Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 127 (Homotopy Theory)

A Serre fibration $p:E\to B$ has the homotopy lifting property for every disc: given $h:D^k\times I\to B$ and ${\tilde{h}}_0:D^k\to E$ with $p{\tilde{h}}_0=h(-,0)$, there is a lift $\tilde{h}$ extending ${\tilde{h}}_0$. Equivalently one may test this property on CW complexes.

For a map $g:E\to B$, its mapping-path fibration replacement isPath spaces have the compact-open topology, with the usual compactly generated convention. The inclusion $s(e)=(e,{\mathrm{const}}_{g(e)})$ satisfies $ps=g$. Forgetting the path is a homotopy inverse to $s$: shorten $\gamma$ to $u\mapsto \gamma (tu)$, from its constant initial path at $t=0$ to the original path at $t=1$. The equivalence here means a commuting triangle over the same base whose map of total spaces is a homotopy equivalence; it does not assert a fibrewise homotopy inverse over the base. Appending the successive segments of a moving base path gives the homotopy lifting property for $p$, even for arbitrary parameter spaces, so $p$ is in particular a Serre fibration. Its ordinary fibre over $b_0$ is the homotopy fibre

For a based Serre fibration, with fibre inclusion $j:F\hookrightarrow E$, the long exact sequence of homotopy groups of a fibration isThe maps $j_{\ast },p_{\ast }$ compose based sphere representatives with $j,p$; the maps on path components do the same with points. To define $\partial$, represent a class by $(I^k,\partial I^k)\to (B,b_0)$ and lift the cube starting at $e_0$ on its bottom face, keeping its vertical sides at $e_0$. The opposite face then gives $(I^{k-1},\partial I^{k-1})\to (F,e_0)$. Relative lifting for a CW pair gives this construction; its based homotopy class is independent of choices. At $k=1$ it records the component of the endpoint of the lifted loop. The last terms are pointed sets rather than groups in general.

Now write $W=S^1\vee S^n$ and let $f:W\to S^n$ be the collapse map. Inclusion of the $n$-sphere is a section, so $f_{\ast }:{\pi }_k(W)\to {\pi }_k(S^n)$ is surjective in every positive degree. Thus all connecting maps in its homotopy fibre sequence vanish. The fibre is path-connected, andThe section also gives the first assertion when $n=2$, by killing the connecting map from ${\pi }_2(S^2)$.

The universal cover of $W$ consists of a line with an $n$-sphere attached at each integer. It is homotopy equivalent to the wedge sum ${\bigvee }_{k\in \mathbb{Z}}{S}^n$. This cover is $(n-1)$-connected, so the Hurewicz theorem identifies its $n$th homotopy group with its $n$th homology. Choose a generator $t$ of the fundamental group whose action shifts the sphere indexed by $k$ to that indexed by $k+1$. The resulting homotopy groups as modules over the fundamental group giveEvery lifted sphere maps to $S^n$ with degree one. Consequently $f_{\ast }$ is the augmentation $\epsilon (\sum a_k{t}^k)=\sum a_k$, with the trivial $R$-action on $\mathbb{Z}$. The inclusion of the fibre respects the fundamental group actions, soThe augmentation ideal of the infinite cyclic group is generated by $t-1$: multiply a Laurent polynomial by a power of $t$ and use divisibility by $t-1$ when its value at one vanishes. Multiplication by $t-1$ is injective, giving the last isomorphism. This is a free module of rank one, whose underlying abelian group has countably infinite rank; its ${\pi }_1$-action is not trivial.

Choose the Hopf map $\eta$ and orientations so its Hopf invariant is one. The cellular cohomology of $Y_d$ is $\mathbb{Z}$ in degrees $0,2,4$ and zero otherwise. Let $x,z$ be generators in degrees two and four. Precomposing $\eta$ with a degree-$d$ self-map of $S^3$ represents $d\eta$. The induced map of mapping cones is the identity on the two-cell and has degree $d$ on the four-cell. Since the Hopf mapping cone has cup square equal to its top generator, naturality of the cup product gives $x^2=dz$. Equivalently, precomposition scales the Hopf invariant by degree. HenceThe coefficient is $d$, not $d^2$: multiplication in ${\pi }_3(S^2)$ is being used, rather than postcomposition by a degree-$d$ map of the target sphere.

For $X=X_{d,f}$, the only nonzero cellular boundary is $C_4\to C_3$, multiplication by $f$. Thus its cellular cohomology givesLet $x$ again generate $H^2$, and let $z$ be the class of the four-cell cochain. Collapsing the three-sphere gives a map $X\to Y_d$ which pulls the degree-two generator back to $x$ and the degree-four generator back to $z$. Hence $x^2=dz$, now interpreted modulo $f$. Every other product of positive-degree classes vanishes by dimension. This is the cohomology ring of a Hopf attachment with a sphere summand. More explicitly, if $f\ne 0$,If $f=0$, there is additionally a generator $y$ of degree three, with $xy=y^2=yz=0$. The displayed description of the groups uses $\mathbb{Z}/0\mathbb{Z}=\mathbb{Z}$, so the zero case is included. In particular, a nonzero $f$ kills the degree-three cohomology but can leave torsion in degree four.

There are no one-cells, and attaching cells of dimension at least three does not change ${\pi }_1$. Thus $X$ is simply connected. The Hurewicz theorem in degree two gives

To compute ${\pi }_3$, choose a map $g:X\to {\mathbb{CP}}^{\infty }=K(\mathbb{Z},2)$ representing $x$. This Eilenberg–MacLane space can be realized by the classifying space of $S^1$. On $S^2$ the map is the standard inclusion; on $S^3$ it is constant, and it extends over the four-cell since ${\pi }_3(K(\mathbb{Z},2))=0$. Its homotopy fibre $T$ is the pullback of the universal circle bundle $S^{\infty }\to {\mathbb{CP}}^{\infty }$. The map on ${\pi }_2$ is an isomorphism, so the long exact sequence of homotopy groups of a fibration gives ${\pi }_1(T)={\pi }_2(T)=0$ and ${\pi }_3(T)\cong {\pi }_3(X)$. Therefore the Hurewicz theorem identifies the latter with $H_3(T;\mathbb{Z})$.

Over $W=S^2\vee S^3$, the circle-bundle total space isThe first summand is the Hopf fibration total space over $S^2$, and the second is the trivial bundle over $S^3$. Their intersection is the fibre over the wedge point. The Mayer–Vietoris sequence gives $H_3(P)=\mathbb{Z}A\oplus \mathbb{Z}B$, with generators from the two three-spheres. These are the lifts of ${\eta }_1$ and $i_2$. The circle bundle over $W$ also has a 2-connected total space, so lifting and the Hurewicz theorem identify $d{\eta }_1+f{i}_2$ with $dA+fB$.

Over the attached four-disc the bundle is trivial. The resulting relative pair has the homology of $(S^1\times D^4,S^1\times S^3)$, so $H_4(T,P)=\mathbb{Z}$ and $H_3(T,P)=0$. Its boundary map sends a generator to the lift of the attaching map, namely $dA+fB$. The long exact sequence in relative homology now givesApplying Smith normal form to this one relation proves the third homotopy group of a Hopf attachment with a sphere summand:Since $d>0$, the greatest common divisor is positive even when $f=0$. This derivation uses the required map to an Eilenberg–MacLane space and determines the extension, rather than merely the orders of its pieces.

Choose a contractible free $G$-space $EG$, with $BG=EG/G=K(G,1)$ the classifying space of a discrete group. The Borel construction gives the fibrationBecause the action on the sphere is free, the map $E\to S^{n-1}/G$ has contractible fibre $EG$ and is a homotopy equivalence. The quotient is a closed $(n-1)$-dimensional manifold, so $H^k(E;\mathbb{F})=0$ for $k>n-1$.

The hypothesis on integral sphere cohomology makes the $G$-action on $H^{n-1}(S^{n-1};\mathbb{F})$ trivial as well. Thus the Serre spectral sequence has just two rows, both copies of $H^{\ast }(BG;\mathbb{F})$, in fibre degrees zero and $n-1$. If $u$ is the fibre generator, the only possible differential is $d_n$, and its transgression definesThe multiplicative differential rule gives $d_n(au)=(-1{)}^{|a|}a⌣\Delta$. Up to the graded sign, this is multiplication by $\Delta$. For $i>0$, the terms $E_{\infty }^{i,n-1}$ and $E_{\infty }^{i+n,0}$ must both vanish because their total degrees exceed $n-1$. The first vanishing says the relevant multiplication map has zero kernel; the second says it has zero cokernel. ThereforeThis is periodic group cohomology from a free sphere action. Degree zero was excluded for a reason: $E_{\infty }^{0,n-1}$ can contribute to the quotient's top cohomology.

For $C_p=\mathbb{Z}/p$, use its periodic free resolution with alternating maps $g-1$ and $N=1+g+\cdots +g^{p-1}$. Applying ${\mathrm{Hom}}_{\mathbb{Z}[C_p]}(-,{\mathbb{F}}_p)$ makes every differential zero, so $H^i(B{C}_p;{\mathbb{F}}_p)$ is one-dimensional in every nonnegative degree. The two-step shift of the resolution supplies a nonzero degree-two class $u$ whose cup product gives the periodicity isomorphisms. For odd $p$, let $v$ be a nonzero degree-one class. Graded commutativity gives $v^2=0$, and the shift gives nonzero $v{u}^k$ and $u^k$ in every degree. For $p=2$, the two resolution maps agree over ${\mathbb{F}}_2[C_2]$, giving a one-step shift whose degree-one class has nonzero powers. Hence the cohomology ring of a finite cyclic group over its prime field isHere the exterior factor is an exterior algebra and the polynomial factor is a polynomial ring. For odd $p$, $u$ may be chosen as the Bockstein homomorphism of $v$.

If $G$ contained $C_p\times C_p$, restriction would give that subgroup the same free, cohomologically trivial sphere action. It would therefore also have period $n$ in positive-degree cohomology. But $B(C_p\times C_p)\simeq B{C}_p\times B{C}_p$, and the Künneth theorem givesThese dimensions strictly increase, so the groups in degrees $i$ and $i+n$ cannot be isomorphic. Thus $G$ contains no subgroup isomorphic to $\mathbb{Z}/p\times \mathbb{Z}/p$ for any prime $p$.

The quaternionic projective space ${\mathbb{HP}}^n$ is the space of one-dimensional right quaternion subspaces of ${\mathbb{H}}^{n+1}$. Equivalently it is the quotient of the unit sphere $S^{4n+3}$ by simultaneous right multiplication by unit quaternions. Its coordinate filtration has one open cell ${\mathbb{H}}^k\cong {\mathbb{R}}^{4k}$ in each dimension $4k$, for $0\le k\le n$. Hence its cellular cohomology is $\mathbb{Z}$ in those dimensions and zero otherwise.

Let $\mathcal{H}$ be the quaternionic tautological line bundle. Its unit sphere bundle is $S^{4n+3}\to {\mathbb{HP}}^n$, with fibre $S^3$. The Gysin sequence of a sphere bundle shows that multiplication by its Euler class $e\in H^4$ is an isomorphism from $H^{4k}$ to $H^{4k+4}$ for $k<n$. Choose the generator $u=-e$. Its powers generate every nonzero positive degree, giving the cohomology ring of quaternionic projective spaceThis also accounts for $n=0$.

First take $n=2$. Under the coordinate inclusion $i:{\mathbb{CP}}^2\to {\mathbb{HP}}^2$, the pulled-back quaternionic line is the quaternionic extension of the complex tautological line $L$. As a complex rank-two bundle it is $L\oplus \overline{L}$: a transition scalar $c\in \mathbb{C}$ acts on the two complex coordinates of a quaternion by $c$ and $\overline{c}$. Put $x=c_1(L^{\ast })$, the degree-two generator of the cohomology ring of complex projective space. The Whitney sum formula for Chern classes givesHere the Euler class of a complex vector bundle is its top Chern class, using the complex orientation. In particular this degree-four pullback has coefficient one; it is not a multiple of larger absolute value. The compatible tautological bundles on the projective filtrations give the same equality for every $n$, and multiplicativity then determines the whole ring map:It is zero whenever $2k>n$. These facts are the complex inclusion into quaternionic projective space.

For an odd prime $p$, the Steenrod reduced powers are natural stable cohomology operationsThey satisfy $P^0=1$, the Cartan formula, $P^{i}a=0$ when $2i>|a|$, and $P^{i}a=a^p$ when $|a|=2i$. In particular, on ${\mathbb{CP}}^{\infty }$ one has $P^{1}x=x^p$ and $P^{i}x=0$ for $i>1$. The Cartan formula and the binomial theorem givePass to the infinite projective spaces, where $i^{\ast }:H^{\ast }({\mathbb{HP}}^{\infty };{\mathbb{F}}_p)\to H^{\ast }({\mathbb{CP}}^{\infty };{\mathbb{F}}_p)$, $u\mapsto x^2$, is injective. The equality just obtained determines the Steenrod powers on quaternionic projective space; restricting to the finite spaces giveswith coefficients modulo $p$ and powers above $n$ set to zero. In particular $P^{1}u=2u^{(p+1)/2}$, $P^{2}u=u^p$, and $P^{i}u=0$ for $i>2$. The infinite-space argument matters: the finite inclusion cannot detect those degrees for which $x^{2k}=0$.

Finally put $Q={\mathbb{HP}}^{15}/{\mathbb{HP}}^{12}$ and $S={\Sigma }^{48}{\mathbb{HP}}^3$. Both have reduced cohomology $\mathbb{Z}$ in degrees $52,56,60$ and zero otherwise. Choose integral generators $a_{13},a_{14},a_{15}$ for $Q$ whose pullbacks under the quotient map are $u^{13},u^{14},u^{15}$, and suspended integral generators $b_1,b_2,b_3$ for $S$ from $u,u^2,u^3$.

Use $p=5$. Naturality for the quotient and the formula above giveStability under the suspension isomorphism instead givesAny homotopy equivalence $h:Q\to S$ would induce isomorphisms on the rank-one integral groups, so $h^{\ast }{b}_1=\epsilon a_{13}$ and $h^{\ast }{b}_3=\delta a_{15}$ with $\epsilon ,\delta \in \{1,-1\}$. Reducing modulo five and commuting $h^{\ast }$ with $P^1$ would require $\epsilon =2\delta$ in ${\mathbb{F}}_5$. Neither $1$ nor $-1$ equals $2$ or $-2$ modulo five. ThereforeThe essential point is that integral generator signs constrain Steenrod comparisons. Arbitrary changes of basis over ${\mathbb{F}}_5$ could rescale these two nonzero coefficients into agreement; a genuine equivalence must also preserve the integral lattices, where only the two signs are available.