Theorem: Annihilator criterion for the null defect

Let $T$ extend $T^{\circ }$. Then $H={\beta }^{-1}(\mathrm{Ann}(T^{\circ }{)}^{M_r})$, $H_T={\beta }^{-1}(\mathrm{Ann}(T{)}^{M_r})$, and $\beta$ induces a linear isomorphism

In particular ${d_0}^T=0$ if and only if $\mathrm{Ann}(T)=\mathrm{Ann}(T^{\circ })$.

Proof. The descriptions of $H$ and $H_T$ are (eq:H)–(eq:HT) read through $\beta$. Since $\ker \beta \subseteq H_T\subseteq H$, $\beta$ induces a well-defined linear map $H/H_T\to \mathrm{Ann}(T^{\circ }{)}^{M_r}/\mathrm{Ann}(T{)}^{M_r}$. It is injective: if $\beta (a)\in \mathrm{Ann}(T{)}^{M_r}$ then $a\in H_T$. It is surjective by Lemma lem:surj: for $g\in \mathrm{Ann}(T^{\circ }{)}^{M_r}$, $\sigma (g)\in H$ and $\beta (\sigma (g))=g$. Finally $\mathrm{Ann}(T^{\circ }{)}^{M_r}/\mathrm{Ann}(T{)}^{M_r}\cong (\mathrm{Ann}(T^{\circ })/\mathrm{Ann}(T){)}^{M_r}$. The last sentence is Proposition prop:recalled.