Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 114 (Algebraic Topology)

Choose the intersection point $p=[({\theta }_0,{\theta }_1)]$ as the zero-cell. Let $a$ be the horizontal one-cell $S^1\times \{{\theta }_1\}$ and $b$ the vertical one-cell $\{{\theta }_0\}\times S^1$. The usual CW complex structure on the torus has one two-cell $c$ attached by the word $ab{a}^{-1}{b}^{-1}$, and the extra disk gives a two-cell $d$ attached along $b$ with degree one. Collapsing $a$ leaves one zero-cell $p$, one one-cell $b$, and the two two-cells $c,d$. The cellular boundary formula givesTherefore

The Excision theorem says that if $Z\subseteq A\subseteq X$ and $\overline{Z}\subseteq \mathrm{int}A$, then inclusion inducesIt lets us compute local homology in arbitrarily small neighborhoods.

Let $C\subseteq X$ be the image of $\{{\theta }_0\}\times S^1$. At a point of $X\setminus C$, a neighborhood is a disk, whose link is a circle. At a point of $C\setminus \{p\}$, three half-disks meet along their diameters, and the link is a Theta graph, with first homology ${\mathbb{Z}}^2$. At $p$, the two folds created by collapsing the horizontal circle give a dumbbell graph as link: two circles joined by an interval. Its first homology is again ${\mathbb{Z}}^2$. The local homology from a link therefore givesBecause local homology is invariant under a homeomorphism, every self-homeomorphism of $X$ preserves the rank-two locus $C$.

It must also preserve $p$. Indeed, $p$ is the unique point of $C$ whose sufficiently small local link is a dumbbell graph; every other point of $C$ has a theta-graph link. The connecting edge of a dumbbell graph is a bridge in a graph, whereas no edge of a theta graph is a bridge, so these local topological types are distinct.

Writing the suspension of a topological space as two cones and applying the Mayer-Vietoris theorem gives the reduced homology of a suspensionThus $H_0(\Sigma X)\cong \mathbb{Z}$ for nonempty $X$, $H_1(\Sigma X)\cong {\tilde{H}}_0(X)$, and the same shift holds in every higher degree. For a space of finite CW type, the reduced Euler characteristic changes sign, soSince $\chi ({\mathbb{CP}}^2)=3$,But the Euler characteristic of a product satisfies $\chi (A\times A)=\chi (A{)}^2$, a nonnegative perfect square. Neither $3$ nor $-1$ is such a square, so no ${\Sigma }^j{\mathbb{CP}}^2$ is homotopy equivalent to $A\times A$.

A homeomorphism $f:{\mathbb{R}}^n\to {\mathbb{R}}^n$ is a proper map, so it extends over the one-point compactification to a homeomorphism $f^{+}:S^n\to S^n$ fixing infinity. Defineusing the degree of a continuous mapping. Since $f^{+}$ acts invertibly on $H_n(S^n;\mathbb{Z})\cong \mathbb{Z}$, its degree is $\pm 1$. Functoriality of induced homology maps givesFor $A\in \mathrm{GL}(n,\mathbb{R})$, path connectedness of each determinant-sign component reduces $A$ to the identity when $\det A>0$ and to one coordinate reflection when $\det A<0$. Hence

Suppose $h:Y\times Y\to {\mathbb{R}}^{2n+1}$ were a homeomorphism, and put $d=2n+1$. Then $Y^4\cong {\mathbb{R}}^d\times {\mathbb{R}}^d\cong {\mathbb{R}}^{2d}$. The square of the cyclic permutationinterchanges the two $Y^2$ factors. Under $h\times h$, it is conjugate to the linear factor swap $(u,v)\mapsto (v,u)$ on ${\mathbb{R}}^d\times {\mathbb{R}}^d$, whose determinant has sign $(-1{)}^{d^2}=-1$. Thus $\mathrm{deg}({\sigma }^2)=-1$. On the other hand, multiplicativity gives $\mathrm{deg}({\sigma }^2)=(\mathrm{deg}\sigma {)}^2=1$, a contradiction. Therefore no such $Y$ exists.

The short exact coefficient sequenceinduces the integral Bockstein homomorphismIf $\rho :H^{i+1}(X;\mathbb{Z})\to H^{i+1}(X;\mathbb{Z}/m)$ is reduction modulo $m$, thenEquivalently, if an integral cochain $a$ lifts a modulo-$m$ cocycle and $\delta a=mb$, then $\tilde{\beta }[a]=[b]$ and $\beta [a]=[b\mathrm{mod}m]$.

For integral lifts $a,c$ of classes $x,y$, the cup product coboundary formula isDividing by $m$ and reducing modulo $m$ proves the Bockstein derivation rule

Now assume that the stated closed five-manifold $M$ exists. Its top integral cohomology makes it connected and orientable. The long exact sequence from a coefficient sequence for multiplication by $p$ shows thatand that reduction $H^3(M;\mathbb{Z})\to H^3(M;\mathbb{Z}/p)$ is an isomorphism. HenceThe same coefficient sequence gives $H^4(M;\mathbb{Z}/p)=0$.

Choose $0\ne x\in H^2(M;\mathbb{Z}/p)$ and put $y=\beta x\ne 0$. By Poincare duality over $\mathbb{Z}/p$, the pairingis nondegenerate. Both factors are one-dimensional, so $x⌣y\ne 0$. But $x⌣x=0$ because $H^4(M;\mathbb{Z}/p)=0$, while the derivation rule and graded commutativity of the cup product giveThis contradicts $p>2$. Therefore no such manifold exists.

For an oriented real rank-$r$ vector bundle $\pi :E\to X$, let $u_E\in H^r(D(E),S(E);\mathbb{Z})$ be its Thom class. If $s:X\to D(E)$ is the zero section, the Euler class of a vector bundle is

Let $u$ generate $H^{2k}(S^{2k};\mathbb{Z})$. The Poincaré-Hopf theorem givesFor every integer $m$, choose a map $f_m:S^{2k}\to S^{2k}$ of degree $m$. Naturality of the Euler class givesThus every even class belongs to ${\mathcal{E}}_{2k}(S^{2k})$.

More generally, every class $a\in H^{2k}(M;\mathbb{Z})$ on a closed $2k$-manifold is $f^{\ast }u$ for some map $f:M\to S^{2k}$, by the realization of top-dimensional cohomology by a sphere map. Pulling back $T{S}^{2k}$ givesso $2H^{2k}(M;\mathbb{Z})\subseteq {\mathcal{E}}_{2k}(M)$. The inclusion can be strict: every complex line bundle on $S^2$ has an underlying oriented real two-plane bundle, and these bundles realize every integral Euler class. Hence

The odd-dimensional analogue fails. The Euler class of an oriented odd-rank vector bundle is two-torsion, whereas $H^{2k+1}(S^{2k+1};\mathbb{Z})\cong \mathbb{Z}$ is torsion-free. Thus every oriented rank-$(2k+1)$ bundle on $S^{2k+1}$ has zero Euler class, and the nonzero subgroup $2\mathbb{Z}$ cannot lie in ${\mathcal{E}}_{2k+1}(S^{2k+1})$.

Finally, if $E$ and $F$ are oriented bundles of ranks $k$ and $l$, orient $E\oplus F$ by the ordered sum. The Whitney product formula for Euler classes statesTherefore cup product restricts to the asserted map ${\mathcal{E}}_k(X)\otimes {\mathcal{E}}_l(X)\to {\mathcal{E}}_{k+l}(X)$.

It need not be injective. For $X=S^2$ and $k=l=2$, one has ${\mathcal{E}}_2(S^2)=\mathbb{Z}$, so the source contains $\mathbb{Z}\otimes \mathbb{Z}\cong \mathbb{Z}$, but $H^4(S^2)=0$ and the map is zero. It need not be surjective either. For $X=S^2$ and $k=l=1$, every oriented real line bundle is trivial, so both degree-one Euler-class sets vanish, while ${\mathcal{E}}_2(S^2)=\mathbb{Z}$.