The UCT splitting is not natural

Integrally $H_1(Y)=0$ and $H_2(Y)=\mathbb{Z}$ by [F2], so ${\mathrm{Ext}}^1(H_1(Y),G)=0$ (use the zero resolution), and evaluation ${\beta }_Y:H^2(Y;G)\to \mathrm{Hom}(\mathbb{Z},G)=G$ is an isomorphism. Write $u:\mathbb{Z}\to G$ for reduction modulo two, the nonzero element of this Hom group. There is a unique nonzero $a={\beta }_Y^{-1}u$. Every section of ${\beta }_Y$ must send $u$ to $a$.

Integrally $H_1(X)=\mathbb{Z}\mathrm{/}2$ and $H_2(X)=0$ by [F1]. The resolution $0\to \mathbb{Z}\stackrel{2}{\to }\mathbb{Z}\to \mathbb{Z}\mathrm{/}2\to 0$ becomes $G\stackrel{0}{\to }G$ on applying Hom into $G$, so ${\mathrm{Ext}}^1(\mathbb{Z}\mathrm{/}2,G)=G$. Thus the UCT sequence for $X$ is $0\to G\stackrel{{\iota }_X}{\to }{H}^2(X;G)\stackrel{{\beta }_X}{\to }0\to 0$. In particular all of $H^2(X;G)$ is in its Ext image.

The actual map $q^{\ast }:H^2(Y;G)\to H^2(X;G)$ is an isomorphism by [F1], so $q^{\ast }a\mathrm{\ne }0$. In contrast, the map of Hom terms is precomposition by the integral homology map $q_{\ast }:H_2(X;\mathbb{Z})=0\to H_2(Y;\mathbb{Z})=\mathbb{Z}$; hence it sends $u$ to the zero homomorphism. The nonzero mod-two cohomology map and the zero map of integral-homology Hom terms are different assertions, with their coefficient conventions fixed.

Suppose a family of group-homomorphism sections were natural for continuous space maps. Its naturality square at $q$ would require $q^{\ast }{s}_Y(u)=s_X(u\circ q_{\ast })$. The left side is $q^{\ast }a\mathrm{\ne }0$ by steps 1.1 and 2.1. The right side is $s_X(0)=0$, since $s_X$ is a group homomorphism from the zero Hom group. This is impossible. Therefore no such natural splitting exists even for fixed coefficients $G={\mathbb{F}}_2$ and degree two, and in particular not for all coefficients and spaces.

The contradiction does not deny an individual split sequence: for $Y$ its section is ${\beta }_Y^{-1}$, while for $X$ the section from its zero Hom term is the zero map. It denies compatibility of these sections with the single explicit continuous quotient $q$. The witness is nonempty and finite dimensional, $u(1)=1$, and the zero Hom group on $X$ is essential, not an omitted endpoint. AC is inherited from [F1] field duality and [F2] UCT and comparison; no new choice is used in the two finite resolutions or the naturality contradiction.