Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 12 (Algebraic Topology)

Unless a coefficient group is displayed, use integral singular cohomology. The standard CW complex structure on infinite-dimensional real projective space has one cell in each nonnegative dimension. Its cellular chain complex has boundary $d_j=2$ for positive even $j$ and $d_j=0$ for odd $j$. The cellular cohomology differential ${\delta }^j$ is therefore zero for even $j$ and multiplication by two for odd $j$. ConsequentlyMore generally, for an abelian group $A$, the positive odd groups are $A[2]=\{a:2a=0\}$ and the positive even groups are $A/2A$. In particular $H^j({\mathbb{RP}}^{\infty };{\mathbb{Z}}_2)={\mathbb{Z}}_2$ in every nonnegative degree.

For the required Bockstein homomorphism, use the short exact sequenceHere $m\ge 2$; for $m=1$ all three coefficient groups are zero. Since singular chains are free abelian groups, applying cochains gives a short exact sequence of cochain complexes. Its connecting homomorphism defines $\beta$ and its long exact sequence is precisely the required one, with the other maps induced by $\iota$ and $q$.

Explicitly, represent a class by a cocycle $u\in C^n(X;{\mathbb{Z}}_m)$ and choose a lift $\tilde{u}\in C^n(X;{\mathbb{Z}}_{m^2})$. Since $q(\delta \tilde{u})=0$, there is a unique cochain $v$ with $\iota (v)=\delta \tilde{u}$. Injectivity of $\iota$ and ${\delta }^2=0$ show $\delta v=0$. DefineChanging the lift by $\iota (w)$ changes $v$ by the coboundary $\delta w$. Changing the representative $u$ by a coboundary can be lifted by a coboundary as well and leaves the resulting class unchanged. Thus this is a well-defined group homomorphism, and the standard cochain lifting argument gives exactness.

Compute the Bockstein homomorphism on infinite-dimensional real projective space using its cellular cohomology complex. A generator with coefficients ${\mathbb{Z}}_2$ is represented by $1$ in degree $n$, lifted to $1\in {\mathbb{Z}}_4$. Its coboundary is $0$ for even $n$ and $2$ for odd $n$. Dividing via $\iota (1)=2$ givesThus the odd-degree maps are isomorphisms. The comparison between cellular cohomology and singular cohomology is natural with respect to coefficient maps, so this computes the same connecting homomorphism constructed above.

Define compactly supported cohomology byFor $K\subseteq L$, the transition map is induced by the identity map of pairs $(X,X\setminus L)\to (X,X\setminus K)$. Equivalently, take the cochain complex of singular cochains that vanish on every chain contained in the complement of some compact set. Directed unions are exact, giving the same definition.

For $\mathbb{R}$, the intervals $[-a,a]$, $a>0$, are cofinal among compact subsets. The complement has two contractible components. The long exact sequence in relative cohomology contains the diagonal map $\mathbb{Z}\to \mathbb{Z}\oplus \mathbb{Z}$, so its cokernel is $H^1(\mathbb{R},\mathbb{R}\setminus [-a,a])\cong \mathbb{Z}$, and all the other relative groups vanish. Enlarging the interval preserves the generator given by the difference of the two ends. Therefore

For the compact-support comparison with a one-point compactification, write $Y=X^{+}$. The assumed Hausdorff one-point compactification is compact; a compact subset $K\subseteq X$ is closed in $Y$. The Excision theorem removes $\{\infty \}$ from the pair $(Y,Y\setminus K)$, because its closure lies inside the open second member. ThusComplements of compact subsets of $X$ are exactly the open neighbourhoods of $\infty$ in $Y$. The hypothesis supplies a cofinal family of contractible such neighbourhoods $U$. For every one, the long exact sequence of the pair identifiesIn degree zero, this is the kernel of evaluation on the component of $\infty$, identified with reduced cohomology by subtracting the constant value there. In degree one the map $H^0(Y)\to H^0(U)\cong \mathbb{Z}$ is surjective; in higher degrees the positive cohomology of $U$ vanishes. These identifications are natural for inclusions of contractible neighbourhoods. Passing to the direct limit proves

For the specified disjoint union of lines, a compact subset meets only finitely many components and is bounded in each. Finite unions $F\times [-a,a]$, with $F\subset \mathbb{Z}$ finite, are cofinal. Applying the preceding relative calculation componentwise givesThe one-point compactification $W$ of this space is the Hawaiian earring: each line becomes a circle by adding the common point $\infty$, and every neighbourhood of $\infty$ contains all but finitely many whole circles. On the remaining finitely many circles it contains neighbourhoods of the common point. This describes exactly the shrinking-circle topology. In particular $W$ is not locally contractible at $\infty$: every such neighbourhood contains a whole circle, whose generator remains nontrivial under the retraction $W\to S^1$ that collapses all the other circles.

For integral singular cohomology, the comparison does not hold. Here is a degree-two obstruction that takes account of the shrinking-circle topology. The standard rational summand in Hawaiian earring homology theorem gives a direct summand $\mathbb{Q}$ in $H_1(W;\mathbb{Z})$. The universal coefficient theorem for cohomology injectsThe summand $\mathbb{Q}$ therefore contributes the nonzero Ext of the rationals with integer coefficients.

For completeness, this last algebraic assertion has an explicit proof. Present $\mathbb{Q}$ using generators $a_n=1/n!$ and relations $a_n-(n+1)a_{n+1}=0$, $n\ge 1$. The corresponding free resolution shows that ${\mathrm{Ext}}_{\mathbb{Z}}^1(\mathbb{Q},\mathbb{Z})$ is the cokernel ofThe constant sequence $(1,1,\dots )$ is not in the image. Otherwise iteration would giveFor large $N$, the factorial sum exceeds $|u_1|$ but is less than $(N+1)!-|u_1|$, making that congruence impossible. Thus the cokernel is nonzero. It follows thatwhich proves the failure of the claimed isomorphism. The ingredient concerning the Hawaiian earring is its singular-homology structure theorem, not the homology of an infinite CW complex wedge of circles; these topologies differ.

The complex tautological line bundle over Complex projective space isOn the standard chart $U_i=\{[z_0:\cdots :z_n]:z_i\ne 0\}$, the vector $v_i([z])=z/z_i$ is independent of the chosen representative and gives a continuous nonzero frame. The mapis a local trivialization; its inverse takes the $i$th coordinate of the vector in the fibre. Hence $\lambda$ is a locally trivial complex line bundle.

Its unit sphere bundle is $S^{2n+1}$: a unit vector $v\in {\mathbb{C}}^{n+1}$ corresponds to $([v],v)$. The projection is the Hopf fibration. Regard $\lambda$ as an oriented real rank-two vector bundle using its complex orientation, and put $t=e(\lambda )\in H^2({\mathbb{CP}}^n;\mathbb{Z})$. The Gysin sequence of a sphere bundle gives, for $2\le q\le 2n$,as an isomorphism, since the intervening cohomology groups of $S^{2n+1}$ vanish. It also gives $H^1({\mathbb{CP}}^n)=0$. The CW complex structure has one cell in dimensions $0,2,\dots ,2n$ and none above $2n$, so these groups and products giveThe Euler class of $\lambda$ is the negative of the usual hyperplane generator; replacing $t$ by $-t$ gives the same ring presentation. For $n=0$ the formula reads $\mathbb{Z}$.

Now let $\alpha$ generate $H^2(S^2;\mathbb{Z})$ and let $u_i$ be its pullback from the $i$th factor of $(S^2{)}^n$. The Künneth theorem gives $H^2((S^2{)}^n;\mathbb{Z})={⨁}_i\mathbb{Z}{u}_i$. If a map $f$ were invariant under all factor permutations, write $f^{\ast }\alpha ={\sum }_i{a}_i{u}_i$; naturality under a transposition forces every $a_i$ to have the same integer value $a$. Let $\Delta :S^2\to (S^2{)}^n$ be the diagonal. ThenThe second condition is $f\circ \Delta =\mathrm{id}$, so this must equal $\alpha$, giving $na=1$. For $n>1$ this is impossible in $\mathbb{Z}$, and $\boxed{\text{no such continuous map exists}.}$ This is the diagonal-degree obstruction to a symmetric sphere retraction.

For a closed oriented $d$-dimensional manifold $M$ and a commutative coefficient ring $R$, Poincare duality says that cap product with the fundamental class is an isomorphismOver a field it equivalently gives a nondegenerate Poincare duality pairing $(a,b)\mapsto ⟨a⌣b,[M]⟩$ between complementary cohomological degrees.

For the six-manifold take rational coefficients and write $b_j={\dim }_{\mathbb{Q}}{H}^j(X;\mathbb{Q})$. Poincare duality gives $b_j=b_{6-j}$, so the Euler characteristic isThe middle-degree Poincare duality pairing on $H^3(X;\mathbb{Q})$ is skew-symmetric by graded commutativity of the cup product, since $(-1{)}^{3\cdot 3}=-1$. It is a nondegenerate alternating bilinear form, so its dimension $b_3$ is even. For example, a nonsingular skew-symmetric matrix of odd size would have $\det A=\det (-A)=-\det A$, impossible over $\mathbb{Q}$. Therefore $\boxed{\chi (X)\in 2\mathbb{Z}}$.

To realize every even integer, use $\chi (S^6)=2$, $\chi ({\mathbb{CP}}^3)=4$ and $\chi (S^3\times S^3)=0$. All three are closed connected orientable six-manifolds. For the connected sum of oriented manifolds, removing a ball from each summand and gluing their boundary spheres givesin dimension six: each removed open ball decreases the Euler characteristic by one, while the gluing sphere $S^5$ has Euler characteristic zero. If $k=2\ell$, putomitting zero copies. The Euler characteristic is $2+2r-2s=k$. This supplies a closed connected orientable example for every $k\in 2\mathbb{Z}$.

Without orientability, evenness need not hold. The Real projective space ${\mathbb{RP}}^6$ is a closed six-manifold with one cell in each dimension $0,\dots ,6$, soIt is nonorientable because the antipodal deck transformation on $S^6$ has degree $(-1{)}^7=-1$ and reverses orientation. Thus it gives the required counterexample.

Use integral cohomology and let $k\ge 1$. On the projective bundle $Y=\mathbb{P}(E)$ define the complex tautological line bundlePut $t=e(\mathcal{L})\in H^2(Y;\mathbb{Z})$, using the canonical complex orientation. On each fibre, $t$ is the Euler class of the tautological line over ${\mathbb{CP}}^{k-1}$, so $1,t,\dots ,t^{k-1}$ restrict to an integral basis of its cohomology.

Here is the finite-cover Leray-Hirsch theorem proof in this case. For each open set $U\subseteq X$, defineIf $E$ is trivial on $U$, its projective bundle is $U\times {\mathbb{CP}}^{k-1}$ and $\mathcal{L}$ is pulled back from the tautological line on the second factor. The Künneth theorem makes ${\Phi }_U$ an isomorphism, since the fibre has finite free integral cohomology. The same holds for every open subset of $U$.

Compactness of $X$ provides a finite trivializing cover $U_1,\dots ,U_N$. Induct on its size. If the result holds on $V=U_1\cup \cdots \cup U_{N-1}$, it holds on $U_N$ and on $V\cap U_N$, both lying in a trivializing chart. Form the diagram of Mayer–Vietoris sequences for the base, with the finitely many degree shifts on the left, and the total space on the right. Naturality of pullback and multiplication by the global even-degree classes $t^i$ makes the diagram commute. The Five lemma gives the isomorphism on $V\cup U_N$. ThusThis is the claimed free module statement, with its graded degree shifts made explicit.

The module basis expresses $t^k$ uniquely using the lower powers, with homogeneous coefficients. Define the Chern classes $c_j(E)\in H^{2j}(X;\mathbb{Z})$ by the unique relationThe pullbacks are suppressed when $f$ is regarded as a polynomial over $H^{\ast }(X)$. Evaluation at $t$ gives a surjective map $H^{\ast }(X)[T]\to H^{\ast }(Y)$. Since $f(T)$ is monic, monic polynomial division over a ring writes any polynomial as $q(T)f(T)+r(T)$ with degree of $r$ below $k$. If its evaluation is zero, module independence forces every coefficient of $r$ to vanish. Hence the kernel is exactly the ideal generated by $f$, provingThe even-degree coefficients are central in the graded commutative algebra, so this division and ideal statement also apply when the base has odd-degree cohomology. Uniqueness of the coefficients proves their naturality under pullback, by pulling back the relation and using the same module basis. With the hyperplane convention $u=-t$, the relation has the usual all-positive Chern coefficients; the alternating signs here correspond to the tautological line itself.

Now suppose $E={⨁}_i{L}_i$. The sections of a projective bundle $s_i:X\to Y$ choose the line $(L_i{)}_x\subset E_x$. They satisfy $s_i^{\ast }\mathcal{L}\cong L_i$, so $s_i^{\ast }t=e_i$ and pulling back the relation gives $f(e_i)=0$.

To obtain the full factorization over the possibly torsion-containing base ring, also use the associated open chartsThey contain the images of the sections $s_i$ and cover $Y$. Projection identifies $\mathcal{L}{|}_{V_i}$ with ${\pi }^{\ast }{L}_i{|}_{V_i}$, so $t-{\pi }^{\ast }{e}_i$ restricts to zero on $V_i$. The long exact sequence of the pair lifts this class to $H^2(Y,V_i)$. The relative cup product of the $k$ lifted classes lies inIts absolute image is ${\prod }_i(t-{\pi }^{\ast }{e}_i)$, so that product vanishes. The polynomial ${\prod }_i(T-e_i)$ is monic of degree $k$ and lies in the kernel of evaluation. Subtracting the monic generator $f(T)$ leaves degree below $k$, and module independence again makes the difference zero. ThereforeThe open-cover argument proves the factorization without a non-zero-divisor assumption on the differences $e_i-e_j$.