Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 114 / 2

A direct system of abelian groups over the directed poset $A$ consists of groups $G_a$ and maps ${\varphi }_{ab}:G_a\to G_b$ for $a\le b$, satisfying ${\varphi }_{aa}=1$ and ${\varphi }_{bc}{\varphi }_{ab}={\varphi }_{ac}$. Its direct limit is the quotient of ${⨁}_a{G}_a$ by the relations $g_a\sim {\varphi }_{ab}(g_a)$.

For the displayed sequence, put $P_0=1$ and $P_n=a_0{a}_1\cdots a_{n-1}$. Map the copy of $\mathbb{Z}$ at stage $n$ to $\mathbb{Q}$ byThis is compatible with the next transition because $a_{n}m/P_{n+1}=m/P_n$. The universal property of the direct limit therefore identifies it withIn reduced form, these are exactly the rationals whose denominator divides one of the finite products $P_n$. This is the sequential direct limit of multiplication maps on the integers.

Every singular simplex has compact image, and a singular chain is a finite sum of simplices. The image of any chain is therefore compact and lies in some $X_a$. Directedness puts any finite collection of chains into one common $X_b$, soBecause filtered colimits of abelian groups are exact, kernels and images commute with this colimit. Taking homology gives the homology of a directed union:

For an open $U\subseteq {\mathbb{R}}^N$, use the directed family of finite unions of closed rational cubes contained in $U$. Every compact subset of $U$ lies in one such finite polyhedron, and each polyhedron has finitely generated cellular homology. There are only countably many of them, so their direct limit is countable. Thus every $H_i(U;\mathbb{Z})$ is countable.

Cohomology behaves differently because $\mathrm{Hom}$ turns a direct sum into a direct product. The connected open sethas one independent loop around each puncture, so $H_1(U;\mathbb{Z})\cong {⨁}_{n\ge 1}\mathbb{Z}$. Since $H_0(U;\mathbb{Z})=\mathbb{Z}$, the universal coefficient theorem for cohomology giveswhich is uncountable. This is the first cohomology of the countably punctured plane.

List the prime numbers without repetition as $p_0,p_1,p_2,\dots$. Choose mapsand let $T$ be their mapping telescope. A finite initial telescope deformation retracts onto its last sphere. The inclusions of successive finite telescopes induce multiplication by $p_i$ on degree-$n$ reduced homology. The homology of a directed union and the sequential direct limit of multiplication maps on the integers therefore giveEvery finite product $p_0\cdots p_r$ is square-free, and every square-free denominator divides one such product. Henceas in the mapping-telescope realization of the rational group with square-free denominators.