Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 16 (Algebraic Topology)

Let $p:\tilde{M}\to M$ be the universal cover. Its number of sheets is $|{\pi }_1(M)|$, which is finite, so compactness of $M$ makes $\tilde{M}$ compact. As a covering space of a manifold without boundary, it is also a three-manifold without boundary. It is connected and simply connected by the definition of universal cover.

A simply connected manifold is orientable: transport of a local orientation around loops gives the orientation character ${\pi }_1(\tilde{M})\to \{\pm 1\}$, and the trivial fundamental group leaves no orientation obstruction. The degree-one Hurewicz theorem, equivalently the identification of $H_1$ with the abelianization of the fundamental group, gives $H_1(\tilde{M};\mathbb{Z})=0$. The universal coefficient theorem for cohomology then givesbecause the additional term $\mathrm{Ext}(H_0(\tilde{M};\mathbb{Z}),\mathbb{Z})$ vanishes for $H_0\cong \mathbb{Z}$. Integral Poincare duality gives $H_2(\tilde{M};\mathbb{Z})\cong H^1(\tilde{M};\mathbb{Z})=0$. Connectedness gives $H_0\cong \mathbb{Z}$, and the orientation fundamental class gives $H_3\cong \mathbb{Z}$; higher groups vanish by dimension. ConsequentlyThus the universal cover of a closed three-manifold with finite fundamental group is an integral homology sphere. No homeomorphism classification of three-manifolds is needed for this homological conclusion.

For the closed orientable surface ${\Sigma }_g$, use the fundamental group presentationSince $g\ge 1$, define a homomorphism to $\mathbb{Z}/n$ by sending $a_1$ to $1$ and every other displayed generator to zero. All commutators map to zero, so the relation is respected, and the map is surjective. Its kernel has index $n$. The classification of connected covering spaces supplies a connected $n$-sheeted covering space, including the identity cover when $n=1$.

Every connected finite covering space of ${\Sigma }_g$ is a closed orientable surface: compactness follows from finite degree, and the base orientation lifts through local covering charts. Lift a finite cell decomposition of the base. Every cell has $n$ lifts, so the Euler characteristic under a finite covering gives $\chi (\tilde{\Sigma })=n\chi ({\Sigma }_g)$. If the covering surface has genus $h$, this reads $2-2h=n(2-2g)$. ThereforeThis genus of a finite cover of a closed orientable surface depends only on the sheet number, not on the particular subgroup used to construct the cover.

Work first over ${\mathbb{F}}_2$. The mod-two cohomology ring of real projective space is $H^{\ast }({\mathbb{RP}}^n;{\mathbb{F}}_2)={\mathbb{F}}_2[a]/(a^{n+1})$, with $|a|=1$. A map missing a point has zero mod-two degree of a map between closed manifolds, so $f^{\ast }(a^n)=0$. The group in degree one has only two elements, so $f^{\ast }a$ is either zero or $a$. The latter would imply $f^{\ast }(a^n)=(f^{\ast }a{)}^n=a^n\ne 0$, a contradiction. Therefore $f^{\ast }a=0$, and multiplicativity makes $f^{\ast }$ zero in every positive cohomological degree. Duality of homology and cohomology over a field consequently makes $f_{\ast }$ zero in every positive degree with mod-two coefficients.

For integral coefficients, cellular homology of real projective space gives $H_k\cong \mathbb{Z}/2$ in odd degrees $0<k<n$, zero in even degrees $0<k<n$, and a top group $\mathbb{Z}$ if $n$ is odd and zero if $n$ is even. For each torsion group, the natural reduction map to mod-two homology is injective by the universal coefficient theorem for homology; naturality and the preceding vanishing imply that its integral induced map is zero. When $n$ is odd, the integral top-degree map is also zero because the nonsurjective map has integer degree of a map between oriented manifolds zero. All groups above dimension $n$ vanish. Hence $f_{\ast }$ is zero on all positive integral homology groups as well.

For $n\ge 2$, ${\pi }_1({\mathbb{RP}}^n)=\mathbb{Z}/2$, and its abelianization is already $H_1({\mathbb{RP}}^n;\mathbb{Z})$. The vanishing just proved therefore implies $f_{\ast }{\pi }_1=0$. By the lifting criterion for a covering space, $f$ lifts through the double covering space $q:S^n\to {\mathbb{RP}}^n$ to a map $\tilde{f}:{\mathbb{RP}}^n\to S^n$ with $q\tilde{f}=f$.

Choose a point $y$ missed by $f$. Both points of $q^{-1}(y)$ are missed by $\tilde{f}$. In particular the lift has image in $S^n\setminus \{z\}\cong {\mathbb{R}}^n$, for either such point $z$. This punctured sphere is contractible, so contraction there gives a homotopy from $\tilde{f}$ to a constant map into $S^n$. Composing with $q$ contracts $f$. ThusThis proves the nonsurjective self-map of real projective space is null-homotopic conclusion, and also shows that all positive-degree induced maps vanish with any coefficients.

Let the integer degree of a map between oriented manifolds be $d\ne 0$, so $f_{\ast }[X]=d[Y]$. For any nonzero $\alpha \in H^k(Y;\mathbb{Q})$, nondegeneracy of the Poincare duality pairing supplies $\gamma \in H^{n-k}(Y;\mathbb{Q})$ with $⟨\alpha ⌣\gamma ,[Y]⟩\ne 0$. Naturality of the cup product and its evaluation givesThus $f^{\ast }\alpha \ne 0$, proving injectivity of pullback on rational cohomology. By the universal coefficient theorem for cohomology, these finite-dimensional rational cohomology groups are the duals of the corresponding homology groups, and the dual of $f_{\ast }$ is $f^{\ast }$. Injectivity of the dual map therefore provesOne can see the surjection constructively. Given $u\in H_k(Y;\mathbb{Q})$, choose its Poincare dual $\eta \in H^{n-k}(Y;\mathbb{Q})$ with $\eta \cap [Y]=u$. Then $v=d^{-1}(f^{\ast }\eta \cap [X])$ satisfies $f_{\ast }v=d^{-1}(\eta \cap f_{\ast }[X])=u$ by naturality of the cap product. Division by $d$ is the reason rational coefficients suffice.

Let $m=\dim X$ be odd. Use mod-two Poincare duality, which applies even when the closed manifold is nonorientable. It gives $b_k^{(2)}=b_{m-k}^{(2)}$, where $b_k^{(2)}={\dim }_{{\mathbb{F}}_2}{H}_k(X;{\mathbb{F}}_2)$. The Euler characteristic can be computed with any field of coefficients: for a finite cell model, the alternating sum of chain dimensions equals the alternating sum of homology dimensions, and the former counts cells independently of the field.

Consequently $\chi (X)={\sum }_{k=0}^m(-1{)}^k{b}_k^{(2)}$. Pair each term with index $m-k$. The paired Betti numbers are equal and their signs opposite because $m$ is odd; there is no unpaired middle index. ThereforeThis argument applies componentwise if the manifold is disconnected and does not require an orientation.

Write the alternating bilinear form as $B$. Alternation gives $B(v,w)=-B(w,v)$ by expanding $B(v+w,v+w)=0$. If $V\ne 0$, choose $v\ne 0$. Nondegeneracy supplies $w$ with $B(v,w)\ne 0$; rescale $w$ so that $B(v,w)=1$. The vectors are independent, and the restriction to $E=\mathrm{span}\{v,w\}$ has matrix $\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, whose determinant is one.

Every $z\in V$ has a unique decomposition into a vector in $E$ and a vector orthogonal to $E$: subtract $B(z,w)v-B(z,v)w$ to obtain the latter. Thus $V=E\oplus E^{\perp }$. If a vector in $E^{\perp }$ annihilates $E^{\perp }$ under $B$, it also annihilates $E$, and hence all of $V$; nondegeneracy makes it zero. The restricted alternating bilinear form on $E^{\perp }$ is therefore again nondegenerate. Induction splits off two dimensions at each step and ends at the zero space. HenceThe proof works in characteristic two as well: alternation, rather than merely skew symmetry, is the needed hypothesis.

Write $\dim X=4r+2=2m$, with $m=2r+1$ odd. Rational Poincare duality gives $b_k=b_{2m-k}$. Pairing every term outside the middle yieldsThe Poincare duality pairing on $H^m(X;\mathbb{Q})$ is $B(\alpha ,\gamma )=⟨\alpha ⌣\gamma ,[X]⟩$. It is nondegenerate, and graded commutativity of the cup product gives $B(\alpha ,\gamma )=-B(\gamma ,\alpha )$ because $m$ is odd. In particular $B(\alpha ,\alpha )=0$ over $\mathbb{Q}$. Part (b) therefore makes $b_m$ even, soFor the failure without orientability, ${\mathbb{RP}}^2$ is a closed nonorientable surface with one cell in each dimension zero, one and two; hence $\chi =1-1+1=1$. More generally, ${\mathbb{RP}}^{4r+2}$ is nonorientable and has Euler characteristic one. Thus the orientation hypothesis in the Euler characteristic parity in dimensions congruent to two modulo four is essential.

Use the short exact sequence of coefficient groupsThe groups of singular chains are free, so applying the cochain construction gives a short exact sequence of cochain complexes. Its long exact sequence in cohomology has the required terms and reduction map $r_{\ast }$. Define the Bockstein homomorphism to be its connecting homomorphism.

Explicitly, represent $x$ by a mod-$p$ cocycle $a$ and lift it to an integral singular cochain $A$. Since $a$ is a cocycle, $\delta A$ is divisible by $p$. ThenThis is exactly the lift-and-coboundary construction for the coefficient sequence above: the mod-$p^2$ coboundary of $A$ lies in the image of $j$. The quotient cochain is a cocycle, since ${\delta }^{2}A=0$ and integral cochains have no $p$-torsion. Replacing $A$ by another lift adds $pC$ and changes the quotient by $\delta C$ modulo $p$; replacing the representative by a coboundary likewise leaves its cohomology class unchanged. Thus the definition is independent of choices.

A degree-zero cocycle is a function on points constant on each path component. Choose an integer representative for its value on each component. This produces an integral zero-cocycle lift, with zero coboundary. Hence the Bockstein vanishes on $H^0(X;{\mathbb{F}}_p)$ for every space.

For a nonzero example in every positive degree $i$, take the Moore space $X=S^i{\cup }_p{e}^{i+1}$, attaching the cell by a map of mapping degree $p$. Its positive-degree integral cellular cochain complex has the differential $\mathbb{Z}\stackrel{p}{\to }\mathbb{Z}$ from degree $i$ to $i+1$; when $i=1$, the preceding differential from degree zero is zero. Modulo $p$ the displayed differential vanishes, and $H^i(X;{\mathbb{F}}_p)=H^{i+1}(X;{\mathbb{F}}_p)={\mathbb{F}}_p$. Lift the degree-$i$ generator to the cellular cochain with value one. Its coboundary has value $p$, so dividing by $p$ gives the generator in degree $i+1$. Naturality of cellular and singular cohomology with coefficient sequences identifies this with the Bockstein on a cyclic Moore space. Therefore $\beta$ is nonzero, indeed an isomorphism, in the requested degree for every prime $p$.

Finally take integral lifts $A,B$ of cocycles representing $x\in H^i(X;{\mathbb{F}}_p)$ and $y\in H^j(X;{\mathbb{F}}_p)$. Write $\delta A=p{A}_1$ and $\delta B=p{B}_1$. The cochain cup product obeys the graded Leibniz identity, soDivide by $p$, reduce modulo $p$ and pass to cohomology. This proves the Bockstein derivation rule

For a singular simplex $\sigma :{\Delta }^i\to Y$, there are exactly $n$ lifts to $X$: choose any of the $n$ points above one vertex, then use the lifting property, since the simplex is simply connected. Defineand extend $R$-linearly to singular chains. The restrictions of these lifts to any face are exactly the $n$ lifts of that face, each appearing once. Indeed a lift is determined uniquely by its value at any point of that face, and every possible value in the corresponding fiber extends to a lift of the whole simplex. Therefore $\partial T_i=T_{i-1}\partial$, so $T$ is a chain map and induces the homological transfer for a finite covering, denoted $f^{\ast }$ in the question.

On chains, every lifted simplex projects back to the original simplex. Thus $f_{\#}{T}_i=n\mathrm{id}$, and on homology this givesOver $\mathbb{Q}$, the map $n^{-1}{f}^{\ast }$ is a right inverse of $f_{\ast }$, proving surjectivity in every degree. Consequently each rational Betti number of the covering space is at least that of its base. This transfer goes in the opposite direction to the usual homological pushforward and is not the ordinary cohomological pullback.

For equality, use the connected double covering space $S^1\to S^1$, $z\mapsto z^2$. Both spaces have $b_0=b_1=1$ and all other Betti numbers zero. For strict increase, take a connected double covering space of the genus-two closed orientable surface, whose existence was proved in 1(b). The genus of a finite cover of a closed orientable surface gives covering genus $1+2(2-1)=3$. Its first Betti number is six, whereas that of the genus-two base is four; their zeroth and second Betti numbers are both one. Thus a connected double cover can preserve every Betti number or strictly increase some of them.

Choose an orientation for each curve and let $c_i=[C_i]\in H_1(X;\mathbb{Q})$. The algebraic intersection number of curves on an oriented surface gives a nondegenerate alternating bilinear form on the $2g$-dimensional space $H_1(X;\mathbb{Q})$, by Poincare duality. Distinct curves are disjoint, so their intersection numbers are zero; each self-pairing is zero by alternation. Thus the span $W=\mathrm{span}\{c_1,\dots ,c_n\}$ is an isotropic subspace of a symplectic vector space. The assumed linear independence gives $\dim W=n$.

Nondegeneracy identifies $H_1(X;\mathbb{Q})$ with its dual. Restricting functionals to $W$ is surjective, so the symplectic orthogonal complement has dimension $\dim W^{\perp }=2g-n$. Isotropy says $W\subseteq W^{\perp }$, giving $n\le 2g-n$. HenceThe bound is sharp: one meridian on each handle of a genus-$g$ surface gives $g$ pairwise disjoint curves with independent homology classes. This is an algebraic bound on independent classes, not on the number of disjoint separating or mutually homologous curves.

Give the smooth curve its complex orientation. The cellular homology of complex projective space gives $H_2({\mathbb{CP}}^2;\mathbb{Z})=\mathbb{Z}$, generated by the class $[L]$ of a projective line. Thus $[C]=D[L]$ for a unique integer $D$, its homological degree of a smooth complex plane curve. Two distinct projective lines meet transversely in one point with positive sign, so $[L]\cdot [L]=1$. Therefore the degree $D$ is the intersection number $[C]\cdot [L]$ with any transverse projective line.

For the displayed curve, take $d\ge 1$ and put $F=x^d+y^d-z^d$. If $d\ge 2$, its three partial derivatives cannot vanish simultaneously at a projective point; if $d=1$, the equation defines a line and is also smooth. Intersect with $L=\{y=0\}$. Every intersection has $z\ne 0$, so in the chart $z=1$ the points areThere are exactly $d$ distinct points. At each, $\partial F/\partial x=d{\zeta }^{d-1}\ne 0$. In affine coordinates the curve is a graph $x=x(y)$, whereas the line is $y=0$; their tangent lines are transverse. For $d\ge 2$ the curve tangent at the point is the $y$-direction, while for $d=1$ it is the direction $dx+dy=0$, still transverse to $dy=0$.

Each intersection contributes $+1$, not merely an unsigned count: both tangent spaces are complex lines, and the real determinant of the complex-linear isomorphism from their direct sum to the ambient tangent space is the squared modulus of its complex determinant, hence positive. Therefore $[C]\cdot [L]=d$, andThis computes the degree by explicit transverse intersections, without using Bézout's theorem to assume the answer.