Why the insertion argument is not an on-shell class

The map constructed below satisfies

which need not vanish. Therefore it is not a map on free on-shell classes in its input. It is a map on the frozen representatives (eq:tests), with output in ${\mathfrak{A}}_0$.

This warning can be witnessed explicitly. For $m=0$, take $h=\psi \eta$, set $r=g\psi$, and choose a compact $r$ with ${\partial }_{t}r\not\equiv 0$. For $u={\partial }_{t}r$, evaluation of the output test against the free solution $t$ gives

Thus even a free-equation-zero insertion can have a nonzero retarded correction. There is no ambiguity here: the geometric prescription supplies the insertion test before the map is applied.