Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 114 (Algebraic Topology)

The standard CW complex structure on Real projective space ${\mathbb{RP}}^n$ has one cell in each dimension $0,1,\dots ,n$. Its cellular chain complex has $C_q\cong \mathbb{Z}$ and differentialConsequentlyAfter tensoring the cellular complex with ${\mathbb{F}}_2=\mathbb{Z}/2$, every differential vanishes, so

The involution $(z_1,z_2)\mapsto (-z_1,{\overline{z}}_2)$ is free. Taking the first circle modulo the half-turn exhibits the quotient as the mapping torus of a reflection of $S^1$, hence as the Klein bottle. It has a finite CW structure with one zero-cell, two one-cells, and one two-cell. With suitable generators its integral cellular differential isThereforewhereas reduction modulo two kills the only nonzero boundary and gives

The quotient is the mapping torus ofA mapping torus of a cellular map has a finite CW structure. On integral homology, ${\tau }_{\ast }$ is $1$ on $H_0(T^2)$ and $H_2(T^2)$ and is $-I$ on $H_1(T^2)\cong {\mathbb{Z}}^2$. The Wang sequence therefore givesOver ${\mathbb{F}}_2$, the map ${\tau }_{\ast }$ is the identity. The Wang sequence splits as vector spaces into one copy of $H_q(T^2;{\mathbb{F}}_2)$ and one copy of $H_{q-1}(T^2;{\mathbb{F}}_2)$, yielding

The long exact sequence in homology of the triple $A\subseteq U\subseteq X$ containsBecause $A\hookrightarrow U$ is a homotopy equivalence, $H_{\ast }(U,A)=0$. Hence the middle map is an isomorphism in every degree:

The Excision theorem says that if $Z\subseteq A\subseteq X$ and $\overline{Z}\subseteq \mathrm{int}A$, then inclusion inducesA good pair $(X,A)$ has $A$ closed and a neighborhood $U$ that deformation retracts onto $A$. The collapsing a pair theorem states that the quotient map givesTo prove it, choose such a neighborhood $U$. The first result identifies $H_{\ast }(X,A)$ with $H_{\ast }(X,U)$. Excision identifies the latter with $H_{\ast }(X/A,U/A)$, while $U/A$ is contractible because the deformation retraction can be chosen relative to $A$ using the homotopy extension property. The long exact sequence of the pair $(X/A,U/A)$ then identifies this relative group with ${\tilde{H}}_{\ast }(X/A)$.

For $f:(D^n,\partial D^n)\to (D^n,\partial D^n)$, define $\mathrm{deg}f$ byin $H_n(D^n,\partial D^n;\mathbb{Z})\cong \mathbb{Z}$. In the long exact sequence of the pair, the connecting homomorphismis an isomorphism. Naturality shows that $f_{\ast }$ and $(f{|}_{\partial D^n}{)}_{\ast }$ multiply the corresponding generators by the same integer, so

Finally,Under this identification, the map induced by $g(x,t)=(f(x),t)$ is the suspension of the map induced by $f$. By degree under suspension,This argument proves the needed product assertion directly from the natural suspension isomorphism in reduced homology.

For a singular $(k+l)$-simplex $\sigma :[v_0,\dots ,v_{k+l}]\to X$, define the cochain-level cup product byIf $\alpha \in C^k(X,A;R)$ and $\sigma$ lies in $A$, its front $k$-face also lies in $A$, so the first factor vanishes. Thus $\alpha ⌣\beta \in C^{k+l}(X,A;R)$.

For the projections from $X\times Y$, the exterior product in cohomology isFor CW pairs over a field $R$, subject to the usual finite-type condition that makes the graded tensor product commute with the relevant direct products, the Künneth theorem says thatMore generally the conclusion holds over a principal ideal domain when one factor has degreewise finitely generated free cohomology. It fails over $\mathbb{Z}$ in general: $H^{\ast }({\mathbb{RP}}^2;\mathbb{Z})$ is $\mathbb{Z}$ in degree zero and $\mathbb{Z}/2$ in degree two, but the integral Künneth and universal coefficient theorem for cohomology calculations giveThe tensor product of the two cohomology groups has no degree-three term, so $\Phi$ is not surjective.

Use the homeomorphismwhich carries the diagonal $\Delta$ to $T^2\times \{0\}$. HenceThe punctured torus deformation retracts onto a wedge of two circles. Let $a,b$ be the degree-one generators from the first torus and $c,d$ those from the punctured torus. Since all cohomology groups are free, the Künneth theorem identifies the integral cohomology ring asThus $ab$, $ac$, $ad$, $bc$, and $bd$ form a basis in degree two; $abc$ and $abd$ form a basis in degree three; and all higher positive degrees vanish.

An $R$-orientation of a rank-$r$ vector bundle $E\to B$ is a coherent choice of generator infor every fiber. Equivalently, it is a Thom class $u_E\in H^r(D(E),S(E);R)$ restricting to those generators. The Euler class of a vector bundle iswhere $j:(D(E),\varnothing )\to (D(E),S(E))$ forgets the subspace and $s$ is the zero section.

The Thom isomorphism theorem isInsert these isomorphisms into the long exact cohomology sequence of $(D(E),S(E))$ and use the deformation retraction $D(E)\simeq B$. This gives the Gysin sequence of a sphere bundleTo verify the labelled maps, represent $a$ under the Thom isomorphism by ${\pi }^{\ast }a⌣u_E$. Its image in $H^{\ast }(D(E))$ pulls back along the zero section toThus multiplication by the Euler class is exactly the map from relative to absolute cohomology in the pair sequence.

Now let $M^m\subset S^n$ and let $V$ be its tubular neighborhood. The normal bundle has rank $r=n-m$. With ${\mathbb{F}}_2$ coefficients it is automatically oriented. Since $n>2m+1$, one has $r>m+1$, so its Euler class lies above the dimension of $M$ and vanishes. The Gysin sequence consequently splits into short exact sequences of vector spaces and givesOn the other hand, Alexander duality givesOver the field ${\mathbb{F}}_2$, the latter is naturally dual to ${\tilde{H}}^{n-q-1}(M;{\mathbb{F}}_2)$, which expresses the complement cohomology entirely in terms of that of $M$.

For unit vectors, the defining inequality for $Y$ is equivalent to $v\cdot w\ge 0$. WriteThen $t=(1-\Vert u{\Vert }^2{)}^{1/2}$, sois a homeomorphism $D(T{S}^{2n})\to Y$. Replacing $v$ by $-v$ gives the same description of $Y^{\prime }$. Their intersection isthe unit tangent bundle, equivalently a Stiefel manifold. Since $e(T{S}^{2n})$ evaluates to $\chi (S^{2n})=2$, its Gysin sequence gives the cohomology of the unit tangent bundle of an even-dimensional sphere:Every product of positive-degree classes vanishes.

Let $G$ consist of the eight signed permutations of the two factors:It is the group of signed permutation matrices in dimension two, hence isomorphic to the dihedral group $D_8$, and every element preserves the equation $v\cdot w=0$. If $a,b$ generate $H^{2n}(X;\mathbb{Z})$ from the two sphere factors, these maps act by the corresponding signed permutation matrix, because the antipodal map of the even-dimensional sphere has degree $-1$. The eight actions are distinct, so only the identity can be homotopic to $1_X$.

Every element of $G$ acts trivially on $H^{\ast }(Z;\mathbb{Z})$. The degree-$2n$ group is $\mathbb{Z}/2$, so its automorphism is forced to be the identity. On the top class, regard $Z$ as the Stiefel manifold of two-frames. The signed permutations are the right action of $O(2)$. The determinant-one component is connected, while a determinant-minus-one element is homotopic within that component to $(v,w)\mapsto (v,-w)$, the antipodal map on the $S^{2n-1}$ fibers. That map has degree $+1$, so the top class is also fixed.

The subgroup preserving $Y$ iswhere $A(v,w)=(-v,w)$, $B(v,w)=(v,-w)$, and $S(v,w)=(w,v)$. It consists exactly of the elements with ${\epsilon }_1{\epsilon }_2=1$. Under the deformation retraction $Y\simeq S^{2n}$ onto the diagonal, $1$ and $S$ act as the identity, while $AB$ and $ABS$ act as the antipodal map. They therefore act on $H^{2n}(Y;\mathbb{Z})$ by $+1,+1,-1,-1$, respectively.

The Thom isomorphism theorem identifies $H^{\ast }(Y,Z)$ with a degree-$2n$ shift of $H^{\ast }(S^{2n})$. On the Thom class in degree $2n$, the same four elements act by $+1,+1,-1,-1$: the factor swap reverses each normal vector, which preserves the orientation because the normal rank $2n$ is even, while the simultaneous antipodal map reverses the oriented tangent fiber. On the degree-$4n$ relative class every element acts by $+1$, since the base and fiber signs for the last two elements cancel.

When $n=1$, $Z=V_2({\mathbb{R}}^3)\cong SO(3)$. Every signed permutation of $(v,w)$ extends, after sending the third frame vector $v\times w$ to the required sign, to right multiplication by an element of $SO(3)$. Since $SO(3)$ is path-connected, every right translation is homotopic to the identity. Therefore