Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 114 (Algebraic Topology)

Write $S^n$ as the union of slightly enlarged northern and southern hemispheres $U$ and $V$. Both are contractible, while $U\cap V$ deformation retracts onto $S^{n-1}$. The reduced Mayer-Vietoris theorem therefore givesStarting from ${\tilde{H}}_0(S^0;\mathbb{Z})\cong \mathbb{Z}$ proves the homology of a sphere:This uses no cellular homology. A reflection of $S^n$ reverses its orientation and has degree of a continuous mapping $-1$, so it induces the identity on $H_0$, multiplication by $-1$ on $H_n$, and the unique map between zero groups in every other degree.

For a CW complex with skeleta $X^k$, its cellular chain complex isThe differential is the connecting map to $H_{k-1}(X^{k-1})$ followed by passage to $H_{k-1}(X^{k-1},X^{k-2})$. Equivalently, the coefficient of a $(k-1)$-cell in the boundary of a $k$-cell is the degree obtained from its attaching map after collapsing the complement of that lower cell. This is the cellular boundary formula.

The quotient $D^k/(x\sim -x\text{ on }{S}^{k-1})$ builds ${\mathbb{RP}}^k$ from ${\mathbb{RP}}^{k-1}$ by one $k$-cell, so ${\mathbb{RP}}^n$ has one cell in each dimension $0,\dots ,n$. The two lifts of the attaching map contribute with relative sign $(-1{)}^k$, and the cellular homology of real projective space has differentialConsequentlyWith ${\mathbb{F}}_2=\mathbb{Z}/2$ coefficients every differential vanishes, and hencewith zero homology outside that range.

Repeated Smith normal form puts a chain complex of finitely generated free abelian groups into its elementary decomposition of a finite free chain complex: one-term summands $\mathbb{Z}$ and two-term summands $0\to \mathbb{Z}\stackrel{m}{\to }\mathbb{Z}\to 0$. Applying ${\mathrm{Hom}}_{\mathbb{Z}}(-,\mathbb{Z})$ reverses a two-term summand but keeps the same multiplication by $m$. Reading its homology gives the universal coefficient theorem for cohomologysplit noncanonically. Thus if $H_n(C)\cong {\mathbb{Z}}^{b_n}\oplus T_n$ with $T_n$ finite, then

The universal coefficient theorem for homology with ${\mathbb{F}}_p$ coefficients givesIf the middle group vanishes for every prime $p$, so does the tensor term. Any nonzero free summand survives for every $p$, and any nonzero finite summand survives for a prime dividing its order. Since integral homology is finitely generated, it must vanish in every degree. This is detection of integral acyclicity modulo primes.

For the displayed complex, writeUsing $f_{\#}{d}_C=d_D{f}_{\#}$, one obtainsThus $M$ is the mapping cone of $f_{\#}$ in this sign convention. Its long exact sequence in homology shows that $f_{\ast }$ is an isomorphism exactly when $H_{\ast }(M)=0$. If $f_{\ast }$ is an isomorphism with ${\mathbb{F}}_p$ coefficients for every prime, then $M\otimes {\mathbb{F}}_p$ is acyclic for every $p$. Prime-field detection makes $M$ integrally acyclic, so the integral long exact sequence gives

An orientation of a vector bundle of real rank $d$ over $R$ is a coherent choice of generator of $H^d(E_b,E_b\setminus \{0\};R)$ in every fibre. For a complex bundle of rank $m$, a complex basis gives the real basis $(v_1,i{v}_1,\dots ,v_m,i{v}_m)$. A complex change of basis has positive real determinant $|{\det }_{\mathbb{C}}A{|}^2$, so these local choices agree and give the canonical orientation of a complex vector bundle over $\mathbb{Z}$, hence over every commutative ring $R$.

Let $D(E)$ and $S(E)$ be the disk and sphere bundles of an $R$-oriented rank-$d$ bundle over compact $B$. A Thom class $u_E\in H^d(D(E),S(E);R)$ restricts to the chosen generator on every fibre. The Thom isomorphism theorem states thatThe Euler class of a vector bundle is $e(E)=s^{\ast }{u}_E\in H^d(B;R)$ for the zero section $s$. Substituting the Thom isomorphism into the long exact sequence of the pair $(D(E),S(E))$ gives the Gysin sequence of a sphere bundle

Apply this to the Hopf fibration and induct on $n$. If $x=c_1(\mathcal{O}(1))$ has degree two, the result is the cohomology ring of complex projective space

The Künneth theorem over a principal ideal domain gives a natural short exact sequence with tensor-product and Tor terms; the sequence splits, though not naturally. Here all groups are free, so the Tor term vanishes andIf a graded-ring automorphism sends $x$ to $ax+by$, then $(ax+by{)}^{n+1}=0$. The coefficients of the nonzero mixed monomials force $ab=0$; the same applies to the image of $y$. Invertibility then forces a signed permutation matrix on the basis $x,y$. Conversely, complex conjugation on either factor changes the sign of its degree-two generator, and swapping the factors exchanges $x$ and $y$. Thus the realizable group isthe group of all signed $2$ by $2$ permutation matrices, as described by the cohomology automorphisms of a product of two complex projective spaces.

A local orientation of a manifold at $x\in M$ is a generator ofAn $R$-orientation is a locally coherent choice of such generators. An R-fundamental class is a class $[M{]}_R\in H_d(M;R)$ whose image in every one of these local homology groups is a generator. Those images vary coherently under the restriction maps between small coordinate balls, so an $R$-fundamental class determines an $R$-orientation.

For a closed $R$-oriented $d$-manifold, Poincare duality says that cap product with its fundamental class is an isomorphismfor every $k$.

Choose a generator $[S^d]\in H_d(S^d;\mathbb{Z})\cong \mathbb{Z}$ from the homology of a sphere. For every $x\in S^d$, the long exact sequence of $(S^d,S^d\setminus \{x\})$, together with the contractibility of $S^d\setminus \{x\}\cong {\mathbb{R}}^d$, shows thatThe chosen generator is therefore a local generator at every point and is a $\mathbb{Z}$-fundamental class.

Put $N=S^d/X$ and let $p=q(X)$. The restrictionis a homeomorphism. At every point of $N\setminus \{p\}$, transport the local orientation of $S^d$ through this homeomorphism. The localization of the global class $q_{\ast }[S^d]$ is therefore a generator at one, and hence every, point of that connected open set. The localizations of a global homology class form a section of the orientation local system; because the manifold $N$ is connected, this generator extends across $p$. Thus $q_{\ast }[S^d]$ is a fundamental class for an orientation of $N$.

Choose $[N]=q_{\ast }[S^d]$. By the degree of a map between oriented manifolds,so $\boxed{\mathrm{deg}q=1}$.

Part 1 makes $q:S^d\to N$ a degree-one map of closed oriented $d$-manifolds. The cohomological injectivity of a degree-one map embeds $H^{\ast }(N;\mathbb{Z})$ into $H^{\ast }(S^d;\mathbb{Z})$. Hence $N$ has no cohomology in degrees $0<i<d$; Poincare duality and the fundamental class giveThus $N$ is an integral homology sphere.

The long exact sequence of the pair $(N,N\setminus \{p\})$ has local relative homology $\mathbb{Z}$ only in degree $d$, and the map $H_d(N)\to H_d(N,N\setminus \{p\})$ is an isomorphism because it sends the fundamental class to the local orientation. Exactness now givesfor every $i$. Since $N\setminus \{p\}\cong S^d\setminus X$, the latter has the integral homology of a point.