Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 114 (Algebraic Topology)

Let $x,y\in H^n(S^n\times S^n;\mathbb{Z})$ be the pullbacks of the two orientation classes. The cohomology ring of a product of two spheres hasand $xy$ generates $H^{2n}\cong \mathbb{Z}$. WriteThe matrix $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ lies in $G{L}_2(\mathbb{Z})$ because a homeomorphism induces a cohomology ring automorphism.

If $n$ is even, graded commutativity givesThus $ab=cd=0$. Invertibility forces $A$ to be diagonal or anti-diagonal, and its two nonzero entries must each be $\pm 1$. Hence there are exactly eight possible actions: the signed permutation matrices.

Evenness is necessary. For $n=1$, the product is the torus, and every $A\in G{L}_2(\mathbb{Z})$ is induced by an integral linear automorphism of the torus. For example, the infinitely many matrices $\left(\begin{array}{cc}1& m\\ 0& 1\end{array}\right)$ give distinct actions on $H^1$.

The cellular chain complex for either space has one cell in dimensions $0,2,3,4$, with the only nonzero differential equal to multiplication by $p$ from degree three to degree two. The universal coefficient theorem for cohomology therefore gives, for both $X$ and $Y$,Every product of positive-degree classes vanishes for dimensional reasons, so

Their modulo-$p$ cohomology rings distinguish them. Let $u\in H^2(X;{\mathbb{F}}_p)$ be the class restricting to the standard generator on ${\mathbb{CP}}^2$. The attaching map has degree $p$, so its cellular coboundary vanishes modulo $p$; the classes in degrees two and four restrict isomorphically to those of ${\mathbb{CP}}^2$. Hence $u^2\ne 0$ in $H^4(X;{\mathbb{F}}_p)$. In $Y=M(\mathbb{Z}/p,2)\vee S^4$, the degree-two class comes from the three-dimensional Moore space, so its square is zero; products between distinct wedge summands also vanish. The mod-p cup-square obstruction to a homotopy equivalence now proves

The standard cellular chain complex of the Klein bottle givesThe universal coefficient theorem for cohomology therefore yields $H^0=\mathbb{Z}$, $H^1=\mathbb{Z}$, and $H^2=\mathbb{Z}/2$. The degree-one generator is pulled back from the base circle in the circle-bundle description of $K$, so its square is zero. Thus every positive-degree product vanishes. The same calculation for ${\mathbb{RP}}^2\vee S^1$ gives the integral cohomology ring of the wedge of the real projective plane and a circle, and hence

There is no map $Y\to K$ inducing this isomorphism. The Klein bottle is an aspherical space, and its fundamental group is torsion-free. Any map ${\mathbb{RP}}^2\to K$ therefore induces the trivial homomorphism $\mathbb{Z}/2\to {\pi }_1(K)$ and is null-homotopic. Its pullback on $H^2$ is zero, while the circle summand has no degree-two cohomology. Consequently every map $Y\to K$ is zero on $H^2(K;\mathbb{Z})$ and cannot be a cohomology isomorphism.

A direct system of abelian groups over the directed poset $A$ consists of groups $G_a$ and maps ${\varphi }_{ab}:G_a\to G_b$ for $a\le b$, satisfying ${\varphi }_{aa}=1$ and ${\varphi }_{bc}{\varphi }_{ab}={\varphi }_{ac}$. Its direct limit is the quotient of ${⨁}_a{G}_a$ by the relations $g_a\sim {\varphi }_{ab}(g_a)$.

For the displayed sequence, put $P_0=1$ and $P_n=a_0{a}_1\cdots a_{n-1}$. Map the copy of $\mathbb{Z}$ at stage $n$ to $\mathbb{Q}$ byThis is compatible with the next transition because $a_{n}m/P_{n+1}=m/P_n$. The universal property of the direct limit therefore identifies it withIn reduced form, these are exactly the rationals whose denominator divides one of the finite products $P_n$. This is the sequential direct limit of multiplication maps on the integers.

Every singular simplex has compact image, and a singular chain is a finite sum of simplices. The image of any chain is therefore compact and lies in some $X_a$. Directedness puts any finite collection of chains into one common $X_b$, soBecause filtered colimits of abelian groups are exact, kernels and images commute with this colimit. Taking homology gives the homology of a directed union:

For an open $U\subseteq {\mathbb{R}}^N$, use the directed family of finite unions of closed rational cubes contained in $U$. Every compact subset of $U$ lies in one such finite polyhedron, and each polyhedron has finitely generated cellular homology. There are only countably many of them, so their direct limit is countable. Thus every $H_i(U;\mathbb{Z})$ is countable.

Cohomology behaves differently because $\mathrm{Hom}$ turns a direct sum into a direct product. The connected open sethas one independent loop around each puncture, so $H_1(U;\mathbb{Z})\cong {⨁}_{n\ge 1}\mathbb{Z}$. Since $H_0(U;\mathbb{Z})=\mathbb{Z}$, the universal coefficient theorem for cohomology giveswhich is uncountable. This is the first cohomology of the countably punctured plane.

List the prime numbers without repetition as $p_0,p_1,p_2,\dots$. Choose mapsand let $T$ be their mapping telescope. A finite initial telescope deformation retracts onto its last sphere. The inclusions of successive finite telescopes induce multiplication by $p_i$ on degree-$n$ reduced homology. The homology of a directed union and the sequential direct limit of multiplication maps on the integers therefore giveEvery finite product $p_0\cdots p_r$ is square-free, and every square-free denominator divides one such product. Henceas in the mapping-telescope realization of the rational group with square-free denominators.

An $R$-orientation of a rank-$d$ real vector bundle $E\to B$ is a locally coherent choice of generator of $H^d(E_b,E_b\setminus \{0\};R)$ in every fibre. Equivalently, it is represented by a Thom class $u_E\in H^d(D(E),S(E);R)$ restricting to the chosen generator on each fibre pair. If $e(E)\in H^d(B;R)$ is the Euler class, the Gysin sequence of its unit sphere bundle contains

The diagonal quotient defining $L(p)$ is the three-dimensional lens space as a circle bundle $S^1\to L(p)\to S^2$ with Euler class $p$ times a generator. With integral coefficients, the only nontrivial Euler-class map is multiplication by $p:H^0(S^2;\mathbb{Z})\to H^2(S^2;\mathbb{Z})$. Exactness givesModulo $p$, the Euler class vanishes, so the same Gysin sequence giveswith all other groups zero.

For the coefficient sequence $0\to \mathbb{Z}\stackrel{p}{\to }\mathbb{Z}\to {\mathbb{F}}_p\to 0$, the long exact sequence from a coefficient sequence containsThe last map is zero and both middle groups have order $p$, so $\delta$ is an isomorphism. Reduction $H^2(L(p);\mathbb{Z})\to H^2(L(p);{\mathbb{F}}_p)$ is likewise an isomorphism. Their composite is the Bockstein isomorphism for a three-dimensional lens space

If $a^{\prime }=na$ is another generator, linearity of the Bockstein homomorphism and bilinearity of the cup product giveMoreover $t(a)\ne 0$: $\beta (a)$ is nonzero, and the Poincare duality pairing $H^1\times H^2\to H^3\cong {\mathbb{F}}_p$ is nondegenerate. If $h:L(p)\to L(p)$ is an orientation-reversing homotopy equivalence and $h^{\ast }a=na$, naturality givesCancelling the nonzero $t(a)$ yields $n^2\equiv -1(\mathrm{mod}p)$. Thus $-1$ must be a quadratic residue modulo $p$.

Let $m=\dim M$. Choose a homogeneous basis $e_{q,i}$ of $H^q(M;\mathbb{Q})$ and its Poincare dual basis $e_{q,i}^{\ast }\in H^{m-q}(M;\mathbb{Q})$, normalized byWith the product orientation, the cohomology class of the diagonal isIndeed, multiplying this class by $\alpha \times \gamma$ and evaluating on $[M\times M]$ gives $⟨\alpha ⌣\gamma ,[M]⟩$, which characterizes the Poincare dual of $\Delta$.

Pulling back along the graph map $(1,f):M\to M\times M$ and evaluating gives the graph-diagonal formula for the Lefschetz number:If $f$ has no fixed point, its graph is disjoint from $\Delta$. Represent ${\epsilon }_{\Delta }$ with support in a tubular neighbourhood disjoint from the graph; its pullback is zero, so $L(f)=0$. Contrapositively, $L(f)\ne 0$ implies that $f$ has a fixed point, the Lefschetz fixed-point theorem.

Now let $M$ be three disjoint circles. A homeomorphism permutes their three components. Its action on $H^1(M;\mathbb{Z})\cong {\mathbb{Z}}^3$ is a signed permutation matrix. A component fixed setwise contributes to the trace by the degree of the corresponding circle homeomorphism. An orientation-reversing circle homeomorphism has a fixed point, so fixed-point-freeness forces that degree to be $+1$. Nonfixed components contribute zero. The trace is therefore the number of fixed points of a permutation of three objects, and

For a compact manifold $N$ with boundary, let $D(N)=N_{+}{\cup }_{\partial N}{N}_{-}$ be its double and define $F$ by applying $f$ on both copies. The Mayer–Vietoris sequence for this decomposition is natural under $F$. Alternating traces in a finite-dimensional exact sequence sum to zero, so the two copies of $N$ contribute twice and their intersection contributes with the opposite sign:This is the Lefschetz number of a doubled map.

Finally let $N$ be a pair of pants. If $f$ is fixed-point-free, so are its double $F$ and its boundary restriction. The Lefschetz fixed-point theorem and the displayed identity give $L(F)=L(f{|}_{\partial N})=0$, hence $L(f)=0$. Since $N$ is connected and $H^i(N;\mathbb{Q})$ is nonzero only for $i=0,1$,Suppose the boundary permutation $\sigma$ had a fixed component. The restriction there is a fixed-point-free circle homeomorphism and thus has degree $+1$, forcing $f$ to preserve the surface orientation. The homology action of a pair-of-pants homeomorphism would then have trace $\#\mathrm{Fix}(\sigma )-1$, which is $2$ for the identity permutation and $0$ for a transposition. Neither is $1$. Therefore $\sigma$ has no fixed component; a permutation of three objects with no fixed point is a three-cycle. Thus