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The submodule of coboundaries in degree $q$. The cohomology group is the corresponding cycle module modulo the boundary module.

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.