Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 16 (Algebraic Topology)

The cellular homology of complex projective space gives one integral generator $[{\mathbb{CP}}^j]$ in each degree $2j$, for $0\le j\le 2$, and zero in odd degrees. Since these groups are free, the Künneth theorem gives the homology cross product basisThus the ranks in degrees $0,2,4,6,8$ are $1,2,3,2,1$, respectively, and every other integral homology group vanishes.

Let $h$ be the normalized hyperplane class, with $⟨h^2,[{\mathbb{CP}}^2]⟩=1$, and write $a=p_1^{\ast }h$, $b=p_2^{\ast }h$. The cohomology ring of complex projective space and the Künneth theorem give $H^{\ast }({\mathbb{CP}}^2\times {\mathbb{CP}}^2;\mathbb{Z})=\mathbb{Z}[a,b]/(a^3,b^3)$. If $d$ is the diagonal embedding, then $d^{\ast }a=d^{\ast }b=h$. Hence each of $a^2,ab,b^2$ evaluates as one on the diagonal's fundamental class. These three classes are dual to the three degree-four product generators, so the diagonal class of complex projective space givesEquivalently its Poincare dual is $a^2+ab+b^2$. The signs are positive with the complex and product orientations.

Use the standard CW complex structure on the real projective plane, with one cell $e_r$ for each $r=0,1,2$. The attaching map of the two-cell winds twice around the one-cell, so the cellular chain complex has $d{e}_2=2e_1$ and $d{e}_1=0$.

The product cells $e_{ij}=e_i\times e_j$, $0\le i,j\le 2$, give nine cells, in dimensions $0,1,2,3,4$ with counts $1,2,3,2,1$. The cellular chains of a product of finite CW complexes have boundary $d(e_i\times e_j)=d{e}_i\times e_j+(-1{)}^i{e}_i\times d{e}_j$. In the ordered basesthe nonzero matrices areIn particular $d_2{d}_3=d_3{d}_4=0$. The degree-one homology is ${\mathbb{Z}}^2/2{\mathbb{Z}}^2$. The degree-two cellular cycles are $\mathbb{Z}{e}_{11}$, and the cellular boundaries there are $2\mathbb{Z}{e}_{11}$. The degree-three cycles are $\mathbb{Z}(e_{21}+e_{12})$, whose double is the boundary of $e_{22}$. Finally $d_4$ is injective. This computes the integral homology of two real projective planes:The degree-three torsion is the Tor functor contribution in the Künneth theorem; it would be missed by retaining only tensor products of the two factors' homology groups.

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.

The space is the mapping torus of the antipodal map $A:S^2\to S^2$. Its mapping degree is $(-1{)}^3=-1$, so transporting a fiber orientation once around the base circle reverses it. $X$ is nonorientable.

On integral fiber homology, $A_{\ast }$ is the identity in degree zero and minus the identity in degree two. The Wang sequence, whose relevant map is $1-A_{\ast }$, therefore givesThe universal coefficient theorem for cohomology shifts the torsion contribution into degree three, because ${\mathrm{Ext}}_{\mathbb{Z}}(\mathbb{Z}/2,\mathbb{Z})=\mathbb{Z}/2$. HenceLet $t\in H^1(X;\mathbb{Z})$ be pulled back from the base circle, and $u$ the nonzero degree-three class. The square of the circle class vanishes, so $t^2=0$. All other products of positive-degree classes land above dimension three and vanish. This gives the cohomology ring of the antipodal two-sphere mapping torusModulo two, $A^{\ast }$ is the identity on fiber cohomology. The cohomological Wang sequence shows that restriction $H^2(X;{\mathbb{F}}_2)\to H^2(S^2;{\mathbb{F}}_2)$ is onto; choose $y$ mapping to its generator. The classes $1,y$ restrict to a fiber basis, so the Leray-Hirsch theorem makes the total cohomology free over $H^{\ast }(S^1;{\mathbb{F}}_2)$ on those two classes. Write $\bar{t}$ for the mod-two base class. The basis is $1,\bar{t},y,\bar{t}y$, and in particular $\bar{t}y\ne 0$. Since ${\bar{t}}^2=0$ by pullback and $y^2=0$ by dimension, the mod-two cohomology ring of the antipodal two-sphere mapping torus isThe argument of Question 3 works over any field, so its use here over ${\mathbb{F}}_2$ is justified. The ring agrees with that of $S^1\times S^2$; a cohomology ring alone does not determine orientability.

The surface group of a closed oriented genus-two surface $Y$ has presentationSend $a_1\mapsto (12)$, $a_2\mapsto (123)$ and $b_1,b_2\mapsto 1$. The relation holds and the images generate $S_3$, giving a surjective group homomorphism. Its normal kernel defines a connected six-sheeted regular covering space $\tilde{Y}\to Y$, with deck transformation group $S_3$.

The covering surface is closed and inherits an orientation. Multiplicativity of the Euler characteristic gives $\chi (\tilde{Y})=6\chi (Y)=6(-2)=-12$, so $2-2g=-12$ and $\boxed{g=7}$. A nonidentity deck transformation of a connected cover fixes no point: if it fixed one, uniqueness of path lifts would force it to be the identity everywhere. Thus this supplies the requested free $S_3$ action. It is a case of finite groups act freely on suitable closed orientable surfaces.

For an $n$-dimensional singular simplex $\sigma$ and a degree-$p$ singular cochain $\alpha$, use the front-face/back-face conventionand set it to zero for $p>n$. Extend linearly over $R$. Passing to homology classes of cycles and cohomology classes of cocycles defines the cap product $H_n(X;R)\times H^p(X;R)\to H_{n-p}(X;R)$.

For a continuous map $f$, the cochain pullback satisfies $(f^{\ast }\alpha )(\tau )=\alpha (f\circ \tau )$, and the induced chain map sends $\sigma$ to $f\circ \sigma$. Since composition commutes with restrictions to faces,The identity holds on chains, so it descends directly to classes and proves the naturality of the cap product: $f_{\ast }(x⌢f^{\ast }\alpha )=f_{\ast }(x)⌢\alpha$. No additional well-definedness verification is required here.