Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 165 (Simplicial Homotopy Theory)

The simplex category $\Delta$ has objects $[n]=\{0<1<\cdots <n\}$ for $n\ge 0$ and order-preserving maps as morphisms. A simplicial set is a functor $X:{\Delta }^{\mathrm{op}}\to \mathbf{Set}$.

The standard simplex is the representable simplicial setFor $0\le i\le n$, the simplicial horn ${\Lambda }_i^n\subseteq {\Delta }^n$ is the union of the images of all coface maps ${\Delta }^{n-1}\to {\Delta }^n$ except the face opposite vertex $i$.

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A quasicategory is a simplicial set having the right lifting property against every inner horn inclusionA Kan complex has the right lifting property against every simplicial horn inclusion, including the two outer horns $i=0,n$.

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The Yoneda lemma and the definition of the nerve of a category giveThus such a map is precisely a diagram of objects and composable morphismsin $\mathcal{C}$. The remaining edges and higher faces record the composites forced by this string.

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Consider a lifting square with ${\Lambda }_1^3\hookrightarrow {\Delta }^3$ on the left, $Q\to N(\mathcal{C})$ on the right, and prescribed bottom simplex $\sigma :{\Delta }^3\to N(\mathcal{C})$. Since $Q$ is a quasicategory, the top horn has some filler $\tilde{\sigma }:{\Delta }^3\to Q$.

The two simplices $p\tilde{\sigma }$ and $\sigma$ of the nerve of a category restrict to the same inner horn. Inner horns in a category's nerve have unique fillers, because composition in a category is defined and unique. Hence $p\tilde{\sigma }=\sigma$, so $\tilde{\sigma }$ is the required lift.

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The contravariant simplicial mapping space functor takes the given pushout to the stated strict pullback. Since $X$ is a Kan complex, all four mapping spaces are Kan complexes. Moreover, the monomorphism $A\hookrightarrow B$ induces a Kan fibrationIndeed, a lifting problem against a horn is adjoint to a lifting problem for $X$ against the pushout-product of $A\hookrightarrow B$ with that horn inclusion; this pushout-product is an anodyne monomorphism, and $X$ fills it.

A strict pullback of fibrant simplicial sets along a fibration computes the homotopy pullback. The displayed pullback square is therefore also a homotopy pullback square.

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For a simplicial abelian group $A$, its normalized chain complex of a simplicial abelian group isThe simplicial identities give ${\partial }^2=0$. Equivalently, $N_{n}A$ is the quotient of $A_n$ by the subgroup generated by degenerate simplices, with differential induced by ${\sum }_i(-1{)}^i{d}_i$.

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The Dold–Kan correspondence says thatis an equivalence from simplicial abelian groups to nonnegatively graded chain complexes of abelian groups. The restriction of the right adjoint $K$ to ${\mathbf{Ch}}_{\ge 0}(\mathbf{Ab})$ is a quasi-inverse: both the unit $C_{\ast }\to N_{\ast }K(C_{\ast })$ and counit $K(N_{\ast }A)\to A$ are natural isomorphisms.

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Under the Dold–Kan correspondence, a $1$-simplex of $K(C_{\ast })$ is a pair $(a,x)\in C_0\oplus C_1$ withThe horn ${\Lambda }_1^2$ consists of the edges $01$ and $12$, joined at vertex $1$. A map from it is therefore a pair of composable $1$-simplices, equivalently a triplewhere the first edge is $(a,x)$ and the second is $(a+{\partial }_{1}x,y)$. Thus

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Let $A[n]$ be the nonnegative chain complex having $A$ in degree $n$, zero in every other degree, and zero differential. The simplicial Eilenberg–MacLane space isIts underlying simplicial set is a Kan complex, with ${\pi }_{n}K(A,n)\cong A$ and all other positive homotopy groups zero.

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Regard the given short exact sequence as a degreewise short exact sequence of chain complexes concentrated in degree $n$. The inverse functor in the Dold–Kan correspondence is exact, so it produces a degreewise short exact sequence of simplicial abelian groupsThe last map is degreewise surjective and hence a Kan fibration. Its strict fiber is $K(A_1,n)$, and a strict fiber of a fibration computes the homotopy fiber. This proves the asserted homotopy fiber sequence of pointed Kan complexes.

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Products of Eilenberg–MacLane spaces satisfywhere the last isomorphism uses the Chinese remainder theorem. This space is $2$-connected, so the Hurewicz theorem identifiesThe assumed surjection on $H_3$ is an isomorphism because the group is finite. Hence $f$ is an isomorphism on ${\pi }_3$; all other homotopy groups of the source and target vanish. Thus $f$ is a weak homotopy equivalence, and the Whitehead theorem for Kan complexes makes it a homotopy equivalence.

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Use the path-loop fibrationand its rational Serre spectral sequence. The total space is contractible, while $H^{\ast }(S^4;\mathbb{Q})$ is $\mathbb{Q}$ in degrees $0$ and $4$ and zero otherwise. The only possible nonzero differential isConvergence to the cohomology of a point first forces $d_4:H^3(\Omega S^4;\mathbb{Q})\to H^4(S^4;\mathbb{Q})$ to be an isomorphism, and then inductively forces an isomorphism from each nonzero vertical group to the group three degrees below it in the other column. Therefore

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Apply the rational Serre spectral sequence to the path-loop fibration of $K(\mathbb{Z},3)$. Its fiber iswhose rational cohomology ring is $\mathbb{Q}[c]$ with $|c|=2$. Since the path space is contractible, $c$ must transgress to a nonzero class $x\in H^3(K(\mathbb{Z},3);\mathbb{Q})$. Multiplicativity givesOver $\mathbb{Q}$ these differentials pair and kill every positive-degree class except $x$, while graded commutativity gives $x^2=0$. Hence

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The loop-space shift of homotopy groups givesThus $\Omega S^4$ is $2$-connected and its first nonzero homotopy group isThe Hurewicz theorem now givesand every positive integral homology group below degree $3$ vanishes. The smallest requested degree is therefore $i=3$.

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The rational cohomology from part i has one generator in every degree divisible by $3$. Its connected graded-commutative Hopf algebra structure isas a graded vector-space-compatible algebra: the odd class has square zero and the degree-six class supplies the even multiples. The rational Hurewicz and Hopf-algebra correspondence for a connected loop space identifies the indecomposable generators with the duals of its rational homotopy groups. Hence, for $0\le i\le 6$,This also agrees with the rational homotopy groups of a sphere and the loop-space shift of homotopy groups.

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By the loop-space shift of homotopy groups,Part iv therefore gives

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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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