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For global cohomology classes satisfying the local basis criterion for Leray-Hirsch classes, multiplication commutes with the Mayer–Vietoris sequence, including its connecting homomorphism: multiply cochains on the right by global cocycle representatives. Induct over a finite trivializing open cover. If $U$ is a union of $r-1$ patches and $V$ is the last patch, $U\cap V$ has a cover by at most $r-1$ trivialized intersections. The induction hypothesis on $U$, $V$ and $U\cap V$, followed by the Five lemma, proves the theorem on $U\cup V$. A compact base supplies the finite cover.

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 $H^1(T^2)={\mathbb{Q}}^2\ne 0$.

For an open subset $U\subseteq B$, let $E_U={\pi }^{-1}(U)$ and define the degree-preserving module mapUsing the prescribed fiber basis identifies its domain with the tensor product in the question. We prove that ${\Phi }_B$ is an isomorphism, giving the Leray-Hirsch theorem in this setting.

First suppose $U$ is a trivializing patch. The cohomological Künneth theorem over a field identifies $H^{\ast }(U\times F;\mathbb{Q})$ with the finite free graded $H^{\ast }(U;\mathbb{Q})$-module on a homogeneous basis of $H^{\ast }(F;\mathbb{Q})$. The required finiteness is supplied by the finite list $c_1,\dots ,c_v$ spanning the fiber's total cohomology. Express the restrictions of the $c_j$ in this product basis. A coefficient multiplying a fiber basis element of degree $k_i$ belongs to $H^{k_j-k_i}(U;\mathbb{Q})$; thus no coefficient can increase fiber degree. The degree-zero blocks are invertible on every path component, because restriction at every point gives a basis. Their inverses are again degree-zero cohomology classes.

After inverting those blocks, the remaining change-of-basis map is $I+N$, where $N$ strictly lowers fiber degree. There are only finitely many fiber degrees, so $N$ is nilpotent and $(I+N{)}^{-1}=I-N+N^2-\cdots$ is a finite sum. This proves the local basis criterion for Leray-Hirsch classes and therefore the isomorphism on every trivializing patch, without requiring the patch to be contractible.

Now use a finite trivializing open cover $U_1,\dots ,U_r$, supplied by compactness of $B$. Induct on $r$. Put $U=U_1\cup \cdots \cup U_{r-1}$ and $V=U_r$. The intersection $U\cap V$ is covered by the at most $r-1$ trivializing open sets $U_i\cap V$, so the induction hypothesis applies to $U$, $V$ and $U\cap V$.

Take the direct sum of shifted Mayer–Vietoris sequences for the base and compare it with the sequence for $E_U\cup E_V$. Restriction commutes with multiplication by each global cohomology class $c_j$. The connecting homomorphism also commutes: multiply the cochain representatives on the right by the restriction of a global cocycle representing $c_j$, and use $d(b⌣c_j)=db⌣c_j$. Consequently the maps $\Phi$ form a morphism of exact sequences. The Five lemma makes ${\Phi }_{U\cup V}$ an isomorphism, completing the finite-cover proof of the Leray-Hirsch theorem.

This is an additive graded module isomorphism. It does not assert that the fiber classes have the same multiplicative relations in the total space.

Cover the projective line by $U_0=\mathrm{Spec}k[t]$ and $U_1=\mathrm{Spec}k[t^{-1}]$. On their overlap the coordinate is invertible. Choose frames $e_0,e_1$ of the twisting sheaf on projective space $\mathcal{O}(n)$ with $e_1=t^n{e}_0$. The vanishing of quasi-coherent cohomology on an affine scheme applies to both charts and their intersection, so the Čech cochain complexcomputes the sheaf cohomology. Its kernel identifies with $k[t]\cap t^{n}k[t^{-1}]$, and its cokernel isFor $n\ge 0$, the intersection has basis $1,t,\dots ,t^n$; for $n<0$ it is zero. In the quotient, powers $t^j$ with $j\ge 0$ or $j\le n$ disappear. The surviving basis for $n\le -2$ is $t^{n+1},\dots ,t^{-1}$; for $n\ge -1$ none survives. HenceThis is the Čech cohomology of twists on the projective line.

By the preceding question, $\omega ={\Omega }_{{\mathbb{P}}^1/k}^1=\mathcal{O}(-2)$. In the $dt$ frame its degree-one Čech cohomology is one-dimensional, represented by $t^{-1}dt$. DefineThis is well defined: differentials regular on $U_0$ have nonnegative powers, and those regular on $U_1$ have powers at most $-2$ when expressed using $dt$. Evaluation and the cup product give natural pairingsFor $E=\mathcal{O}(n)$ with $n\ge 0$, the basis element $t^a$ in $H^0(E)$ pairs with $t^{-a-1}dt$ in $H^1(E^{\vee }\otimes \omega )$ and with no other element of this dual Laurent basis. The pairing matrix is a permutation matrix, hence invertible. For $n\le -2$ the same computation exchanges degrees zero and one. For $n=-1$ both sides vanish. Thus the pairings are perfect for every line bundle on ${\mathbb{P}}^1$.

Here we used the classification of line bundles on the projective line: they are $\mathcal{O}(a)$ for integers $a$. One direct justification is that every line bundle has a nonzero rational section and hence is associated with a divisor on an algebraic curve. The finite closed point defined by an irreducible polynomial $p(t)$ is linearly equivalent to $(\mathrm{deg}p)\infty$, using the principal divisor of $p(t)$. Thus every divisor is equivalent to its degree times $\infty$, proving the classification over arbitrary $k$.

To extend the pairing to every locally free coherent sheaf $E$, induct on rank, without assuming the splitting theorem of the next question. A one-dimensional subspace of the generic fibre, saturated in $E$, gives an exact sequencewith $L$ a line bundle and $Q$ a lower-rank vector bundle. This saturated line subbundle on a smooth curve exists because its stalk and its torsion-free quotient are free over the discrete valuation rings of the curve. The dual sequence, tensored with $\omega$, is exact as well, since the original sequence splits locally.

The two long exact sequences in sheaf cohomology, after dualizing one, are compatible with the evaluation pairings. In Čech cohomology, this compatibility follows by lifting a section locally: the differences of its lifts represent its connecting class, and contracting those differences with a dual cocycle gives the transpose connecting map, with the usual sign. Thus the connecting maps are adjoint; changing one vertical map's sign gives a commuting diagram. Its rows areThe maps for $L$ and $Q$ are isomorphisms by the line-bundle calculation and induction. Exactness then makes the maps for $E$ isomorphisms, by the Five lemma (or by comparing the relevant kernels and cokernels). All groups are finite-dimensional because the curve is projective and the sheaves coherent. Therefore residue duality on the projective line provesThese isomorphisms are natural in $E$. Higher sheaf cohomology vanishes by the same two-chart complex.

In an abelian category, for $f:A\to B$ set $\mathrm{coim}f=\mathrm{coker}(\ker f)$ and $\mathrm{im}f=\ker (\mathrm{coker}f)$. The canonical map from coimage to image is an isomorphism. Thus the image factorization in an abelian category isunique up to a unique compatible isomorphism.

The short five lemma states that in a commuting diagram of short exact sequences $0\to L\to M\to R\to 0$ and $0\to L^{\prime }\to M^{\prime }\to R^{\prime }\to 0$ in an abelian category, if the vertical maps $L\to L^{\prime }$ and $R\to R^{\prime }$ are isomorphisms, then so is $M\to M^{\prime }$.

For the Five lemma, consider commuting exact rowswith vertical maps $v_i:A_i\to B_i$. We prove the usual stronger version: $v_1$ epic, $v_2$ and $v_4$ invertible, and $v_5$ monic imply that $v_3$ is invertible. In particular the conclusion holds if all four outer vertical maps are isomorphisms.

Put $L=\mathrm{im}{d}_2$ and $R=\mathrm{im}{d}_3$, and define $L^{\prime },R^{\prime }$ similarly. Exactness gives short exact sequencesThe functoriality of abelian image factorization supplies their vertical outer maps. We show that both are isomorphisms.

First, $L\cong \mathrm{coker}{d}_1$ and $L^{\prime }\cong \mathrm{coker}{d}_1^{\prime }$: exactness at $A_2$ identifies $\mathrm{im}{d}_1$ with $\ker d_2$, and the image-coimage isomorphism for $d_2$ gives the claimed cokernel. Write $q:A_2\to L$ and $q^{\prime }:B_2\to L^{\prime }$ for these categorical cokernels. To construct an inverse to the induced map $\ell :L\to L^{\prime }$, observeCancellation of the epimorphism $v_1$ gives $q{v}_2^{-1}{d}_1^{\prime }=0$, so $q{v}_2^{-1}$ factors uniquely as $\bar{\ell }{q}^{\prime }$ with $\bar{\ell }:L^{\prime }\to L$. From $\ell q=q^{\prime }{v}_2$, composing with the epic $q,q^{\prime }$ gives $\bar{\ell }\ell =1_L$ and $\ell \bar{\ell }=1_{L^{\prime }}$.

Second, $R\cong \ker d_4$ and $R^{\prime }\cong \ker d_4^{\prime }$ by exactness at the fourth objects. Write their inclusions as $j:R\to A_4$ and $j^{\prime }:R^{\prime }\to B_4$. The equality $d_4^{\prime }{v}_4=v_5{d}_4$ gives the induced map $\rho :R\to R^{\prime }$. Sinceand $v_5$ is a monomorphism, $d_4{v}_4^{-1}{j}^{\prime }=0$. Kernel universality gives $\bar{\rho }:R^{\prime }\to R$ with $j\bar{\rho }=v_4^{-1}{j}^{\prime }$. Cancelling the monic $j,j^{\prime }$ proves $\bar{\rho }\rho =1_R$ and $\rho \bar{\rho }=1_{R^{\prime }}$.

Apply the short five lemma to the two short exact sequences: their outer maps $\ell ,\rho$ are invertible, so $v_3$ is an isomorphism. This five lemma via image factorization argument uses only universal properties and therefore works in any abelian category, without treating its objects as literal sets of elements.

Use integral cohomology and let $k\ge 1$. On the projective bundle $Y=\mathbb{P}(E)$ define the complex tautological line bundlePut $t=e(\mathcal{L})\in H^2(Y;\mathbb{Z})$, using the canonical complex orientation. On each fibre, $t$ is the Euler class of the tautological line over ${\mathbb{CP}}^{k-1}$, so $1,t,\dots ,t^{k-1}$ restrict to an integral basis of its cohomology.

Here is the finite-cover Leray-Hirsch theorem proof in this case. For each open set $U\subseteq X$, defineIf $E$ is trivial on $U$, its projective bundle is $U\times {\mathbb{CP}}^{k-1}$ and $\mathcal{L}$ is pulled back from the tautological line on the second factor. The Künneth theorem makes ${\Phi }_U$ an isomorphism, since the fibre has finite free integral cohomology. The same holds for every open subset of $U$.

Compactness of $X$ provides a finite trivializing cover $U_1,\dots ,U_N$. Induct on its size. If the result holds on $V=U_1\cup \cdots \cup U_{N-1}$, it holds on $U_N$ and on $V\cap U_N$, both lying in a trivializing chart. Form the diagram of Mayer–Vietoris sequences for the base, with the finitely many degree shifts on the left, and the total space on the right. Naturality of pullback and multiplication by the global even-degree classes $t^i$ makes the diagram commute. The Five lemma gives the isomorphism on $V\cup U_N$. ThusThis is the claimed free module statement, with its graded degree shifts made explicit.

The module basis expresses $t^k$ uniquely using the lower powers, with homogeneous coefficients. Define the Chern classes $c_j(E)\in H^{2j}(X;\mathbb{Z})$ by the unique relationThe pullbacks are suppressed when $f$ is regarded as a polynomial over $H^{\ast }(X)$. Evaluation at $t$ gives a surjective map $H^{\ast }(X)[T]\to H^{\ast }(Y)$. Since $f(T)$ is monic, monic polynomial division over a ring writes any polynomial as $q(T)f(T)+r(T)$ with degree of $r$ below $k$. If its evaluation is zero, module independence forces every coefficient of $r$ to vanish. Hence the kernel is exactly the ideal generated by $f$, provingThe even-degree coefficients are central in the graded commutative algebra, so this division and ideal statement also apply when the base has odd-degree cohomology. Uniqueness of the coefficients proves their naturality under pullback, by pulling back the relation and using the same module basis. With the hyperplane convention $u=-t$, the relation has the usual all-positive Chern coefficients; the alternating signs here correspond to the tautological line itself.

Now suppose $E={⨁}_i{L}_i$. The sections of a projective bundle $s_i:X\to Y$ choose the line $(L_i{)}_x\subset E_x$. They satisfy $s_i^{\ast }\mathcal{L}\cong L_i$, so $s_i^{\ast }t=e_i$ and pulling back the relation gives $f(e_i)=0$.

To obtain the full factorization over the possibly torsion-containing base ring, also use the associated open chartsThey contain the images of the sections $s_i$ and cover $Y$. Projection identifies $\mathcal{L}{|}_{V_i}$ with ${\pi }^{\ast }{L}_i{|}_{V_i}$, so $t-{\pi }^{\ast }{e}_i$ restricts to zero on $V_i$. The long exact sequence of the pair lifts this class to $H^2(Y,V_i)$. The relative cup product of the $k$ lifted classes lies inIts absolute image is ${\prod }_i(t-{\pi }^{\ast }{e}_i)$, so that product vanishes. The polynomial ${\prod }_i(T-e_i)$ is monic of degree $k$ and lies in the kernel of evaluation. Subtracting the monic generator $f(T)$ leaves degree below $k$, and module independence again makes the difference zero. ThereforeThe open-cover argument proves the factorization without a non-zero-divisor assumption on the differences $e_i-e_j$.

A Serre fibration has the homotopy lifting property for disks: for every $k\ge 0$, a map $D^k\to E$ and a homotopy $D^k\times I\to B$ starting at its composite with $p:E\to B$ admit a compatible lift $D^k\times I\to E$. Equivalently, it has that lifting property for CW complexes. Relative lifting for CW pairs follows by attaching cells.

Let $i:F\hookrightarrow E$ be the inclusion of the chosen fiber. The long exact sequence of homotopy groups of a fibration isIts low-dimensional end isThe maps $i_{\ast }$ and $p_{\ast }$ are induced by inclusion and projection on based maps; at degree zero they send a path component to its containing or image component. The last part is an exact sequence of pointed sets, not in general a sequence of groups.

For $\partial :{\pi }_q(B)\to {\pi }_{q-1}(F)$, represent a class by a based cube $a:I^{q-1}\times I\to B$, constant at $b_0$ on its boundary. Lift it from the constant map $e_0$ on the bottom and side faces, using relative disk lifting. Its top face lands in $F$ and is constant on that face's boundary; its class is $\partial [a]$. The choice of lifts does not change the resulting class. For $q=1$, lift a based loop starting at $e_0$ and take the path component of its endpoint in $F$. This fixes the boundary-map convention and defines every map in the displayed sequence.

For the splitting assertion, first take $n\ge 1$. We use the cohomological Serre spectral sequence with coefficient group $G$:Since $B$ is simply connected, these coefficient systems are constant. The Eilenberg–MacLane space $F=K(G,n)$ has $H^t(F;G)=0$ for $0<t<n$, andFor $n=1$ this uses $H_1(F)=G$, with $G$ abelian; for $n>1$ it follows from the Hurewicz theorem and the universal coefficient theorem for cohomology. Let $u$ correspond to ${\mathrm{id}}_G$, the universal cohomology class of an Eilenberg–MacLane space.

In bidegree $(0,n)$ there are no incoming differentials. For $2\le r\le n$, an outgoing differential lands in a vanishing fiber-cohomology row. The only remaining possible differential isThus $u$ survives. The edge map, which is restriction to $F$, supplies a classBy representability of cohomology by Eilenberg–MacLane spaces, choose a based map $q:E\to K(G,n)$ representing $c$. On $F$, $q$ induces the identity on ${\pi }_n=G$ and is an isomorphism on every homotopy group, since the other positive groups vanish.

Consider $h=(p,q):E\to B\times K(G,n)$. This is a map of fibrations over $B$, whose map on the fiber is the weak equivalence just obtained. The map on the base is the identity. Comparing the two instances of the long exact sequence of homotopy groups of a fibration and applying exactness, or the Five lemma in the group-valued range, shows that $h$ induces all homotopy isomorphisms. The spaces are path-connected because the base and fiber are. Hence the splitting of a simply connected Eilenberg–MacLane fibration givesThe crucial use of $H^{n+1}(B;G)=0$ is to lift the fiber's identity class; existence of a section alone was not assumed to prove a product splitting.

If $K(G,0)$ is allowed, its components have trivial positive homotopy groups. The simply connected base has no monodromy on this discrete set. Obstruction theory on the CW base gives a section in each fiber component, since all positive-dimensional fiber obstructions vanish and the component system is constant. Together these sections give $B\times G\to E$, a weak equivalence on each component. Thus the same conclusion also covers that interpretation.

Here is an intrinsic definition that also proves well-definedness. For a complex line bundle $L$, define its First Chern class to be the Euler class of its canonically oriented underlying real rank-two bundle: pull its integral Thom class back along the zero section after forgetting relative support. The complex orientation fixes the sign, so this construction makes no arbitrary choice of generator.

For a rank-$r>0$ complex vector bundle $E$, let $\pi :\mathbb{P}(E)\to X$ be its projective bundle of lines, let $S\subset {\pi }^{\ast }E$ be the tautological bundle, and put $h=-c_1(S)=c_1(S^{\ast })$. On every fiber ${\mathbb{CP}}^{r-1}$, $h$ is the positive degree-two generator. The Leray-Hirsch theorem says that if globally defined cohomology classes restrict to a free basis of the cohomology of every fiber, then multiplication by those classes identifies the cohomology of the total space with a free module over the base. Applied here, it givesThe finite trivializing cover in the question is sufficient for this application: the assertion holds on each trivializing open set by the Künneth theorem, since the fiber has finite free cohomology, and the Mayer–Vietoris sequence and the Five lemma glue it over the finite cover. The same local argument constructs the oriented Thom class used for line bundles. It does not require a choice of classifying map.

There are therefore unique classes $c_i(E)\in H^{2i}(X;\mathbb{Z})$ such thatDefine $c_0(E)=1$ and $c_i(E)=0$ for $i>r$. For rank zero the Total Chern class is $1$. This is the projective bundle definition of Chern classes. Existence and uniqueness follow by expressing $h^r$ in the displayed free module basis, with degrees determining each coefficient. The projective bundle and tautological bundle are intrinsic to $E$, and the line Thom class is uniquely fixed by orientation. Hence the resulting Chern classes do not depend on trivializations or other auxiliary choices. For a line bundle the relation is $h+c_1(E)=0$, agreeing with the original normalization. Pulling back this unique relation also proves naturality. This standard construction and the sum theorem are treated in Vector Bundles and K-Theory, Section 3.1.

The requested result is the Whitney sum formula for Chern classes:Here $c(E)=1+c_1(E)+\cdots +c_r(E)$, and the formula holds for complex vector bundles over a common base. In particular, a trivial bundle has total Chern class $1$.

Now take $B={\mathbb{CP}}^{n-1}$ and let $L\subset B\times {\mathbb{C}}^n$ be its tautological bundle. The standard Hermitian inner product gives the rank-$(n-1)$ complex bundle $Q=L^{\perp }$, with $L\oplus Q\cong {\underline{\mathbb{C}}}^n$. The orthogonal complex line flag manifold in the question is precisely $\mathbb{P}(Q)$: over a first line $\ell$, the second line is any line in ${\ell }^{\perp }$. Local orthonormal frames give this identification as a fiber bundle, with fiber ${\mathbb{CP}}^{n-2}$.

Let $x=c_1(L^{\ast })$ on $B$, also writing $x$ for its pullback to $\mathbb{P}(Q)$. By part 3(a), $H^{\ast }(B;\mathbb{Z})=\mathbb{Z}[x]/(x^n)$. The Whitney sum formula for Chern classes givesLet $S_2$ be the second tautological line on $\mathbb{P}(Q)$, and set $y=c_1(S_2^{\ast })$. Thus $x,y$ are exactly the pullbacks of the positive hyperplane classes from the two projective factors. The projective bundle definition of Chern classes for $Q$ givesTogether with $x^n=0$, this gives a surjective graded ring mapThere are no additional relations. Indeed the second relation is monic of degree $n-1$ in $y$, so polynomial division makes its source free over $\mathbb{Z}[x]/(x^n)$ on $1,y,\dots ,y^{n-2}$. The projective bundle formula for complex vector bundles gives exactly the same free basis on the target. The map sends each basis element to its corresponding basis element, and is therefore an isomorphism. ConsequentlyThis is the cohomology ring of the orthogonal complex line flag manifold. Its additive basis is $x^i{y}^j$ with $0\le i<n$, $0\le j<n-1$. As a symmetry check, multiplying the second relation by $y-x$ gives $y^n-x^n=0$, so $y^n=0$ as expected from the second projection. For $n=2$, every line has a unique orthogonal line, and the relations become $x^2=0$, $x+y=0$, giving the ring of ${\mathbb{CP}}^1$. If $n=1$, the space is empty and the second relation is $1=0$, so the printed formula still gives the zero cohomology ring.

The Bott isomorphism is multiplication by the Bott element $\beta \in {\tilde{K}}^0(S^2)$:Together with the suspension isomorphism andit gives the Complex K-theory of a sphere

Letbe a CW filtration in which each quotient $Y_r/Y_{r-1}$ is a wedge of even-dimensional spheres. The six-term exact sequence in Topological K-theory, the sphere calculation, and induction giveand a short exact sequence whose new summand in $K^0(Y_r)$ is free abelian on the newly attached cells. Every such extension splits as an extension of free abelian groups, so $K^0(Y)$ is free, with one generator for each cell. This proves the Complex K-theory of an even-cell complex result.

The exterior product definesFor a point it is the identity. Attaching one layer of even cells gives corresponding exact sequences on the source and target; the sphere case is the suspension isomorphism, and induction with the Five lemma proves that the product map remains an isomorphism. This is the Künneth theorem for complex K-theory with an even-cell factor.

For a mapping torus $T_f$, the K-theory Wang sequence of a mapping torus containsWhen $K^{-1}(Z)=0$, exactness gives

For $Z={\mathbb{CP}}^2\times {\mathbb{CP}}^2$, the Complex K-theory of complex projective space and the K-theory Künneth isomorphism giveThe factor swap interchanges $x$ and $y$. Its invariant subgroup has the basisIt follows that the K-theory of the mapping torus of the factor swap on two complex projective planes is

A complex in an abelian category is a sequence $(A_n,d_n)$ with $d_n{d}_{n+1}=0$. A sequence is exact when the image of every incoming map equals the kernel of the outgoing map.

The Five lemma says that in a morphism between exact five-term sequences, suitable epimorphism assumptions on the left and monomorphism assumptions on the right, together with isomorphisms in the four surrounding positions, force the middle map to be an isomorphism.

The Snake lemma associates to a morphism of short exact sequences the exact sequence

The commutative diagram of short exact sequences of complexes induces a commutative diagram between the two long exact homology sequences from part (b). If any two vertical chain maps induce isomorphisms in every degree, then in each five-term window four of the five vertical homology maps are isomorphisms. The Five lemma makes the remaining map an isomorphism. Rotating the window handles each of the three possible missing columns, proving the two-out-of-three assertion.

The Five lemma has a short form: in a commutative diagram of short exact sequences, if the maps on both outer terms are isomorphisms, then the map on the middle terms is an isomorphism.