abelian groups. Moreover $H^2(N;\mathbb{Z}N)\ne 0$. Here is a useful proof of that last point: if top cohomology of a finite projective resolution were zero, its last dual differential would be onto. Its target is projective, so that differential splits. Dualizing back splits the last injection of the original resolution, shortening it and contradicting integral cohomological dimension two.

It follows that $H^3(H;\mathbb{Z}H)\ne 0$, and hence $\mathrm{cd}H=3$. Strebel's theorem forces $H$ to have finite index in $G$. It is therefore itself a $P{D}^3$ group, by finite-index invariance of Poincare duality for torsion-free groups. The displayed shift now makes $H^j(N;\mathbb{Z}N)$ zero except for $j=2$, where it is $\mathbb{Z}$. Together with its finite projective resolution, this provesThe classification of two-dimensional Poincare duality groups identifies the latter with the fundamental group of a closed aspherical surface. This classification includes orientation-twisted surface groups; it does not include the two-sphere or real projective plane, whose fundamental groups do not have integral cohomological dimension two.

In this latter case we have also proved that $G/N$ is virtually infinite cyclic, since $H/N=⟨\bar{t}⟩$ has finite index. If necessary replace the surface group by its characteristic index-two orientable subgroup and take the subgroup generated by it and $t$. This still has finite index in $G$ and is a semidirect product of an orientable closed aspherical surface group by $\mathbb{Z}$. The Dehn-Nielsen-Baer theorem for closed orientable surfaces realizes the outer automorphism defined by conjugation by $t$ as a surface homeomorphism $f$. For the torus this is just realization of $G{L}_2(\mathbb{Z})$ by linear homeomorphisms. Changing the chosen lift accounts for an inner automorphism. The mapping torusis a surface bundle over $S^1$ with fundamental group ${\pi }_1(S){⋊}_{f_{\ast }}\mathbb{Z}$. Thus a finite-index subgroup of $G$ is the fundamental group of a surface fibration over the circle. Taking a further double cover of the base if required makes the monodromy orientation-preserving.

Let $R=M{U}^{\ast }(\mathrm{pt})$ be the coefficient ring of complex cobordism. We use its standard coefficient theorem: $R\cong \mathbb{Z}[b_1,b_2,\dots ]$ with $|b_j|=-2j$. In particular $R$ is concentrated in nonpositive even degrees. Let $H$ be the quaternionic tautological line bundle on ${\mathbb{HP}}^2$, regarded as a complex rank-two bundle by right multiplication by a fixed imaginary quaternion. SetWe will proveThe argument must establish the multiplication as well as the additive groups, since collapse of a spectral sequence alone does not settle ring extensions.

The quaternionic projective space has one cell in each dimension $0,4,8$, giving integral cohomology $\mathbb{Z}$ in exactly these three degrees. The sphere bundle of $H$ is the quaternionic Hopf fibrationWrite $u=e(H_{\mathbb{R}})=c_2(H)$ for its ordinary degree-four class, fixing its sign by this equality. The Gysin sequence of a sphere bundle gives surjections, in fact isomorphisms,Indeed the corresponding neighboring cohomology groups of $S^{11}$ vanish. Both source and target groups are $\mathbb{Z}$, so these maps make $u$ and $u^2$ generators. There are no higher cohomology groups. This proves the cohomology ring of quaternionic projective space in the form $H^{\ast }({\mathbb{HP}}^2;\mathbb{Z})=\mathbb{Z}[u]/(u^3)$.

Apply the multiplicative Atiyah-Hirzebruch spectral sequence for complex cobordism:The skeletal construction obtains this sequence from the long exact sequences of successive skeleta; the quotients are wedges of spheres, and the first differential is the ordinary cellular differential with coefficient groups $R^t$. Its multiplication on the associated graded group is the ordinary cup product with these coefficients. The finite eight-dimensional skeleton makes the filtration finite, so there is no convergence ambiguity.

Only columns $s=0,4,8$ and even rows $t$ are nonzero. If $r$ is odd, the target column $s+r$ is odd and zero. If $r$ is even, the target coefficient degree $t-r+1$ is odd and zero. The same observations exclude incoming differentials. Consequently $E_{\infty }=E_2$, with associated graded $R$-module freely generated by $1,u,u^2$ in filtrations $0,4,8$.

The canonical map from complex cobordism to integral cohomology preserves the chosen complex orientation and hence sends $q$ to $u$. The class $q$ vanishes on the zero-dimensional skeleton since $R^4=0$, and its leading skeletal term is $u$ in filtration four. Multiplicativity makes the leading term of $q^2$ the generator $u^2$ in filtration eight. Thus $1,q,q^2$ lift the associated graded basis.

This resolves all module extensions explicitly. For a class of total degree $d$, subtract its coefficient $a_0\in R^d$ in filtration zero. From the remainder subtract $a_{1}q$, where $a_1\in R^{d-4}$ is its leading coefficient in filtration four, and then subtract $a_2{q}^2$ with $a_2\in R^{d-8}$. The remaining filtration is beyond dimension eight and is zero. Conversely a relation $a_0+a_{1}q+a_2{q}^2=0$ has zero leading coefficient in each of the three successive filtrations, forcing all three $a_j$ to vanish. ThereforeFinally $q^3$ has degree twelve. Every associated graded contribution in that total degree would have coefficient degree $12-s>0$ for $s=0,4,8$, and all such coefficients vanish. Thus $M{U}^{12}({\mathbb{HP}}^2)=0$ and $q^3=0$. The polynomial map $R[t]/(t^3)\to M{U}^{\ast }({\mathbb{HP}}^2)$ taking $t$ to $q$ is surjective by the basis argument and injective by its uniqueness. This proves the complex cobordism ring of the quaternionic projective plane as a graded ring over the stated coefficient ring, including the absence of any additional coefficient-dependent relation.

For a cohomological spectral sequence, $d_r:E_r^{p,q}\to E_r^{p+r,q-r+1}$ and the next page is its cohomology. In the bounded setting, convergence of a spectral sequence to $H^{\ast }$ means that each $H^n$ has a finite decreasing, exhaustive and separated filtration withHere $E_{\infty }^{p,q}$ is the eventual stable value. This determines the associated graded module of $H^n$, rather than automatically a canonical direct-sum decomposition of $H^n$. For unbounded filtrations additional completeness/convergence conditions are necessary; boundedness removes those issues here.

The bounded filtered-complex convergence theorem states the following. If $F^{p}C$ is a decreasing filtration by cochain subcomplexes, preserved by the differential, finite in each cochain degree (uniform bounds $F^{a}C=C$, $F^{b}C=0$ are sufficient), then there is a spectral sequenceThe abutment filtration is $F^p{H}^n(C)=\mathrm{im}[H^n(F^{p}C)\to H^n(C)]$. Its finite length ensures stabilization and the displayed limiting-page identification. The filtered cochain complex need not itself be bounded in cochain degree. In the degreewise finite version, the bounds in degrees $n-1,n,n+1$ suffice to stabilize the terms of total degree $n$.

For a double cochain complex $D^{p,q}$ bounded in both indices, form the total cochain complex ${\mathrm{Tot}}^{n}D={⨁}_{p+q=n}{D}^{p,q}$ with anticommuting differentials $d_h,d_v$ and total differential $d_h+d_v$. If the original differentials commute, inserting the usual sign in one of them gives this convention. Filtering by the first index gives $E_1^{p,q}=H_v^q(D^{p,\ast })$ and then $E_2^{p,q}=H_h^p(H_v^{q}D)$. Filtering by the second index gives the spectral sequence taking horizontal cohomology first and vertical cohomology next. Both filtrations are finite, so both converge to $H^{\ast }(\mathrm{Tot}D)$. This is the two spectral sequences of a bounded double complex construction.

The printed left/left tensor expression needs a handedness repair. Over an arbitrary ring, a tensor product of modules over $R$ pairs a right module with a left module. To retain the order of the printed formula, take $P$ to be a bounded cochain complex of projective right $R$-modules and $M$ a left $R$-module. Alternatively, keep $P$ left, take $M$ right, and write $M{\otimes }_{R}P$ and ${\mathrm{Tor}}_{-p}^R(M,H^{q}P)$. For a commutative ring no repair is needed. We prove the first, correctly typed formulation.

Finite projective dimension gives a finite projective resolution $Q_{\bullet }\to M$ by left modules. Form $D^{p,q}=P^q{\otimes }_R{Q}_{-p}$ for $-d\le p\le 0$. On $x\otimes y$ take $d_v=d_P\otimes 1$ and $d_h=(-1{)}^{q}1\otimes d_Q$. These anticommute, and the double cochain complex is bounded in both directions. Projective modules are flat modules, so taking vertical cohomology first givesThe first identification uses flatness of $Q_{-p}$; the second is the Tor functor computed by resolving its left-module argument $M$. Balancedness of the Tor functor allows either correctly sided projective resolution to compute it, by the double-resolution argument.

Taking horizontal cohomology first instead uses flatness of $P^q$. The resolution $Q_{\bullet }$ then leaves only $P^q{\otimes }_{R}M$ in column $p=0$. Its remaining differential is $d_P\otimes 1$, so the augmentation $\mathrm{Tot}D\to P{\otimes }_{R}M$ is a quasi-isomorphism. Apply the convergence theorem to this underlying abelian-group double complex. The two spectral sequences therefore prove the Künneth spectral sequence:These are abelian groups in general; an extra module structure requires appropriate bimodule hypotheses. The index $p$ is nonpositive, so $-p$ is the nonnegative homological index of the Tor functor.

Finally write the cycle modules as $Z^q=\ker d^q$ and the boundary modules as $B^q=\mathrm{im}{d}^{q-1}$. If every boundary module $B^q$ is projective, the short exact sequence $0\to Z^q\to P^q\to B^{q+1}\to 0$ splits. Thus $Z^q$ is projective. The short exact sequence $0\to B^q\to Z^q\to H^{q}P\to 0$ is now a length-one projective resolution, so ${\mathrm{Tor}}_i^R(H^{q}P,M)=0$ for $i\ge 2$. The $E_2$ page occupies only columns $p=-1,0$. Every $d_r$ for $r\ge 2$ goes $r$ columns to the right, hence has zero source or zero target. ThereforeThis two-column degeneration of a Künneth spectral sequence gives the edge short exact sequencesDegeneration itself does not supply a canonical splitting. Projective boundaries also do not force $H^{q}P$ to be projective: a complex with differential multiplication by $2$ over $\mathbb{Z}$ has projective boundary but a $\mathbb{Z}/2\mathbb{Z}$ cohomology group.

Filtering the total cochain complex by each index yields two spectral sequences. One first takes vertical cohomology and the other horizontal cohomology; boundedness ensures both converge to total cohomology.