Non-naturality of the Künneth splitting

Let \[ M=M(\mathbb Z_m,n)=S^n\cup_m e^{n+1}. \] Its reduced cellular chain complex has \[ 0\longrightarrow \mathbb Z\langle a\rangle \xrightarrow{m}\mathbb Z\langle b\rangle\longrightarrow0, \qquad |a|=n+1,\ |b|=n. \] Let \[ f:M\longrightarrow M/S^n=S^{n+1} \] be the quotient map. On cellular chains \(f(a)=u\), the top cell of \(S^{n+1}\), and \(f(b)=0\). In particular \(f_*\) is zero on all reduced homology groups.

Consider \(M\times M\). In degrees \(2n+2,2n+1,2n\) its cellular chain complex is \[ \mathbb Z\langle a\otimes a\rangle \longrightarrow \mathbb Z\langle a\otimes b,b\otimes a\rangle \longrightarrow \mathbb Z\langle b\otimes b\rangle, \] with \[ \partial(a\otimes a)=m\bigl(b\otimes a+(-1)^{n+1}a\otimes b\bigr), \] \[ \partial(a\otimes b)=m(b\otimes b), \qquad \partial(b\otimes a)=(-1)^n m(b\otimes b). \] Thus the kernel in degree \(2n+1\) is generated by \[ a\otimes b+(-1)^{n+1}b\otimes a, \] and the image from degree \(2n+2\) is \(m\) times this generator. Hence \[ H_{2n+1}(M\times M)\cong\mathbb Z_m. \] This is precisely the Tor term \[ \operatorname{Tor}(H_n(M),H_n(M))\cong\mathbb Z_m \] in the Künneth formula.

For \(S^{n+1}\times M\), the relevant cellular chains are \[ \mathbb Z\langle u\otimes a\rangle \xrightarrow{\pm m} \mathbb Z\langle u\otimes b\rangle, \] so \[ H_{2n+1}(S^{n+1}\times M)\cong\mathbb Z_m. \] The map \(f\times1\) sends \[ a\otimes b+(-1)^{n+1}b\otimes a \longmapsto u\otimes b, \] therefore it induces an isomorphism \[ H_{2n+1}(M\times M)\xrightarrow{\cong} H_{2n+1}(S^{n+1}\times M). \]

If the splitting of the Künneth short exact sequence were natural, the Tor summand in the source would map to the corresponding summands using only the induced maps on the homology of the factors. But \(f_*\) is zero on reduced homology, so naturality would force the displayed map on the Tor contribution to be zero. It is in fact an isomorphism. This contradiction proves that the Künneth splitting cannot be natural.