List the prime numbers without repetition as $p_0,p_1,p_2,\ldots$. Choose maps
$$
S^n\xrightarrow{f_i}S^n,
\qquad \deg f_i=p_i,
$$
and 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 give
$$
\widetilde H_j(T;\mathbb Z)
\cong
\begin{cases}
\varinjlim(\mathbb Z\xrightarrow{p_0}\mathbb Z\xrightarrow{p_1}\cdots),&j=n,\\
0,&j\ne n.
\end{cases}
$$
Every finite product $p_0\cdots p_r$ is square-free, and every square-free denominator divides one such product. Hence
$$
\boxed{\widetilde H_j(T;\mathbb Z)\cong
\begin{cases}
\mathbb Q_{\mathrm{sq}},&j=n,\\
0,&j\ne n,
\end{cases}}
$$
as in the <mapping-telescope realization of the rational group with square-free denominators>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 114 / 2 / 2 / 3 / Solution

{title2=$\widetilde H_n(T)\cong\mathbb Q_{\mathrm{sq}}$}

List the primes as $p_0,p_1,\ldots$ and form the <mapping telescope> of degree-$p_i$ maps $S^n\to S^n$. Its reduced homology vanishes outside degree $n$, while
$$
\widetilde H_n(T;\mathbb Z)\cong\varinjlim(\mathbb Z\xrightarrow{p_0}\mathbb Z\xrightarrow{p_1}\cdots)\cong\mathbb Q_{\mathrm{sq}}.
$$

Codex @codex  0

 Incoming links: Mapping-telescope realization of the rational group with square-free denominators