Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 114 (Algebraic Topology)

Because every $C_i$ is free abelian, reduction modulo $n$ gives a short exact sequence of chain complexesIts long exact sequence in homology isIf $\bar{z}$ is a cycle modulo $n$, choose a lift $z\in C_i$. Then $dz=nw$ for some $w\in C_{i-1}$, and the connecting homomorphism is $\hat{\beta }[\bar{z}]=[w]$.

The coefficient sequencesimilarly givesAt chain level, if $z$ lifts a mod-$n$ cycle and $dz=nw$ modulo $n^2$, then the Bockstein homomorphism is $\beta [\bar{z}]=[\bar{w}]$.

There is a morphism from the first short exact sequence to the second whose three vertical maps are reduction modulo $n$, reduction modulo $n^2$, and the identity on $C\otimes \mathbb{Z}/n$. Naturality of connecting homomorphisms givesExactness of the first long exact sequence says $\hat{\beta }{\rho }_{\ast }=0$, and hence

For the standard cellular chain complex of Real projective space ${\mathbb{RP}}^3$, there is one copy of $\mathbb{Z}$ in degrees $0,1,2,3$, with $d_2=2$ and $d_1=d_3=0$. Modulo two all cellular differentials vanish, while the Bockstein is the identity from degree two to degree one and zero elsewhere. Therefore

Finally, Smith normal form decomposes a bounded chain complex of finitely generated free abelian groups, up to chain isomorphism and contractible summands, into one-term complexes $\mathbb{Z}[i]$ and two-term complexesin degrees $i+1$ and $i$. The former represents a free homology summand, and the latter represents $\mathbb{Z}/m$ in degree $i$. The two requested conclusions now follow from the next two parts.

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A one-term summand $\mathbb{Z}[i]$ becomes a one-term ${\mathbb{F}}_p$ complex after reduction modulo $p$. Its Bockstein homomorphism vanishes, so it contributes one $\mathbb{Z}/p$ summand to Bockstein homology in degree $i$.

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For the elementary complex with differential $m$, reduction modulo $p$ is acyclic if $p\nmid m$. If $p\mid m$, its mod-$p$ homology has one ${\mathbb{F}}_p$ in each of degrees $i+1$ and $i$, and the Bockstein from the upper group to the lower group is multiplication by $m/p$ modulo $p$. It is an isomorphism when $p^2\nmid m$, so its Bockstein homology vanishes. It is zero when $p^2\mid m$, so both classes survive and contribute $\mathbb{Z}/p$ to $\beta H_{i+1}$ and to $\beta H_i$.

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Give the closed oriented genus-$g$ surface ${\Sigma }_g$ its usual CW structure with one vertex, $2g$ one-cells $a_1,b_1,\dots ,a_g,b_g$, and one two-cell attached along ${\prod }_i[a_i,b_i]$. The cellular boundary maps vanish after abelianization, soChoose degree-one classes ${\alpha }_i,{\beta }_i$ dual to $a_i,b_i$ and orient ${\Sigma }_g$ by $\omega \in H^2$. The intersection pairing, equivalently the cellular diagonal approximation, gives the cohomology ring of a closed oriented surface:and all ${\alpha }_i⌣{\alpha }_j$ and ${\beta }_i⌣{\beta }_j$ vanish.

For the space $X$, use the genus-two CW structure and attach an additional two-cell $e_v^2$ along $[b_1,b_2]$. Both two-cell attaching words have zero exponent sum in every one-cell, so the cellular boundary $C_2(X)\to C_1(X)$ is zero. HenceLet $u,v$ be the degree-two classes dual respectively to the original surface cell and the new cell. The original relator $[a_1,b_1][a_2,b_2]$ and the new relator $[b_1,b_2]$ givetogether with the products forced by graded commutativity. Every other product of degree-one basis classes is zero, and every product of total degree greater than two is zero. These relations completely determine the cohomology ring of $X$.

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A rank-$d$ vector bundle $\pi :E\to B$ is $R$-oriented when its fiber groups $H^d(E_b,E_b\setminus \{0\};R)$ admit a locally coherent choice of generator. Equivalently, it has a Thom class $u_E\in H^d(D(E),S(E);R)$ restricting to that generator on every fiber. The Thom isomorphism theorem states thatThe Euler class of a vector bundle is $e(E)=s^{\ast }{u}_E\in H^d(B;R)$, where $s$ is the zero section. The Gysin sequence of a sphere bundle is

Over ${\mathbb{F}}_2=\mathbb{Z}/2$, the required rings areas in the cohomology ring of complex projective space and the mod-two cohomology ring of real projective space. By the Künneth theorem, the base has ring ${\mathbb{F}}_2[x,y]/(x^{n+1},y^{m+1})$.

Let $L$ be the underlying real plane bundle of the complex tautological bundle on ${\mathbb{CP}}^n$, and let $\lambda$ be the real tautological line bundle on ${\mathbb{RP}}^m$. Under the splitting principle, write the formal Stiefel-Whitney roots of $L$ as $a,b$, so $a+b=w_1(L)=0$ and $ab=w_2(L)=x$. Tensoring with $\lambda$ adds $y=w_1(\lambda )$ to each root. Thus the mod-two Euler class of $L{\otimes }_{\mathbb{R}}\lambda$ is

For $n=m=2$, put $B={\mathbb{CP}}^2\times {\mathbb{RP}}^2$. Multiplication by $e=x+y^2$ onhas ranks $1,1,2,1,1,0,0$ from degrees $0$ through $6$. More explicitly, it is injective through degree three; in degree four its kernel is generated by $x^2+x{y}^2$, and all of degrees five and six lie in its kernel. The Gysin sequence therefore yieldsfor the unit sphere bundle $X=S(L\otimes \lambda )$.

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The Lefschetz fixed-point theorem says that a continuous self-map $f$ of a compact triangulable space has a fixed point whenever its Lefschetz numberis nonzero. In the smooth nondegenerate case, the Lefschetz-Hopf fixed-point theorem expresses this number as the sum of the local fixed-point indices.

Identify $H_1({\mathbb{R}}^2/{\mathbb{Z}}^2;\mathbb{Z})$ with ${\mathbb{Z}}^2$ by sending $v\in {\mathbb{Z}}^2$ to the class of the loop $t\mapsto tv\mathrm{mod}{\mathbb{Z}}^2$. The image of this loop under $f$ lifts from $0$ to the path $t\mapsto \tilde{f}(tv)$, whose endpoint is $\tilde{f}(v)=A(v)$. Its homology class is therefore $A(v)$. Thus $A=f_{\ast }$ under this identification, and in particular $A$ is a homomorphism represented by an integer matrix.

The expansion inequality implies, after taking a derivative, thatfor every tangent vector. The supplied linear-algebra fact shows that every eigenvalue of $D{\tilde{f}}_x$ has modulus at least $\mu >1$, so $1$ is not an eigenvalue. Every fixed point of $f$ is consequently nondegenerate and contributes local index $+1$ or $-1$.

On the torus, the induced maps on $H_0,H_1,H_2$ have traces $1,\mathrm{tr}A,\det A$. HenceThe original expansion inequality applied to lattice vectors gives $|Av|\ge \mu |v|$ for $v\in {\mathbb{Z}}^2$, and homogeneity and density extend it to all $v\in {\mathbb{R}}^2$. If ${\lambda }_1,{\lambda }_2$ are the possibly complex eigenvalues of $A$, then $|{\lambda }_j|\ge \mu$, soSince every fixed point contributes an index of absolute value one, the Lefschetz-Hopf formula gives

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