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Use the standard integral cohomology ring $H^{\ast }({\mathbb{CP}}^2)=\mathbb{Z}[u]/(u^3)$, $|u|=2$. Any self-homeomorphism $h$ sends $u$ to $u$ or $-u$, and hence sends $u^2$ to $u^2$. Its Lefschetz number is therefore $1\pm 1+1$, which is either $3$ or $1$. By the Lefschetz fixed-point theorem, every such homeomorphism has a fixed point.

Now ${\mathbb{CP}}^2$ is simply connected: its CW structure has no one-cells and only two- and four-cells. If it covered a connected space with degree $d\ge 2$, it would be the universal cover. The usual covering space hypotheses hold for the base because it is locally homeomorphic to this manifold. Its deck transformation group would act transitively on every fibre and have $d$ elements. A nonidentity deck transformation has no fixed point: uniqueness of lifts makes one fixing a point the identity everywhere on the connected cover. This contradicts the preceding calculation. ThusThus the complex projective plane has no nontrivial covering quotient.

For integral singular cochains $a\in C^p(X)$ and $b\in C^q(X)$, define the cup product on an ordered simplex byThe same construction works over any commutative coefficient ring. Expanding the coboundary as the alternating sum of faces gives the Leibniz identityIndeed faces deleting vertices before the split occur in the first sum, faces after the split occur in the second, and the two terms at the common vertex cancel with opposite signs. Thus the product of cocycles is a cocycle. Replacing $a$ by $a+\delta c$ changes the product with a cocycle $b$ by $\delta (c⌣b)$; replacing $b$ by $b+\delta d$ changes the product with a cocycle $a$ by $(-1{)}^p\delta (a⌣d)$. Hence the product is independent of representatives and defines a well-defined product on cohomology classes.

With ${\mathbb{F}}_2={\mathbb{Z}}_2$ coefficients, the real-projective-space cohomology ring isFor $n>m\ge 1$, the pullback of the degree-one generator is either zero or $t$. The second choice would send the zero class $t^{m+1}$ to the nonzero class $t^{m+1}\in H^{m+1}({\mathbb{RP}}^n;{\mathbb{F}}_2)$, contradicting the naturality of the cup product. Thus the generator pulls back to zero, and so do all its positive powers. For $m=0$ the target's reduced cohomology is already zero. Therefore every such map induces zero on reduced mod-two cohomology.

Suppose a continuous $f:S^n\to {\mathbb{R}}^n$ had $f(x)\ne f(-x)$ everywhere. Thenis a continuous odd map to $S^{n-1}$ and descends to $\bar{h}:{\mathbb{RP}}^n\to {\mathbb{RP}}^{n-1}$. For $n\ge 2$, a path from $x$ to $-x$ projects to a loop detected by the degree-one generator of the double cover. Its image lifts to a path from $h(x)$ to $-h(x)$, so is detected by the target's double cover as well. Thus ${\bar{h}}^{\ast }$ is nonzero in degree one, contradicting the preceding calculation. For $n=1$, an odd continuous map from the connected circle to $S^0$ is already impossible. The case $n=0$ is immediate. HenceThis proves the Borsuk-Ulam theorem rather than assuming it.

For a closed cover $A_1,\dots ,A_{n+1}$, assume no member contains an antipodal pair. Apply the just-proved theorem to $F(x)=(d(x,A_1),\dots ,d(x,A_n))$, using any metric on the sphere. Empty sets can be assigned a constant positive coordinate. It gives equal distances at $x,-x$. If either point belongs to an $A_i$ with $i\le n$, zero distance and closedness place both points in it, a contradiction. Hence neither lies in any of the first $n$ sets, and both must lie in $A_{n+1}$, another contradiction. This proves the Lusternik-Schnirelmann-Borsuk theorem for closed sets.

Four closed sets on $S^2$ need not contain an antipodal pair. Take the four unit vertices $v_i$ of a regular tetrahedron centered at zero and define $C_i=\{x:x\cdot v_i\ge x\cdot v_j\text{ for every }j\}$. These closed sets cover the sphere by maximizing the four scalar products. If $x$ and $-x$ belonged to $C_i$, both inequalities would force all four scalar products to be equal. Since $\sum v_i=0$, they would all be zero, and because the vertices span ${\mathbb{R}}^3$, $x$ would be zero. This is impossible on the sphere. Thus these four sets give the antipodal-free closed cover from simplex Voronoi cells counterexample.

An immersion $M^4\to {\mathbb{R}}^5$ would inject $TM$ into the trivial rank-five bundle. Taking its orthogonal complement gives a normal real line bundle $\nu$ with $TM\oplus \nu \cong {\epsilon }^5$. The Whitney product formula for Stiefel–Whitney classes requiresThe real line bundle need not be orientable, but it has no Stiefel–Whitney classes above degree one.

Let $a,b$ be the pulled-back degree-one generators from the two real projective planes. Their mod-two cohomology ring, by the Künneth theorem, is ${\mathbb{F}}_2[a,b]/(a^3,b^3)$. To compute the tangent classes, at a line $L$ the projective tangent space is $\mathrm{Hom}(L,L^{\perp })$. Adding $\mathrm{Hom}(L,L)\cong \mathbb{R}$ gives $\mathrm{Hom}(L,{\mathbb{R}}^3)\cong 3L$, with the real line identified with its dual using a metric. Thus the stabilized tangent bundle of real projective space gives $T{\mathbb{RP}}^2\oplus {\epsilon }^1\cong 3\gamma$ andThe tangent bundle of the product is the direct sum of the pulled-back tangent bundles. Since $(1+a+a^2)(1+a)=1$ in this truncated ring, and similarly for $b$,The degree-two class $ab$ is nonzero by the product-ring basis. It cannot be a class of the rank-one bundle $\nu$. Equivalently $w_1(TM{)}^2+w_2(TM)=ab\ne 0$, violating the codimension-one Stiefel–Whitney immersion obstruction. Hence ${\mathbb{RP}}^2\times {\mathbb{RP}}^2$ admits no immersion in ${\mathbb{R}}^5$.

Let $Z={\mathbb{CP}}^5/{\mathbb{CP}}^2$. Its quotient cell structure has one cell in each of dimensions $0,6,8,10$. In particular $Z$ is five-connected. The relative cohomology sequence, or its cellular cochain complex, givesThe subscripts denote degrees. Under the quotient map back to ${\mathbb{CP}}^5$, the three positive classes pull back to $x^3,x^4,x^5$. All positive products vanish, since their degree is at least twelve. Thus the cohomology ring of a collapsed projective subspace in this case is the square-zero algebra $R$ on these three classes.

Now compute its Sullivan minimal model $(\Lambda V,d)$, using cohomological grading and the convention that even generators generate a polynomial ring and odd generators an exterior algebra. Five-connectivity gives $V^j=0$ for $j\le 5$. Since a minimal differential is decomposable, it cannot have a nonzero value below degree twelve. Matching cohomology through degree ten therefore forceswith no other generators in degrees at most ten. The first products must then be killed by generatorsThese follow by killing respectively the unwanted degree-twelve, degree-fourteen, and two degree-sixteen classes. Subsequent generators kill $bc,c^2$ and the resulting syzygies. For example $av-bu$ is closed of degree nineteen, so it requires a further generator of degree eighteen. Thus the full minimal model is not just the algebra on the first seven generators.

There is nevertheless a simple map from the full model to $R$: send $a,b,c$ to the three displayed classes and every other generator to zero. It is a chain map, since every generator differential is a sum of products of positive-degree elements, and all positive products in $R$ vanish. It induces an isomorphism in cohomology: the entire cohomology of $\Lambda V$ is concentrated in degrees $0,6,8,10$, and its indicated basis maps to the indicated basis of $R$. This proves formality of a topological space and identifies the full minimal model as a minimal resolution of $R$, rather than only matching the initial part of its cohomology.

An explicit description of that full resolution is useful. Take the free graded Lie algebraIts Chevalley–Eilenberg cochain algebra $C^{\ast }(L)$ has one generator of degree $|\ell |+1$ for every homogeneous Lie basis word $\ell$, and its quadratic differential is dual to the bracket, with the usual graded signs. Rescaling the bracket-word duals gives exactly the generators and differentials listed above. Positive Lie degrees make this algebra Sullivan minimal. Its cohomology is $R$: the universal enveloping algebra is $T(p,q,r)$, and the augmentation module has the exact free resolutionExactness is immediate because every nonempty tensor word has a unique last letter. Applying ${\mathrm{Hom}}_{T(p,q,r)}(-,\mathbb{Q})$ gives just the unit and the three dual classes in total degrees $6,8,10$; products of positive classes land in an absent Ext degree two and are zero. Chevalley–Eilenberg cochains of a graded Lie algebra compute this Ext, so this also verifies the full model's cohomology.

The wedge sum $W=S^6\vee S^8\vee S^{10}$ has the same connectivity, cohomology ring, and initial closed minimal generators. The same full-model-to-$R$ argument applies to $W$, so its minimal model is also $C^{\ast }(L)$, by uniqueness of Sullivan minimal models of quasi-isomorphic simply connected rational cochain algebras. The Sullivan minimal-model classification identifies isomorphic minimal models with the same rational homotopy type. ConsequentlyThe explicit infinite minimal resolution above is shared by both spaces.

Use cohomological degree, and write the Sullivan minimal models asandThe subscripts indicate degrees. The even generators generate a polynomial ring and the odd generators an exterior algebra. Their cohomology rings are respectively $\mathbb{Q}[x]/(x^{n+1})$ and $\mathbb{Q}[u]/(u^2)$, since each odd differential kills exactly the indicated non-zero-divisor polynomial relation. These are the usual projective-space and even-sphere minimal models.

Orient the top classes so that the degree-one condition is $f^{\ast }[u]=[x^n]$. A representative model map must send $u$ to $x^n$. It must send $v$ to an element of degree $4n-1$ with differential $x^{2n}$. The element $x^{n-1}y$ has exactly these properties. ThusThe differential check is $d(x^{n-1}y)=x^{n-1}{x}^{n+1}=x^{2n}=(f^{\ast }u{)}^2$. In these degrees there is no other available monomial: even powers of $x$ have even degree, and an odd monomial is a power of $x$ times $y$. Hence the coefficient is forced by the differential. For $n=1$ the two models coincide and the map is the identity model map.

For finite-dimensional Complex projective space ${\mathbb{CP}}^r$, use the filtration by its linear projective subspaces. The difference ${\mathbb{CP}}^j\setminus {\mathbb{CP}}^{j-1}$ is ${\mathbb{C}}^j$, so there is one topological cell in every even dimension $0,2,\dots ,2r$, and none in odd dimensions. Every cellular boundary vanishes for degree reasons. The integral cohomology is therefore $\mathbb{Z}$ in those even degrees and zero elsewhere.

For the cohomology ring, let $x\in H^2({\mathbb{CP}}^r;\mathbb{Z})$ be the Poincare dual of a projective hyperplane, using the complex orientation. Generic hyperplanes meet transversely, and a collection of $j$ of them intersects in ${\mathbb{CP}}^{r-j}$. The geometric interpretation of the cup product identifies $x^j$ with the dual of that intersection. Restricting to a linear ${\mathbb{CP}}^j$, the $j$ hyperplanes meet in one point, with positive sign because the transverse normal directions are complex lines. HenceThus $x^j$ is a primitive generator in degree $2j$, not merely a nonzero multiple. There are no classes beyond degree $2r$, soThe point case $r=0$ has ring $\mathbb{Z}$. Taking the union of the finite projective spaces gives $H^{\ast }({\mathbb{CP}}^{\infty };\mathbb{Z})=\mathbb{Z}[x]$, since each fixed cohomological degree stabilizes. This proves the cohomology ring of complex projective space computation.

For the rank bound for a nonsingular complex bilinear map, write $a={\dim }_{\mathbb{C}}U$, $b={\dim }_{\mathbb{C}}V$, $R=\mathrm{im}\varphi$, and $s={\dim }_{\mathbb{C}}R$. Assume $a,b\ge 1$. Injectivity with either nonzero factor fixed implies $\varphi (u\otimes v)\ne 0$ whenever $u,v\ne 0$. Thereforeis well defined and continuous. Scaling either vector only scales the output, and in local projective charts the formula is given by bilinear coordinates with no simultaneous zero. Let $h$ be the target's degree-two generator and $x,y$ the pullbacks of the two factor generators. The Künneth theorem givesFixing $[v]$ or $[u]$ makes $F$ a projective linear embedding induced by an injective linear map. Such an embedding pulls back the hyperplane class to the factor's hyperplane class. Since the two restrictions determine $H^2$ of the product, $F^{\ast }h=x+y$, with the generator of a point factor interpreted as zero.

Put $d=a+b-2$. In the source cohomology ring, every term in $(x+y{)}^d$ vanishes except the one with exponents $a-1,b-1$, andThe binomial coefficient is a positive integer and the final monomial generates the top integral cohomology. If $s\le d$, the class $h^d$ in ${\mathbb{CP}}^{s-1}$ would be zero. This contradiction provesWhen $a=b=1$, the nonzero pure-tensor image already gives $s\ge 1$. A positivity convention is genuinely necessary: if $U=0$, $V={\mathbb{C}}^2$, and $W=0$, all restrictions to the zero tensor subspaces are injective, but the literal bound would read $0\ge 1$. The displayed theorem is therefore the nonzero-factor version. If a nonzero factor is infinite-dimensional, either factor restriction already bounds the image dimension below by that factor dimension; the sum of two positive cardinal dimensions, with at least one infinite, is their maximum, giving the corresponding cardinal bound.

Take ${\mathbb{CP}}^3$. It is a closed smooth manifold and is simply connected: its CW complex structure has no one-cells, and attaching cells of dimension at least two to a point cannot create a fundamental group. Its cohomology ring is $\mathbb{Z}[x]/(x^4)$ with $|x|=2$, so $x^3\ne 0$.

The theorem used is the Lusternik–Schnirelmann critical point theorem: every smooth real-valued function on a closed connected smooth manifold has at least its unnormalized Lusternik–Schnirelmann category many distinct critical points, without a nondegeneracy assumption. Here category is the minimum number of open subsets covering the manifold whose inclusions are null-homotopic. The cup length bound gives category at least four. To justify that bound, a cover by three categorical sets would make each positive-degree class vanish on its assigned set. The long exact sequence of each pair lifts three copies of $x$ to relative classes, and their relative cup product would lie in $H^6(M,M;\mathbb{Z})=0$, contradicting $x^3\ne 0$. Thus every smooth function has at least four critical points, including functions with degenerate critical points. The critical-point theorem in this formulation is described in Conley pairs in geometry—Lusternik–Schnirelmann theory and more.

False. Take $M={\mathbb{CP}}^2\#{\mathbb{CP}}^2$ and $N=S^2\times S^2$, with their standard product and connected-sum orientations. They are closed simply connected four-manifolds. For the connected sum, removing a four-ball leaves the fundamental group unchanged, and the Seifert-van Kampen theorem across the separating three-sphere preserves simple connectivity. The Mayer–Vietoris sequence computes integral cohomology $\mathbb{Z},{\mathbb{Z}}^2,\mathbb{Z}$ in degrees zero, two and four, and zero in other degrees. The Künneth theorem gives the same groups for $S^2\times S^2$. Thus their Betti numbers agree and all their integral cohomology is torsion-free.

Their cohomology rings nevertheless differ. In $M$, let $u,v$ come from the two projective summands and let $z$ generate degree four. The cohomology ring of the connected sum of two complex projective planes hasThese relations follow by using cycles in the separate summands and the degree-one pinch projections onto them. In $N$, the sphere-factor generators $x,y$ satisfyConsequently every degree-two class of $N$ has squarean even multiple of the top generator, whereas $u^2$ in $M$ is a primitive generator. An integral graded-ring isomorphism cannot change this property, since it sends a generator of $H^4$ to its positive or negative. A homotopy equivalence induces a graded-ring isomorphism, so none exists. Equivalently, their intersection forms are $\mathrm{diag}(1,1)$ and $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, an odd form and an even form respectively.

Choose orientations on the two sphere factors whose product is the specified orientation of $X$. Give the four factors of $Y=(S^2{)}^4$ the corresponding orientations in the order $1,2,3,4$. The Künneth theorem has no Tor functor terms here and givesLet $u_i\in H^2(Y;\mathbb{Z})$ be the pullback of the positively normalized cohomology class from factor $i$. The cohomology ring iswith $⟨u_1{u}_2{u}_3{u}_4,[Y]⟩=1$. Its generators commute by graded commutativity of the cup product.

Fix a point in each sphere. For each pair $i<j$, let $S_{ij}$ be the four-dimensional submanifold in which factors $i,j$ vary and the remaining factors equal their fixed points. Orient $S_{ij}$ by the ordered factors $i,j$. The six homology classesgive a coordinate sphere intersection basis by the homology cross product description in the Künneth theorem. Moving a fixed point in one of the other two factors produces a disjoint homologous representative of each $S_{ij}$. Hence every basis element has zero self-intersection number.

More precisely, the intersection pairing isFor complementary pairs the representatives intersect transversely at one point. The sign is positive because swapping two-dimensional factors contributes $(-1{)}^{2\cdot 2}=1$. For pairs with a common index, a factor fixed in both representatives can be displaced to make them disjoint. In the paired order $(12,34),(13,24),(14,23)$, the intersection form is three copies of $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$.

For the diagonal map $\delta :X\to Y$, write $a,b\in H^2(X;\mathbb{Z})$ for the two sphere generators. ThenThe coefficient of $[S_{ij}]$ in $[\Delta ]$ is $⟨u_i{u}_j,[\Delta ]⟩$. Pullback and the preceding cup product relations show that these coefficients are one for $12,14,23,34$ and zero for $13,24$. ThereforeThere are exactly two complementary pairings among these four terms, each counted twice in the square. ThusThis computes both the explicit homology class and its intersection pairing, including all orientation signs.

Use the standard complex orientations and the product orientation. Let $a,b\in H^2(S^2\times S^2;\mathbb{Z})$ satisfy $a^2=b^2=0$ and $⟨ab,[S^2\times S^2]⟩=1$, and let $x\in H^2({\mathbb{CP}}^2;\mathbb{Z})$ satisfy $⟨x^2,[{\mathbb{CP}}^2]⟩=1$. These are the cohomology rings of the product of two spheres and the Complex projective plane.

For a map from the Complex projective plane, write $f^{\ast }a=rx$ and $f^{\ast }b=sx$. Naturality of the cup product gives $r^2{x}^2=f^{\ast }(a^2)=0$ and $s^2{x}^2=f^{\ast }(b^2)=0$. Since $x^2$ has infinite order, $r=s=0$. It follows that $f^{\ast }(ab)=0$, so every such map has degree $0$.

In the reverse direction, write $g^{\ast }x=ra+sb$. ThenEvery even mapping degree really occurs. To construct it, identify $S^2$ with ${\mathbb{CP}}^1$ and use the Segre embeddingThe generator $x_3\in H^2({\mathbb{CP}}^3;\mathbb{Z})$ pulls back to $a+b$: restricting to either factor gives a projective line and hence coefficient $1$. By the cellular approximation theorem, $\sigma$ is homotopic to a cellular map. Its four-dimensional domain then maps into the four-skeleton ${\mathbb{CP}}^2$ of ${\mathbb{CP}}^3$. Denote that map by $g_0$; restriction of $x_3$ to ${\mathbb{CP}}^2$ is $x$, so $g_0^{\ast }x=a+b$ and $\mathrm{deg}{g}_0=2$.

For any integer $m$, choose a map $d_m:S^2\to S^2$ of mapping degree $m$. For $m>0$ one may use $z\mapsto z^m$ on the Riemann sphere, for $m<0$ use $z\mapsto {\overline{z}}^{|m|}$, and for $m=0$ use a constant map. Then $g_m=g_0\circ (d_m\times \mathrm{id})$ satisfies $g_m^{\ast }x=ma+b$, so the possible degrees are exactly all even integers, with $g_m$ realizing $2m$. The cellular approximation here deforms this particular map into the skeleton; it does not require a retraction of ${\mathbb{CP}}^3$ onto ${\mathbb{CP}}^2$.

Finally, take the connected sum of oriented manifolds with both summands carrying their standard complex orientations. Its degree-two cohomology has generators $u,v$ withThis follows by choosing the generators supported away from the two balls used to form the connected sum, so mixed cup products vanish while each square gives the common orientation class. WriteIf $\mathrm{deg}h=n$, the three cup product relations sayTaking determinants gives $-(\det P{)}^2=n^2$, which forces $n=0$. Therefore the only possible degree to the connected sum is $0$, realized by a constant map. This is a degree constraint from intersection forms: the indefinite intersection form of the sphere product cannot pull back a definite form with a nonzero degree multiplier.

Integral groups for $M_4$. Put $r=n-1$. The space $M_n$ is the inversion mapping torus of a torus, fibred over $S^1$ with fiber $T^r$. The cohomology ring of a torus is the exterior algebra on $r$ degree-one generators. Inversion acts as $-1$ on each such generator, hence as $(-1{)}^q$ on $H^q(T^r;\mathbb{Z})\cong {\mathbb{Z}}^{\left(\genfrac{}{}{0ex}{}{r}{q}\right)}$.

The Wang sequence therefore givesThe right-hand term is free, so this short exact sequence splits as a sequence of groups. In even fiber degree, $F^{\ast }-1=0$; in odd fiber degree it is multiplication by $-2$. For $r=3$ this computes the integral cohomology of an inversion mapping torus:The top torsion group is consistent with $M_4$ being nonorientable: inversion of its three-dimensional fiber reverses orientation.

Mod-two groups. Over ${\mathbb{F}}_2$, inversion acts as the identity on the fiber cohomology. The Wang sequence givesThuswith out-of-range binomial coefficients interpreted as zero. These are exactly the dimensions of the graded groups $H^{\ast }(T^n;{\mathbb{F}}_2)$.

The intersection pairing for $M_2$. The mapping torus of reflection of the circle is the Klein bottle. Let $a$ be the section loop through a fixed point of the reflection and $b$ a fiber circle. They form a basis of $H_1(M_2;{\mathbb{F}}_2)$. The section has a normal neighborhood homeomorphic to a Möbius band, so it is one-sided and a transverse displacement meets it once modulo two. The fiber is two-sided and can be displaced disjointly. The two loops meet once. The mod-two intersection pairing of the Klein bottle therefore has matrixTake the evaluation-dual basis $t,x\in H^1(M_2;{\mathbb{F}}_2)$, with $t(a)=1,t(b)=0$ and $x(a)=0,x(b)=1$. By Poincare duality, the cup product matrix in this dual basis is $Q^{-1}$, not $Q$:Writing $w$ for the nonzero top class gives $t^2=0$, $tx=w$ and $x^2=w$. Hence the cohomology ring isThese relations already force all degrees above two to vanish.

The ring for general $n$. For each of the $r$ fiber coordinates, projection induces $M_n\to M_2$. Pull back the classes above, calling the common base class $t$ and the fiber-coordinate classes $x_1,\dots ,x_r$. Naturality of the cup product gives $t^2=0$ and $x_i^2=t{x}_i$. The square-free products $x_I={\prod }_{i\in I}{x}_i$ restrict to the exterior algebra basis of the fiber cohomology. The Leray-Hirsch theorem then says that $x_I$ and $t{x}_I$ form an additive basis. This proves the full mod-two cohomology ring of an inversion mapping torus:For $n\ge 2$, $x_i^2=t{x}_i$ is nonzero by this basis description. In the cohomology ring of a torus every degree-one class squares to zero: the generators square to zero and the cross terms occur twice in characteristic two. The degree-one cup-square obstruction to ring isomorphism therefore proves that the rings are not isomorphic for any $n\ge 2$, despite their isomorphic graded groups. For $n=1$, both spaces are circles and the rings are isomorphic. Even if grading is forgotten, the rings differ for $n\ge 2$: every element of the torus ring has square either zero or one, whereas $x_i^2=t{x}_i$ is nonzero and is not the unit.

Integral cup products in $M_4$. To specify $H^{\ast }(M_4;\mathbb{Z})$ also as a ring, let $a\in H^1$ be the pullback of the positive generator of $H^1(S^1;\mathbb{Z})$. Let $\rho$ denote reduction modulo two and set ${\tau }_i=\hat{\beta }(x_i)$ using the integral Bockstein homomorphism. The degree-one Bockstein square identity gives $\rho {\tau }_i=x_i^2=t{x}_i$, so the ${\tau }_i$ are the three independent order-two classes in degree two. Choose free classes $b_{12},b_{13},b_{23}\in H^2$ which restrict to the corresponding two-fold fiber products and have reductions $\rho b_{ij}=x_i{x}_j$. Such choices exist: reduction in degree two is surjective because $H^3(M_4;\mathbb{Z})$ is free, and adding the ${\tau }_i$ removes any $t{x}_i$ terms from an initial lift.

The Wang sequence identifies $a{b}_{12},a{b}_{13},a{b}_{23}$ as a free basis of $H^3$. Let $\kappa$ generate $H^4\cong \mathbb{Z}/2$, with $\rho \kappa =t{x}_1{x}_2{x}_3$. Reduction in degree four is an isomorphism. The integral cup products in the four-dimensional inversion mapping torus are consequently determined byTogether with the unit, graded commutativity and vanishing above degree four, these give every product. For example, $\rho (b_{12}{b}_{13})=x_1^2{x}_2{x}_3=t{x}_1{x}_2{x}_3$, whereas $\rho (b_{12}^2)=t^2{x}_1{x}_2=0$. The products $a{\tau }_i$ vanish because they are torsion in the free group $H^3$.

For nonzero finite-dimensional $U$ and $V$, the intended assertion is true. Write $a=\dim U$, $b=\dim V$, and replace $W$ by $\mathrm{im}\varphi$, of dimension $r$. Injectivity on the stated slices implies $\varphi (u\otimes v)\ne 0$ whenever $u,v\ne 0$. Consequently the bilinear map defines a continuous map of Complex projective spacesThe rank $r$ is at least one, since one such nonzero tensor has nonzero image.

Let $L_U,L_V,L_W$ be the respective tautological bundles. Fiberwise, $\varphi$ identifies $L_U\otimes L_V$ with $F^{\ast }{L}_W$. For the positive hyperplane classes $x=c_1(L_U^{\ast })$, $y=c_1(L_V^{\ast })$ and $z=c_1(L_W^{\ast })$ (zero when the projective space is a point), the first Chern class of a tensor product of complex line bundles givesThe Künneth theorem, together with part (a), identifies the product cohomology ring withThere are no Tor terms because the factor groups are free. In particular, its monomials $x^i{y}^j$ with $0\le i<a$, $0\le j<b$ form an integral additive basis. In top degree,If $r\le a+b-2$, the target relation $z^r=0$ would imply $(x+y{)}^r=0$ and hence contradict this nonzero top power. ThusWhen $a=b=1$, the already-established inequality $r\ge 1$ gives the same conclusion. The complex bilinear dimension bound is sharp: multiplication of complex polynomials of degrees less than $a$ and $b$ has target dimension $a+b-1$, is injective in either nonzero fixed factor, and its image spans every monomial in that target.

Literal zero-space qualification. The printed assertion does not explicitly exclude zero vector spaces. If $U=0$, $V={\mathbb{C}}^2$ and $W=0$, its slice-injectivity hypothesis is vacuous, while the claimed inequality would be $0\ge 1$. Thus, with zero spaces permitted, this is a counterexample to the assertion exactly as written; the proof above supplies the usual nonzero finite-dimensional 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.

Use ${\mathbb{F}}_2$ coefficients throughout. Write $h=c_1(\mathcal{O}(1))\mathrm{mod}2$ on Complex projective space and $a=w_1({\gamma }_d)$ on Real projective space. We use the standard cohomology ring of complex projective space ${\mathbb{F}}_2[h]/(h^{d+1})$, with $|h|=2$, and the real projective calculation from the preceding solution. The complex projective calculation follows, for example, from its one cell in each even dimension together with the fact that transverse projective hyperplanes represent the powers of $h$; $d$ such hyperplanes have one intersection point.

For the standard inclusion $i:{\mathbb{RP}}^d\hookrightarrow {\mathbb{CP}}^d$, the restricted complex tautological line is ${\gamma }_d{\otimes }_{\mathbb{R}}\mathbb{C}$. As a real vector bundle it is ${\gamma }_d\oplus {\gamma }_d$, whose total Stiefel–Whitney class is $(1+a{)}^2=1+a^2$. The Stiefel–Whitney class of the underlying real bundle of a complex line identifies its degree-two class with its First Chern class reduced modulo two. Passing to the dual line defining $h$ changes only a sign, which disappears modulo two. Thus mod-two restriction from complex to real projective space gives $i^{\ast }h=a^2$. In particular the answer for dimension two is the ring mapIt sends $1$ to $1$, is an isomorphism in degree two, and sends $h^2$ to zero.