The homology universal-coefficient sequence splits nonnaturally

By Boundaries and cycles in a free complex over a PID are free under AC, $B_{n-1}$ and $B_{n-2}$ are free abelian. Choose a basis of $B_{n-1}$ and a preimage in $C_n$ of each basis element. AC supplies these choices, which extend to a homomorphism $s:B_{n-1}\to C_n$ with $d_{n}s=\mathrm{id}$. Hence $C_n=Z_n\oplus s(B_{n-1})$. The same argument gives $C_{n-1}=Z_{n-1}\oplus s^{\mathrm{\prime }}(B_{n-2})$, so the inclusion $Z_{n-1}\otimes G\to C_{n-1}\otimes G$ is injective.

The free presentation $0\to B_{n-1}\stackrel{i}{\to }{Z}_{n-1}\to H_{n-1}(C)\to 0$ identifies $T:=\ker (i\otimes 1:B_{n-1}\otimes G\to Z_{n-1}\otimes G)\cong {\mathrm{Tor}}_1^{\mathbb{Z}}(H_{n-1}(C),G)\mathrm{.}$ For $t\in T$, the element $(s\otimes 1)(t)$ is a cycle in $C_n\otimes G$: its differential is the image of $(i\otimes 1)(t)=0$ in $C_{n-1}\otimes G$. Thus $\sigma :T⟶H_n(C\otimes G),t⟼[(s\otimes 1)(t)]$ is a well-defined homomorphism. The UCT quotient map of the cited theorem first takes a cycle to its $B_{n-1}\otimes G$ component under $C_n\otimes G=(Z_n\otimes G)\oplus (B_{n-1}\otimes G)$ and then identifies that component with $T$. Consequently its composite with $\sigma$ is the identity on $T$, giving the asserted splitting. The chosen section $s$ need not be preserved by chain maps, so this construction makes no naturality claim.