have that $\phi(m) s(f) \phi(n)^* = t(m \otimes f \otimes n^*)$ and that $\phi(n)^* t(e) \phi(m) = s(n^* \otimes e \otimes m)$. Consequently we obtain $${\left[ \phi(M) s(F) \phi(M)^* \right]^{-\|\cdot\|}} \subseteq {\left[ t(E) \right]^{-\|\cdot\|}}$$ and $${\left[ \phi(M)^* t(E) \phi(M) \right]^{-\|\cdot\|}} \subseteq {\left[ s(F) \right]^{-\|\cdot\|}}.$$ Since $A$ and $B$ act non-degenerately we obtain equalities.

To avoid technical notation we will show the second item just when $\pi$ is injective (and thus $t$ is isometric). For the general case substitute $E$ by ${\left[ t(E) \right]^{-\|\cdot\|}}$ and $F$ by ${\left[ s(F) \right]^{-\|\cdot\|}}$ in what follows. We will also make the simplifications $$t(E) = E \subseteq {{\mathfrak{L}}}(E) \subseteq {\mathrm{C}^*}(\pi,t).$$ Suppose that $(\pi,t)$ admits a gauge action $\{\be_z\}$. Then every $\be_z$ induces a representation $(\be_z|_A, \be_z|_E)$ on $E$. Since $F \simeq M^* \otimes_A E \otimes_A M$ we can define the mapping $$\ga_z = {{\operatorname{id}}}_{M^*} \otimes \be_z \otimes {{\operatorname{id}}}_{M} \colon F \to F$$ where for simplicity we don’t write the unitary of the equivalence. Then $({{\operatorname{id}}}_B, \ga_z)$ induces an injective representation of $F$. We have to show that it induces a gauge action on the entire ${\mathrm{C}^*}(\si, s)$. For the TRO $M$ fix the net $n_\la^* = [(n_1^\la)^*, \dots, (n_{k_\la}^\la)^*]$ such that $$\lim_\la m n_\la^* n_\la 