Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 114 (Algebraic Topology)

For a CW complex $X$, the cellular chain complex isthe free abelian group generated by the oriented $n$-cells. Its differential is the compositeNaturality of the long exact sequences of the triples $X^{n-2}\subset X^{n-1}\subset X^n$ makes two successive connecting maps compose to zero, so $d_{n-1}{d}_n=0$.

The cellular boundary formula says that the coefficient of an $(n-1)$-cell $e_{\alpha }^{n-1}$ in the boundary of $e_{\beta }^n$ is the degree of

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The quotient of the equator is $S^1/(z\sim -z)\cong S^1$, giving one zero-cell and one one-cell. The interiors of the upper and lower hemispheres give two two-cells. Each boundary circle maps to the quotient equator by the degree-two covering, so, after choosing orientations,is the cellular chain complex; changing one orientation only changes one sign. Therefore

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The point $x$ on the quotient equator has two preimages. Small discs around them become four half-discs glued along their common diameter, so a neighborhood of $x$ is the cone on a graph with two vertices joined by four edges. The Excision theorem and the local homology from a link identifywhere $L$ is this four-edge graph. It is connected and has first Betti number $4-2+1=3$. Hence

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No. If a finite CW complex $Y$ had exactly one two-cell, then $C_2^{\mathrm{cell}}(Y)\cong \mathbb{Z}$. The isomorphism $H_2(Y)\cong H_2(X)\cong \mathbb{Z}$ forces $d_2=0$ and $\mathrm{im}{d}_3=0$: a subquotient of $\mathbb{Z}$ can remain infinite cyclic only in this way. Consequentlyis a subgroup of the free abelian group $C_1(Y)$ and is therefore torsion-free. This contradicts the homotopy-invariant calculation $H_1(Y)\cong H_1(X)\cong \mathbb{Z}/2$.

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For $n>0$,The CW structure with one cell in every even dimension gives the additive groups, and the generator $x=c_1({\gamma }^{\ast })$ restricts compatibly along ${\mathbb{CP}}^{n-1}\subset {\mathbb{CP}}^n$; its powers generate each even group.

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The integral Künneth theorem gives a natural short exact sequence containing tensor and Tor terms. Here all cohomology groups are free abelian, so the Tor terms vanish and the external cup product is a ring isomorphism. Thus

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With the indicated points as basepoints, $X$ is the smash product ${\mathbb{CP}}^2\wedge {\mathbb{CP}}^2$. The quotient map identifies its positive-degree cohomology with the relative cohomology of the product modulo the wedge, which under the Künneth isomorphism is the ideal generated by $xy$. Thusin degrees $4,6,6,8$. The only nonzero product of positive-degree basis elements isTogether with the unit in degree zero, this determines the cohomology ring.

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Suppose such maps existed, and let $u$ generate $H^4(S^4;\mathbb{Z})$. Since $f^{\ast }{g}^{\ast }=1$, write $g^{\ast }u=ma$ and $f^{\ast }a=nu$; then $mn=1$, so $m=\pm 1$. Naturality of the cup product givescontradicting the nonzero degree-eight class computed in part c.

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A singular cochain has compact support when it vanishes on every simplex whose image lies outside some compact set. Such cochains form a subcomplex because the faces of a simplex outside that compact set also lie outside it. Its cohomology is compactly supported cohomology $H_c^{\ast }(X)$.

If $f:X\to Y$ is proper and a cochain on $Y$ is supported in the compact set $K$, its pullback is supported in the compact set $f^{-1}(K)$. Pullback commutes with the coboundary, so it induces a well-defined map

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There is a natural descriptionClosed balls form a cofinal family of compact subsets of ${\mathbb{R}}^n$. Excision and radial deformation identify every sufficiently large-ball term, and the resulting stable group is the local group at the origin:It is $\mathbb{Z}$ in degree $n$ and zero otherwise.

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Properness implies $f(v)\ne 0$ for $v\ne 0$: otherwise the whole noncompact complex line $\mathbb{C}v$ would lie in $f^{-1}(0)$. Homogeneity therefore definesOn the fiber over $[v]$, defineThe identity $f(\lambda v)=\lambda f(v)$ makes this independent of the representative $v$, and it is a nonzero complex-linear map on every fiber. It is therefore the bundle isomorphism

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Normalize on the unit sphere:This is an $S^1$-equivariant map of the Hopf fibration covering $\bar{f}$, and it has degree one on every circle fiber. The bundle isomorphism in part c givesSince this first Chern class generates the cohomology ring of ${\mathbb{CP}}^{n-1}$, $\bar{f}$ acts identically on its cohomology. Naturality of the oriented Gysin sequence of a sphere bundle then gives $\mathrm{deg}F=1$.

The original homogeneous map is properly homotopic to the cone on $F$ by radial normalization; the norms $\Vert f(v)\Vert$ are bounded above and away from zero on the unit sphere, so this homotopy is proper. Thus $f$ has proper degree one. It fixes the generator of $H_c^{2n}({\mathbb{C}}^n)$, and all other compactly supported groups vanish. Therefore $f^{\ast }:H_c^{\ast }({\mathbb{C}}^n)\to H_c^{\ast }({\mathbb{C}}^n)$ is the identity.

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The cap product is the chain operationobtained by evaluating the cochain on the front $k$-face of a singular simplex and retaining its back $(p-k)$-face, with the standard sign convention. It descends to homology and cohomology.

A fundamental class $[M]\in H_d(M;\mathbb{Z})$ restricts at every $x\in M$ to the local generator of $H_d(M,M\setminus \{x\};\mathbb{Z})$ selected by the orientation. Poincare duality states thatis an isomorphism for every $k$ and coefficient ring $R$.

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The small simplex theorem says that the inclusion of the subcomplex generated by singular simplices whose images lie wholly in $U$ or wholly in $V$,is a chain-homotopy equivalence. Repeated barycentric subdivision supplies the inverse up to chain homotopy.

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Let $0<k<d$ and $\alpha \in H^k(M;\mathbb{Z})$. Its restrictions to the contractible sets $U$ and $V$ vanish. The cap-product support lemma for a two-set cover, proved by representing $[M]$ with small simplices and replacing the restricted cocycles by coboundaries, therefore givesPoincare duality makes cap product with $[M]$ injective, so $H^k(M;\mathbb{Z})=0$ for every intermediate degree. Applying duality again gives $H_i(M;\mathbb{Z})=0$ for $0<i<d$. Finally the connected oriented closed manifold has

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