Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 165 / 4

Write $F=K(\mathbb{Z}/4,1)$, $E=K(\mathbb{Z},2)$, and $B=K(\mathbb{Z},2)$. In total degree at most two, the $E_2$ page of the mod-two Serre spectral sequence hasand $E_2^{1,0}=E_2^{1,1}=0$. A periodic free resolution of the cyclic group gives $H_1(F;\mathbb{Z})=\mathbb{Z}/4$ and $H_2(F;\mathbb{Z})=0$. Hence $H^1(F;{\mathbb{F}}_2)\cong \mathrm{Hom}(\mathbb{Z}/4,{\mathbb{F}}_2)$, while the universal coefficient theorem for cohomology gives $H^2(F;{\mathbb{F}}_2)\cong \mathrm{Ext}(\mathbb{Z}/4,{\mathbb{F}}_2)$; both are one-dimensional.

The edge map $H^2(B;{\mathbb{F}}_2)\to H^2(E;{\mathbb{F}}_2)$ is induced by multiplication by $4$ and is therefore zero modulo two. Consequentlyis an isomorphism. The differential out of $E_2^{0,2}$ is zero, because the total space has a one-dimensional $H^2$ which must survive in filtration zero. Thus for every $r\ge 3$ and $p+q\le 2$,and all other groups in that range vanish.

It follows thatThe surviving filtration-zero class is the restriction of the degree-two class of $E$, so

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Let $P$ be the homotopy fiber of the map representingThe long exact sequence of homotopy groups shows that every ${\pi }_i(P)$ except ${\pi }_1(P)$ vanishes and thatHence $P$ is either $K({\mathbb{F}}_2\oplus {\mathbb{F}}_2,1)$ or $K(\mathbb{Z}/4,1)$.

The extension is classified by the degree-two class represented by the original map. If $t\in H^1(K({\mathbb{F}}_2,1);{\mathbb{F}}_2)$ is the standard generator, that class isin $H^{\ast }({\mathbb{RP}}^{\infty };{\mathbb{F}}_2)={\mathbb{F}}_2[t]$. It therefore classifies the non-split extension, whose middle group is $\mathbb{Z}/4$. Thus

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The group $H^1(K(\mathbb{Z}/4,1);{\mathbb{F}}_2)$ is one-dimensional, so it suffices to consider its nonzero class $x$. This class is represented by the quotient homomorphism $\mathbb{Z}/4\to \mathbb{Z}/2$, which lifts to the identity homomorphism with coefficients in $\mathbb{Z}/4$. Its Bockstein homomorphism fortherefore vanishes. Since the first Steenrod square is this Bockstein and ${\mathrm{Sq}}^{1}x=x^2$ for every degree-one class,The zero degree-one class plainly has square zero as well.

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