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Hypercohomology applies the derived global-section functor to a complex of sheaves. One may resolve the complex by an injective double complex, take global sections, and take the cohomology of its total complex. For a complex concentrated in degree zero it is ordinary sheaf cohomology. The relative de Rham complex uses this construction to compute fibre cohomology and its Hodge filtration.

The degree-zero normalized Deligne intersection complex extends the constant sheaf from the regular stratum by successive derived direct images and cohomological truncations at degree $\bar{p}(k)$ when codimension-$k$ strata are added. Its hypercohomology defines intersection cohomology. On a cone with vertex of codimension $k$, its local cohomology is the link intersection cohomology in degrees at most $\bar{p}(k)$ and zero above.

For an inclusion $A\subseteq X$, the restriction cochain map fits into an exact triangle with the relative cochain complex. Its cohomology gives $\cdots \to H^i(X,A)\to H^i(X)\to H^i(A)\stackrel{\delta }{\to }{H}^{i+1}(X,A)\to \cdots$. The same construction with derived sections of a sheaf complex gives relative hypercohomology and is natural under restrictions of pairs.

Put $B=\partial M$ and let $v$ be the cone vertex. Specify the convention away from the Witt case: first take intersection cohomology to mean the hypercohomology of the intersection complex $I{C}_{\bar{m}}$, normalized in degree zero on the regular stratum. The lower middle perversity at the vertex is $\bar{m}(2m+1)=m-1$. The standard local cone calculation for this convention saysand restriction to $U\simeq B$ is the identity below the cutoff. Here this can also be read directly from the construction $I{C}_{\bar{m}}={\tau }_{\le m-1}R{j}_{\ast }\mathbb{Q}$, with $j:\hat{M}\setminus \{v\}\hookrightarrow \hat{M}$: punctured conical neighbourhoods retract onto $B$, and the truncation discards the degrees above $m-1$.

Define relative intersection cohomology as the hypercohomology of the homotopy fibre of restriction, namely $R\Gamma (\hat{M},\hat{U};IC)=\mathrm{Cone}(R\Gamma (\hat{M};IC)\to R\Gamma (\hat{U};IC))[-1]$. Restricting to the pair $(M,U)$ gives a map to the corresponding ordinary relative cohomology, since $IC{|}_M={\mathbb{Q}}_M$. The induced map of exact triangles gives the requested commutative diagram:Both rows are long exact cohomology sequences of a pair. Commutativity includes the connecting homomorphisms, because they arise from the same map of restriction cones, with a common sign convention. The relative vertical maps are isomorphisms: excision removes the common conical neighbourhood, leaving precisely the manifold pair outside it, where $IC$ is constant. Also the collar retracts onto $B$, so $H^{\ast }(M,U)\cong H^{\ast }(M,B)$.

The only possible nonmanifold stratum is $v$, with link $B$ of dimension $2m$. Since intersection homology of a manifold is ordinary homology, the exact necessary and sufficient condition isThe equivalence with the defining $H_m(B;\mathbb{Q})=0$ follows from the universal coefficient theorem for cohomology over $\mathbb{Q}$. In the Witt case the cone restriction is an isomorphism through degree $m$, since both degree-$m$ groups are zero. Comparing the two exact rows, or applying the Five lemma to each five-term segment, gives $I{H}^i(\hat{M})\cong H^i(M)$ for $i\le m$. For $i>m$, both $I{H}^{i-1}(\hat{U})$ and $I{H}^i(\hat{U})$ vanish, so the top row identifies the absolute group with the relative one. HenceWithout the Witt hypothesis, comparison still gives $I{H}_{\bar{m}}^i(\hat{M})\cong H^i(M)$ for $i<m$. At degree $m$ the upper row readsThe lower row identifies the quotient by $\mathrm{im}\delta$ with the image of the relative-to-absolute map. In higher degrees both cone terms vanish. Thus the full lower-middle Deligne answer isFor clarity, if $I{H}^i$ is instead defined as the linear dual of the lower-middle intersection chains of question 1, it corresponds to the complementary upper middle perversity Deligne sheaf. Its cone group retains $H^m(B)$ as well. The same diagram then givesEquivalently the exceptional group is $H^{m+1}(M,B)/\mathrm{im}(H^m(B)\stackrel{\delta }{\to }{H}^{m+1}(M,B))$. These conventions agree under the stated Witt condition, explaining why that condition removes the ambiguity.

Take the solid torus $M=S^1\times D^2$, whose boundary is the two-dimensional torus $B=T^2$. Its ordinary cohomology is $H^0(M)=H^1(M)=\mathbb{Q}$, zero otherwise; Poincare duality for a manifold with boundary gives $H^2(M,B)=H^3(M,B)=\mathbb{Q}$ and all other relative groups zero. The lower-middle Deligne formula givesThus degrees one and two have different dimensions, so there can be no three-dimensional Poincare duality. In the dual-lower-chain convention the dimensions are $(1,1,0,1)$, which likewise fail duality. Both computations exhibit the obstruction $H^1(T^2)={\mathbb{Q}}^2\ne 0$.

All cohomology below has rational coefficients. First describe the incidence space geometrically. The image of $g$ is $\mathbb{P}(W)\cong {\mathbb{CP}}^1$, inside $\mathrm{Gr}(1,4)\cong {\mathbb{CP}}^3$. For a line $U\subset W$, choosing a plane $V$ containing $U$ is equivalent to choosing the line $V/U$ in ${\mathbb{C}}^4/U$. If $\mathcal{L}$ is the tautological line bundle on $\mathbb{P}(W)$, thenThis is the projective bundle of a rank-three quotient bundle, with fibre ${\mathbb{CP}}^2$. Local frames of the quotient bundle give local product charts. In particular $F_W$ is an oriented real six-dimensional manifold.

For $V\in S_W\setminus \{W\}$, the intersection $V\cap W$ is exactly a line, so $f^{-1}(V)$ is a single point and the inverse map is $V\mapsto (V\cap W,V)$. The intersection line varies continuously, indeed holomorphically, wherever its dimension is constant: in local Grassmannian charts it is the kernel of a matrix of constant rank, whose kernel bases can be chosen locally. Thus $f$ is an isomorphism over the open stratum. At the remaining point,Set $K=R{f}_{\ast }{\mathbb{Q}}_{F_W}$, a derived direct image of sheaves. The proper base change for sheaves theorem identifies its derived stalks with the cohomology complexes of the fibres. Consequently $K$ is the constant rational sheaf in degree zero on $S_W\setminus \{W\}$, andThe cohomology sheaves are therefore finite-rank local systems on the two strata, with no other nonzero degrees. This proves that $K$ is a constructible complex of sheaves and verifies normalization. The open stratum has real dimension six, so the isolated point has codimension six. Its middle perversity is $\bar{m}(6)=2$, and the intersection-complex support and cosupport axioms require stalk vanishing in degrees above two. The fibre calculation verifies the support axiom, as well as vanishing in negative degrees.

For an oriented real six-dimensional manifold, Verdier duality gives $D_{F_W}{\mathbb{Q}}_{F_W}\cong {\mathbb{Q}}_{F_W}[6]$. For a proper map, Verdier duality commutes with derived direct image. Applying these statements here yieldsThe duality identity $i_W^{!}{D}_{S_W}K\cong D_{\{W\}}{i}_W^{\ast }K$ converts this intoHere $i_W^{!}K$ is the costalk, and dualization of rational complexes reverses degrees. Thus its only nonzero groups are $\mathbb{Q}$ in degrees four and six. In particular they vanish for $a\le 3$, which is precisely the cosupport axiom at this isolated six-dimensional singularity. The uniqueness characterization by constructibility, normalization, support and cosupport now identifies $K$ with the unshifted intersection complex of $S_W$. Taking hypercohomology, and using the composition of derived global sections with $R{f}_{\ast }$, givesFinally apply the fibration result allowed in the question to $g:F_W\to \mathbb{P}(W)$. Its base is the simply connected complex projective line, its fibre is the Complex projective plane, and the projective-bundle charts above verify the fibration hypothesis. Hence, as graded rational vector spaces,The resulting Betti numbers and Poincaré polynomial areThis uses $H^{\ast }({\mathbb{CP}}^r)=\mathbb{Q}[h]/(h^{r+1})$ with $\mathrm{deg}h=2$. No assertion that an arbitrary projective fibration has the product cohomology ring is needed.

For a cochain complex $E$, define the shift of a cochain complex byFor a cochain map $a:E\to F$, put $(\Sigma a{)}^i=a^{i+1}$. This is again a cochain map, so the construction is a functor. The sign ensures the standard cone identities; the shifted differential squares to zero. Iteration gives $E[r{]}^i=E^{i+r}$ and differential $(-1{)}^r{d}_E$, and $H^i(E[r])=H^{i+r}(E)$.

For $\varphi :E\to F$, its algebraic mapping cone isApplying this differential twice gives first component $d_F^{2}f+d_F\varphi e-\varphi d_{E}e=0$, since $\varphi$ is a cochain map, and second component $d_E^{2}e=0$. The mapsare cochain maps: $d_{C}u(f)=(d_{F}f,0)=u(d_{F}f)$ and $v(d_C(f,e))=-d_{E}e=d_{\Sigma E}v(f,e)$. They are maps of complexes of sheaves, because every displayed operation is a sheaf morphism in each degree.

The bounded below derived category of sheaves $D(X)=D^{+}(X)$ can be constructed by first quotienting the category of bounded below cochain complexes by cochain homotopy and then formally inverting all quasi-isomorphisms. A morphism may be represented by a roof $E\stackrel{\sim }{\leftarrow }{E}^{\prime }\to F$. Alternatively, take injective resolutions of sheaves and use homotopy classes of maps between the resulting injective complexes. The cohomology sheaf functors descend to the localization, because they already identify homotopic maps and turn quasi-isomorphisms into isomorphisms. If $E\cong 0$ in $D(X)$, applying these functors gives ${\mathcal{H}}^i(E)=0$ for every $i$. Conversely, if all these sheaves vanish, the unique map $E\to 0$ is a quasi-isomorphism, which becomes invertible in $D(X)$. HenceAn exact triangle in a derived category is a triangle isomorphic to a cone triangleIn particular the last arrow is part of the structure; a mere sequence of three objects with successive composite zero does not suffice.

Suppose $0\to E\stackrel{\varphi }{\to }F\stackrel{\psi }{\to }G\to 0$ is degreewise exact. DefineThen $r(d_C(f,e))=\psi (d_{F}f)+\psi \varphi (e)=d_G\psi (f)=d_{G}r(f,e)$, so $r$ is a cochain map. It is natural under maps of short exact sequences. It is degreewise surjective as a sheaf map, and its kernel can be identified with $E^i\oplus E^{i+1}$ using $(a,e)\mapsto (\varphi (a),e)$. In these coordinates the kernel differential isThere is a contracting homotopy $h^i(a,e)=(0,a)$ in degree $i-1$:Thus the kernel has zero cohomology sheaves. The cohomology-sheaf exact sequence is the sequence of the short exact sequence of complexes $0\to \ker r\to C(\varphi )\to G\to 0$, or equivalently its ordinary complex-cohomology sequence at every stalk. It follows that $r$ induces isomorphisms of all cohomology sheaves, provingUnder this isomorphism the cone triangle becomes the exact trianglewhere $\delta =v\circ r^{-1}$ in the derived category.

For the final construction write $U=S^2\setminus S^1$ and set $A=i_{!}{\mathbb{Q}}_U$, $B={\mathbb{Q}}_{S^2}$, $C=j_{\ast }{\mathbb{Q}}_{S^1}$, each initially placed in degree zero. The stated sequence $0\to A\to B\to C\to 0$ is exact on stalks: on $U$ its first map is an isomorphism and its last sheaf is zero; on the equator its first sheaf is zero and the second map is an isomorphism. Take its connecting morphismThe source has only ${\mathcal{H}}^0(C)=C$, whereas the target has only ${\mathcal{H}}^{-1}(A[1])=A$. Therefore every induced map ${\mathcal{H}}^i(\mu )$ is zero: in any degree at least one of its source and target is zero. This is a ghost morphism in a derived category.

To prove the derived morphism itself is nonzero, apply derived global sections, giving the hypercohomology sequence. The extension by zero identifies $H^{\ast }(S^2;A)$ with the compactly supported cohomology of $U$, and closed direct image identifies $H^{\ast }(S^2;C)$ with $H^{\ast }(S^1;\mathbb{Q})$. The complement $U$ is two open disks, each homeomorphic to ${\mathbb{R}}^2$, soThe degree-one part of the resulting exact sequence isThus ${\mu }_{\ast }$ is injective, in particular nonzero. With the hemispheres oriented from $S^2$, the last map is $(a,b)\mapsto a+b$, so an appropriately oriented equator generator maps to $(1,-1)$. A zero morphism in the derived category of sheaves would induce a zero map after derived global sections and taking cohomology. Consequently $\mu \ne 0$, although ${\mathcal{H}}^i(\mu )=0$ for every $i$. This establishes explicitly that the collection of cohomology-sheaf functors detects zero objects but is not faithful on morphisms.

Here a smooth family is understood as a smooth proper family of compact complex manifolds: $f:\mathcal{X}\to S$ is a proper holomorphic submersion, with compact Kähler fibres. Properness and compactness are essential for the usual statement. Indeed, the projection $({\mathbb{C}}_t\times {\mathbb{C}}_z)\setminus \{(0,0)\}\to {\mathbb{C}}_t$ is a holomorphic submersion with Kähler fibres, but its fibres are $\mathbb{C}$ for $t\ne 0$ and ${\mathbb{C}}^{\ast }$ for $t=0$. Their first Betti numbers are zero and one, so their cohomology cannot give a fixed-rank variation. This supplies a counterexample if “smooth family” is read without the usual proper compact-fibre hypothesis. The Ehresmann fibration theorem gives local differentiable trivializations, sois a local system with fibre $H^k(X_s,\mathbb{Z})/\text{torsion}$. Its associated flat bundle has the Gauss–Manin connection, given by differentiating cohomology classes in topologically constant frames.

The Kähler Hodge decomposition givesThe Betti numbers are locally constant by differentiable triviality. The dimensions of $H^q(X_s,{\Omega }_{X_s}^p)$ are upper semicontinuous, and their sums for fixed $p+q=k$ equal the fixed Betti number. Near any fixed fibre, every summand can only decrease, while their sum cannot change; hence none decreases. Thus the Hodge numbers are locally constant. The relative de Rham complex and the Grauert base change theorem then makethe fibres of holomorphic subbundles $F^p\mathcal{H}$. More intrinsically these are the images of the hypercohomology of the truncated relative de Rham complex ${\Omega }_{\mathcal{X}/S}^{\ge p}$ in the full relative de Rham cohomology. Kähler degeneration of the spectral sequence ensures that these images have the required ranks.

For the transversality proof, let $A^2$ denote the ideal in the absolute holomorphic de Rham complex generated by two pulled-back base one-forms. Keeping zero or one base differential gives a short exact sequence of cochain complexes, with the usual shifted differential in the left-hand complexThe connecting morphism in degree $k$ is the Gauss–Manin connectionTo compute it, lift a relative de Rham cocycle to the middle complex, apply the absolute differential, and retain the terms with one base differential. A cocycle in filtration degree $p$ can be represented using absolute forms of degree at least $p$. In a term containing one base differential, removing that factor leaves a relative form of degree at least $p-1$. The connecting construction therefore lands in $F^{p-1}$:This is the filtration shift behind Griffiths transversality. It explains the inequality rather than merely naming a theorem. The unfiltered construction agrees with topological parallel transport by the de Rham comparison, and its square is zero, as required for the flat connection.

Consequently the fibre cohomology gives a pure variation of Hodge structure. To claim a polarized integral variation one needs the additional compatible flat polarization. In a smooth projective family, a relative ample line bundle supplies a flat integral Kähler class; its primitive pieces are polarized by the Hodge–Riemann bilinear relations, and the full cohomology is polarized using the appropriate primitive-piece signs. For a general compact Kähler family, the unpolarized variation above is automatic, but a flat integral polarization is not supplied merely by knowing that each fibre is Kähler. This distinction is particularly relevant to unrestricted K3 families.