Proposition: Universal relations carry no defect

For every $T\in {\mathcal{D}}^{\prime }(X)$, $a^{(j)}\in H_T$. Consequently $a^{(j)}\in H$ and ${d_0}^T(a^{(j)})=0$ for every extension $T$ of every $T^{\circ }$. More generally, if $a\in {{\mathcal{P}}_r}^{\mathrm{jet}}$ satisfies ${\beta }_{\gamma }(a)=0$ in $C^{\infty }(X)$ for all $|\gamma |\le r$, then $a\in H_T$ for every $T$ and ${d_0}^T(a)=0$.

Proof. From (eq:beta), ${\beta }_0(a^{(j)})={a^{(j)}}_0(x,0)-{\partial }_{q_j}{a^{(j)}}_{e_j}(x,0)=1-1=0$, ${\beta }_{e_j}(a^{(j)})={a^{(j)}}_{e_j}(x,0)=q_j{|}_{q=0}=0$, and all other ${\beta }_{\gamma }(a^{(j)})$ vanish because the corresponding $a_{\alpha }$ vanish. Thus ${\beta }_{\gamma }(a^{(j)})T=0$ for every $T$ and (eq:Hfull) gives $a^{(j)}\in H_T$. The general statement is the same argument.