Gaussian transfer is not strictly functorial on arbitrary cochain maps

Retract data for $X^{\bullet }$ onto $Y^{\bullet }$. Write elements of $X^j=k\oplus k$ as pairs with the first coordinate in $Y^j$ and the second in $K^j$, and let $\mathrm{ı}=\left(\genfrac{}{}{0px}{}{1}{0}\right)$, $p=(10)$ in both degrees, with $h^1=\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)$, $h^0=0$ and $h^2=0$. Then $p\mathrm{ı}=1_k$ and $1-\mathrm{ı}p=\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)$. Here $d^0=1_{k^2}$ and $d^1=0$: in degree $1$, $d^0{h}^1+h^2{d}^1=1_{k^2}{h}^1+0=\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)$, and in degree $0$, $d^{-1}{h}^0+h^1{d}^0=0+h^1{1}_{k^2}=\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)$. Thus $1-\mathrm{ı}p=dh+hd$, and the displayed data is a strong deformation retract of $X^{\bullet }$ onto $Y^{\bullet }$.

Two cochain maps. In both nonzero degrees set $f=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$ and $g=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$. At the only nonzero differential $d^0=1_{k^2}$, $f^1=f^0$ and $g^1=g^0$, so $f^1{d}^0=d^0{f}^0$ and $g^1{d}^0=d^0{g}^0$; at $d^1=0$ the cochain-map equations hold trivially. Thus $f$ and $g$ are cochain maps. Here $f$ sends the $Y$-coordinate isomorphically onto the $K$-coordinate and kills the $K$-coordinate, while $g$ sends the $K$-coordinate isomorphically onto the $Y$-coordinate and kills the $Y$-coordinate.

The individual transfers vanish. Since $\mathrm{ı}=\left(\genfrac{}{}{0px}{}{1}{0}\right)$, one has $f\mathrm{ı}=\left(\genfrac{}{}{0px}{}{0}{1}\right)$ and $g\mathrm{ı}=0$; applying $p=(10)$ gives $\bar{f}=pf\mathrm{ı}=0$ and $\bar{g}=pg\mathrm{ı}=0$ as cochain maps $Y^{\bullet }\to Y^{\bullet }$.

The transfer of the composite does not vanish. Since $gf=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)=\mathrm{ı}p$, one has $\overline{(gf)}=p(gf)\mathrm{ı}=p\mathrm{ı}p\mathrm{ı}=p\mathrm{ı}=1_{Y^{\bullet }}$, the identity cochain map of $Y^{\bullet }$, which is nonzero. Hence $\overline{(gf)}-\bar{g}\bar{f}=1_{Y^{\bullet }}\mathrm{\ne }0$, so transfer is not strictly functorial on cochain maps, and the statement refuted is false.

The failure is consistent with the proposition. The complex $Y^{\bullet }$ is contractible, with contracting homotopy ${\bar{h}}^1=1_k$ and ${\bar{h}}^0={\bar{h}}^2=0$, because in degree $0$ one has $d^{-1}{\bar{h}}^0+{\bar{h}}^1{d}^0=1_k$ and in degree $1$ one has $d^0{\bar{h}}^1+{\bar{h}}^2{d}^1=1_k$. Consequently every endomorphism $u$ of $Y^{\bullet }$, in particular the discrepancy $\overline{(gf)}-\bar{g}\bar{f}$ of step 3.1, is null-homotopic: from $1_{Y^{\bullet }}-0=d\bar{h}+\bar{h}d$ and $ud=du$ one obtains $u=u(d\bar{h}+\bar{h}d)=d(u\bar{h})+(u\bar{h})d$, so $u\simeq 0$ with homotopy $u\bar{h}$. This is exactly the up-to-homotopy functoriality asserted by [L1], so the counterexample refutes strictness only, not the homotopy-class statement. ∎