Universal coefficients split naturally

Suppose there are splitting homomorphisms $s_G:\mathrm{Hom}(M(G),A)\to H^2(G;A)$ natural in $G$, with $p_G{s}_G=\mathrm{i}\mathrm{d}$. For every subgroup inclusion $i:L\hookrightarrow V$ with $L\cong C_2$, [L2] gives $M(L)=0$, so naturality forces $i^{\ast }{s}_V(\lambda )=s_L(\lambda \circ M(i))=s_L(0)=0$ for every $\lambda :M(V)\to A$.

The map $b((x_1,x_2),(y_1,y_2))=x_1{y}_2+x_2{y}_1$ is alternating and bilinear to $A$. It takes value $1$ on the standard basis pair, so [L3] supplies a nonzero homomorphism $\lambda :M(V)\to A$. Put $\alpha =s_V(\lambda )$. Then $p_V(\alpha )=\lambda \mathrm{\ne }0$.

Represent $\alpha$ by a central extension $1\to A\to E\stackrel{\pi }{\to }V\to 1$. Its restriction to each of the three order-two subgroups $L\le V$ is zero by step 1.1. Hence each preimage ${\pi }^{-1}(L)$ is a split central extension, isomorphic to $C_2\times C_2$, and every element of that preimage has square $1$.

Every element of $E$ either belongs to the kernel $A$, or maps to a nonzero vector of $V$ and therefore belongs to one of these three preimages. Thus every element of $E$ has square $1$. For $x,y\in E$, this gives $xy=(xy{)}^{-1}=y^{-1}{x}^{-1}=yx$, so $E$ is abelian. Choose lifts $u,v$ of the two standard basis vectors of $V$. Since $u^2=v^2=1$ and $uv=vu$, the map $(a,b)\mapsto u^a{v}^b$ is a homomorphic section of $\pi$.

This section makes $\alpha =0$, contradicting $p_V(\alpha )=\lambda \mathrm{\ne }0$. Therefore no splitting can be natural in the group variable even for the fixed coefficient group $C_2$, and in particular none is natural in both variables. Under the Axiom of Choice, the individual sequences split after choices by the separate splitting theorem cited in [L1].