The verdict

XI-F's $\pm \mathrm{d}t$ was the conormal of the step $\theta (y^0)$ in the frozen kernel. The class forgets the step and keeps its two null halves $k_{1♭}+k_{2♭}=2\mathrm{d}t$. Representatives in ${\mathcal{I}}_{\mathrm{sol}}+u$ can move the halves off the diagonal, but by Theorem thm:criterion(a) they cannot remove them.