Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 114 (Algebraic Topology)

The quotient is obtained from its image $Z\cong S^1$ by attaching the interior of $D^2$. The boundary map winds $n$ times around $Z$, so this is the cyclic Moore space in dimension one $M(\mathbb{Z}/n,1)$. Its cellular chain complex isTherefore

Suppose that $f$ is not surjective, and choose $y\notin f(Y)$. If $y$ lies in the open two-cell, then $Y\setminus \{y\}$ deformation retracts onto the target circle of the degree-$n$ attaching map, so its first integral homology is $\mathbb{Z}$. If $y\in Z$, the punctured space deformation retracts onto a finite graph and again has free abelian first homology. The factorizationtherefore makes $f_{\ast }:H_1(Y)\to H_1(Y)$ factor through a free abelian group. Every homomorphism from the finite group $H_1(Y)\cong \mathbb{Z}/n$ to a free abelian group is zero, contradicting the assumed surjectivity of $f_{\ast }$.

The converse is false. The radial coordinate descends to a surjection $r:Y\to [0,1]$. The finite CW complex $Y$ is a Peano continuum, so the Hahn-Mazurkiewicz theorem supplies a continuous surjection $g:[0,1]\to Y$. Then $g\circ r:Y\to Y$ is surjective, but its induced map on $H_1$ is zero because it factors through the contractible interval.

At a point of the open two-cell, the local link is a circle, so local homology in degree two has rank one. At a point of $Z$, a neighborhood consists of $n$ half-discs sharing their boundary interval; its link is the Theta graph ${\Theta }_n$. The local homology from a link calculation givesWhen $n>2$, this local rank characterizes the points of $Z$. Since local homology is invariant under a homeomorphism, every homeomorphism satisfies $f(Z)=Z$, and in particular $f(Z)\subset Z$.

For $n=2$, the two pages form an ordinary disc neighborhood, and $Y$ is ${\mathbb{RP}}^2$. The distinguished circle $Z$ is a projective line, but a projective linear homeomorphism can carry it to a different projective line. Thus the conclusion does not hold.

Write ${\mathbb{F}}_2=\mathbb{Z}/2$. A double covering has a deck transformation $\tau$ that exchanges the two points in every fibre. Every singular simplex $\sigma :{\Delta }^q\to Y$ has exactly two lifts $\tilde{\sigma }$ and $\tau \tilde{\sigma }$. Define the mod-two transfer chain map of a double covering byand let $f_{\#}$ send a simplex of $X$ to its composite with $f$. Uniqueness of lifted faces shows that $T$ commutes with the boundary operator, so both maps are chain maps.

For each base simplex, its two lifts span a copy of ${\mathbb{F}}_2^2$. On this summand, $T$ is $a\mapsto (a,a)$ and $f_{\#}$ is $(a,b)\mapsto a+b$. Hence the sequence is exact in every degree:

Assume for contradiction that the involution $\varphi$ is fixed-point-free. The finite group action is then a covering space action, so the quotient mapis a double covering and $Y$ is an $n$-manifold.

Apply the long exact sequence in homology to the transfer chain map of a double covering. Since ${\mathbb{R}}^n$ is contractible, its positive-dimensional mod-two homology vanishes. The degree-zero portion, together with the fact that $q_{\ast }:H_0({\mathbb{R}}^n;{\mathbb{F}}_2)\to H_0(Y;{\mathbb{F}}_2)$ is an isomorphism, givesIn every higher degree the same exact sequence givesThus $H_i(Y;{\mathbb{F}}_2)\cong {\mathbb{F}}_2$ for every $i\ge 0$. This contradicts homology above the dimension of a manifold, which gives $H_i(Y;{\mathbb{F}}_2)=0$ for $i>n$. Therefore $\varphi$ has a fixed point.

A closed manifold of odd dimension has Euler characteristic zero by the Euler characteristic of an odd-dimensional closed manifold. Decomposing $M$ into $P$ and the removed closed ball along $S^{2k}$ givesso $\chi (P)=1$.

If the antipodal boundary map extended to a fixed-point-free involution of $P$, the quotient map $P\to P/⟨\tau ⟩$ would be a double covering. The formula for Euler characteristic under a finite covering would implywhich is impossible. Hence no such extension exists.

The standard CW complex structure on ${\mathbb{CP}}^n$ has one cell in every even dimension from $0$ to $2n$, and ${\mathbb{CP}}^k$ is its $2k$-skeleton. Collapsing that subcomplex leaves one zero-cell and one cell in each dimension $2i$ for $k+1\le i\le n$. All cellular boundaries vanish, so

Let $x\in H^2({\mathbb{CP}}^n;\mathbb{Z})$ be the usual generator. For $i>k$, choose $u_i$ whose pullback under the quotient map is $x^i$. Naturality of the cup product and the cohomology ring of complex projective space giveTogether with the unit, this determines the ring; equivalently, its reduced part is the ideal $(x^{k+1})/(x^{n+1})$ with the inherited multiplication. This is the cohomology ring of a collapsed projective subspace.

If $n=k+1$, the quotient has just a zero-cell and a $2n$-cell, so it is $S^{2n}$ and is a compact manifold. Conversely, suppose $n\ge k+2$ and the quotient is homotopy equivalent to a compact manifold. Its top cohomology is $\mathbb{Z}$, so that manifold must be closed, orientable, and $2n$-dimensional. But $H^{2n-2}\cong \mathbb{Z}$ while $H^2=0$, contradicting Poincare duality. Therefore

Set $X={\mathbb{CP}}^4/{\mathbb{CP}}^1$ and $A={\mathbb{CP}}^2/{\mathbb{CP}}^1\cong S^4$. Let $u\in H^4(A;\mathbb{Z})$ be a generator and let $a\in H^4(X;\mathbb{Z})$ be the class restricting to $u$. The ring computation in part (a) gives

If $r\circ \iota \simeq {\mathrm{id}}_A$, then homotopy invariance of cohomology gives ${\iota }^{\ast }{r}^{\ast }(u)=u$, hence $r^{\ast }(u)=a$. But $u^2=0$ because $H^8(S^4;\mathbb{Z})=0$, whereas naturality of the cup product would givea contradiction. No such map $r$ exists.

Let $a$ and $b$ be the pullbacks of the standard generator of $H^n(S^n;\mathbb{Z})$ from the two factors. The Künneth theorem and graded commutativity of the cup product give

Each homeomorphism acts invertibly on $H^n\cong {\mathbb{Z}}^2$. To respect ordinary composition, send $h$ to $(h^{-1}{)}^{\ast }$; functoriality of induced map on cohomology then defines a homomorphism

For $n=1$, the space is the torus. Every matrix in $GL(2,\mathbb{Z})$ induces a linear homeomorphism ${\mathbb{R}}^2/{\mathbb{Z}}^2\to {\mathbb{R}}^2/{\mathbb{Z}}^2$, so the image is all of $GL(2,\mathbb{Z})$.

For $n=2$, write $h^{\ast }a=pa+qb$. Since $a^2=0$ and $ab=ba$,so $pq=0$; the same argument applies to $h^{\ast }b$. Invertibility then forces the matrix to be a signed permutation matrix. Every such matrix is realized by swapping the two sphere factors and applying an orientation-reversing homeomorphism to either factor. Thus the image consists exactly of the eight signed permutation matrices.

Fibrewise evaluation identifiesThe identity endomorphism is a nowhere-zero continuous section, giving the canonical trivialization of a line bundle tensored with its dual.

Give a complex line bundle its natural orientation as a real plane bundle. Then its Euler class equals its First Chern class. If $a,b$ are the standard generators of $H^2(S^2\times S^2;\mathbb{Z})$, choose complex line bundles $L_1,L_2$ pulled back from the Hopf fibration on the two factors, with $e(L_1)=a$ and $e(L_2)=b$. The triviality of $L_i^{\ast }\otimes L_i$ gives $e(L_i^{\ast })=-e(L_i)$. Hence, for anythe tensor product $L_1^{\otimes p}\otimes L_2^{\otimes q}$, with negative powers interpreted using dual bundles, has Euler class $\alpha$. Its underlying oriented real rank-two bundle is the required $E_{\alpha }$.

Write $\alpha =pa+qb$. In the Gysin sequence of a sphere bundle for the oriented circle bundle $S(E_{\alpha })\to S^2\times S^2$, multiplication by $\alpha$ isin degrees $0$ to $2$, andin degrees $2$ to $4$. If $\alpha \ne 0$ and $d=\gcd (|p|,|q|)$, taking the relevant kernels and cokernels givesWhen $\alpha =0$, the bundle is trivial and the groups in degrees $0$ through $5$ have ranks $1,1,2,2,1,1$, respectively. These are precisely the groups recorded in integral cohomology of a circle bundle over a product of two spheres.

The additive cohomology for nonzero $\alpha$ depends only on $d$. On the other hand, part (a) shows that the homeomorphism group acts on $(p,q)$ only by signed permutations. For example, $(1,0)$ and $(1,1)$ both have $d=1$, so their sphere bundles have isomorphic additive cohomology, but no signed permutation carries one Euler class to the other. The cohomology therefore does not determine the homeomorphism-group orbit of $\alpha$.