Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 127 (Homotopy Theory)

For a based space $(X,x_0)$, the homotopy groupis the set of based homotopy classes, with its usual concatenation operation. A map $f:X\to Y$ is a weak homotopy equivalence when it induces a bijection on path components and an isomorphismfor every $k\ge 1$ and every basepoint $x$. It is an n-connected map when it is bijective on ${\pi }_k$ for $k<n$ and surjective on ${\pi }_n$; equivalently, every homotopy fiber is $(n-1)$-connected.

A CW complex is built from a discrete set of zero-cells by successively attaching $k$-discs along maps from their boundary spheres, with the weak topology and closure-finiteness conditions. Its filtration by skeleta is the CW filtration.

For any space $X$, form its singular simplicial set $\mathrm{Sing}X$. Its geometric realization of a simplicial set is a CW complex, with one cell for each nondegenerate singular simplex, and evaluation givesThe Simplicial approximation theorem identifies based maps and homotopies from finite simplicial spheres into $|\mathrm{Sing}X|$ with singular simplices in $X$. Consequently $ϵ$ induces a bijection on components and isomorphisms on all homotopy groups. Thus every space admits a CW approximation.

The vanishing assumptions do not permit removal of all $n$-cells. Take $n=2$ andThen ${\pi }_2(X)=0$, and homology of a finite cyclic group gives $H_2(X;\mathbb{Z})=0$. If a connected CW complex $Y$ had no two-cells, attaching cells of dimension at least three would not change the fundamental group of its one-skeleton. Hence ${\pi }_1(Y)$ would be a free group. A weak equivalence $Y\to X$ would instead give ${\pi }_1(Y)\cong \mathbb{Z}/r$, which is nontrivial and finite and therefore not free. No such $Y$ exists.

A map $p:E\to B$ is a Serre fibration when it has the homotopy lifting property for every disc: given $g:D^k\to E$ and $H:D^k\times I\to B$ with $pg=H(-,0)$, there is a lift $\tilde{H}:D^k\times I\to E$ satisfying $\tilde{H}(-,0)=g$ and $p\tilde{H}=H$.

Fix $b_0\in B$, $e_0\in F=p^{-1}(b_0)$, and write $i:F\hookrightarrow E$. The long exact sequence of homotopy groups of a fibration isThe first two maps are induced by inclusion and projection. To define $\partial$, represent a class of ${\pi }_k(B)$ by a map of pairs $(D^k,S^{k-1})\to (B,b_0)$, lift it beginning at $e_0$ along radial paths, and restrict the lift to $S^{k-1}$; that restriction lies in $F$. At the bottom, exactness continues through the pointed sets ${\pi }_1(B)\to {\pi }_0(F)\to {\pi }_0(E)\to {\pi }_0(B)$.

Every fiber bundle is a Serre fibration. Pull a bundle back along $H:D^k\times I\to B$; a lift of $H$ is the same as a section of this pullback extending the section over $D^k\times \{0\}$ supplied by $g$. Since $D^k\times I$ is compact, finitely many bundle charts cover it. A Lebesgue-number subdivision of $I$, followed by a finite subdivision of $D^k$, makes each resulting prism lie in one chart. In a trivialization $U\times F\to U$, extend the section across a prism by keeping its $F$-coordinate constant along the interval direction. Proceed prism by prism and time-slab by time-slab; on an already treated face use its prescribed coordinate, and the transition functions ensure agreement on overlaps. The resulting sections glue to the required lift $\tilde{H}$.

Now consider $i:{\mathbb{RP}}^n\hookrightarrow {\mathbb{CP}}^n$. The target is simply connected, so $i_{\ast }$ is zero on ${\pi }_1$. For $k\ge 2$, every based map $S^k\to {\mathbb{RP}}^n$ lifts through the double cover to $S^n$. The compositelifts through the Hopf fibration $S^{2n+1}\to {\mathbb{CP}}^n$ to the real-coordinate inclusion $S^n\hookrightarrow S^{2n+1}$. This inclusion is null-homotopic because ${\pi }_n(S^{2n+1})=0$. Hence $i_{\ast }$ is zero on every ${\pi }_k$, including the cases $n=1$ and $k\ge 2$ where the source groups already vanish.

Because $S^n$ is simply connected for $n\ge 2$, the homological Serre spectral sequence has constant coefficients and only two nonzero columns:Its only possible nonzero differential isUsing the orientation generator of $H_n(S^n)$ to identify both columns with $H_{\ast }(F)$ defines the Wang homomorphismThe kernel and cokernel descriptions of the two surviving columns splice with the filtration of $H_{\ast }(E)$ to give the Wang sequence over a sphere

Let $F_m$ be the homotopy fiber of a degree-$m$ map $f_m:S^n\to S^n$. Apply this sequence to the fibrationAt the bottom, the map between the two copies of $H_n(S^n)$ is multiplication by $m$. It follows thatThe first torsion group can also be seen from ${\pi }_{n-1}(F_m)\cong \mathbb{Z}/m$ and the Hurewicz theorem; the Wang map then propagates it periodically.

The universal coefficient theorem for cohomology givesEvery product of two positive-degree classes is zero. For $n>2$, this follows immediately because the sum of two degrees of the form $k(n-1)+1$ is not of that form. For $n=2$, degree counting does not suffice, since $H^j(F_m)\cong \mathbb{Z}/m$ for every $j\ge 2$. Use instead the multiplicative cohomological Serre spectral sequence forIts transgression in degree one is multiplication by $m$. Every positive integral cohomology class that survives lies in filtration two, while the product of two such classes lies in filtration four; the base has dimension two, so filtration four is zero. Thus the reduced cohomology is a square-zero ideal, andas a graded ring, with zero multiplication on the second summand.

The space $K(\mathbb{Z}/n,1)$ is a classifying space $B(\mathbb{Z}/n)$. The periodic resolution of a finite cyclic group givesThus positive odd cohomology vanishes and every positive even group is $\mathbb{Z}/n$.

Let $p$ be odd. If $p\nmid n$, transfer makes multiplication by $n$ both zero and invertible on positive-degree cohomology, soIf $p\mid n$, restriction to the cyclic Sylow $p$-subgroup and transfer giveFor $n=p$, one may take $v=\beta u$, where $\beta$ is the mod-$p$ Bockstein homomorphism.

For the second part put $A=\mathbb{Z}/p$ and $X=K(A,1)\times K(A,1)$. The long exact homotopy sequence of the homotopy fibre of $f:X\to K(A,2)$ givesand ${\pi }_k(F)=0$ for $k>1$. Therefore $F$ is a $K(G,1)$ and $|G|=p^3$.

Regard it as the fibrationWriteand write $w,t$ for the degree-one and degree-two generators of the fibre. The fibration is classified by $u_1{u}_2$, so in its cohomological Serre spectral sequenceThe Kudo transgression theorem and the mod-$p$ Bockstein giveConsequently $H^1(F;{\mathbb{F}}_p)$ has basis $u_1,u_2$, while $H^2(F;{\mathbb{F}}_p)$ has the four surviving classes represented byIf $G$ were abelian, an abelian group of order $p^3$ mapping onto $A^2$ would be either $A^3$ or $\mathbb{Z}/p^2\times A$. The first has three-dimensional $H^1(-;{\mathbb{F}}_p)$; the second has two-dimensional $H^1$ but three-dimensional $H^2$. Both contradict the dimensions just calculated. Hence $G$ is nonabelian.

Let $i:F\hookrightarrow E$. A pairis transgressive pair when, in the long exact sequence of the pair $(E,F)$,where $H^{r+1}(B,b_0)$ is identified with reduced cohomology. In the Serre spectral sequence, this says that $x$ survives to the transgression andunder the edge identifications, modulo the usual earlier-differential indeterminacy.

The Kudo transgression theorem says that if $(x,y)$ is transgressive and $0\le j\le r$, thenis transgressive. To prove it, use relative Steenrod squares. Naturality givesand stability, equivalently compatibility with the suspension isomorphism, makes squares commute with the connecting map:Applying ${\mathrm{Sq}}^j$ to $\delta x={\pi }^{\ast }y$ proves the theorem. The properties used are naturality, stability, additivity, and the instability conditions ${\mathrm{Sq}}^{j}z=0$ for $j>|z|$ and ${\mathrm{Sq}}^{|z|}z=z^2$.

Let $x\in H^1({\mathbb{RP}}^{\infty };{\mathbb{F}}_2)$ be the generator. Instability givesThe Cartan formula says the total square is multiplicative, soComparing components yields the complete formulawith the binomial coefficient reduced modulo two.

Finally, ${\mathrm{Sq}}^1$ is the Bockstein homomorphism associated withIf a mod-two cocycle representing $y$ is lifted to an integral cochain $A$, write $dA=2B$. Then $B$ modulo two represents ${\mathrm{Sq}}^{1}y$. But $dB=0$, because integral cochains are torsion-free and $2dB=d^{2}A=0$. Thus $B$ itself is a cocycle lift, so its Bockstein vanishes. Thereforefor every space $Y$ and every $y\in H^{\ast }(Y;{\mathbb{F}}_2)$.