,\, u_{\pi, \lambda^0})$ is a twisted coaction of $H^0$ on $A$.

This is immediate by routine computations.

We call the above pair $(\rho_{\pi, \lambda^0}, u_{\pi, \lambda^0})$ the twisted coaction of $H^0$ on $A$ induced by $(\rho, u)$ and $\pi$, $\lambda^0$.

Let $(\sigma, v)$ be a coaction of $K^0$ on $B$ and $(\sigma_{\pi, \lambda^0}, v_{\pi, \lambda^0})$ the twisted coaction of $H^0$ on $A$ induced by $(\sigma, v)$ and $\pi$, $\lambda^0$.

\[lem:exterior\]With the above notations, if $(\rho, u)$ is exterior equivalent $(\sigma, v)$, then $(\rho_{\pi, \lambda^0} ,\, u_{\pi, \lambda^0})$ is exterior equivalent to $(\sigma_{\pi, \lambda^0}, v_{\pi, \lambda^0})$.

Let $\Delta_H^0$ and $\Delta_K^0$ be the comultiplications of $H^0$ and $K^0$, respectively. Since $(\rho, u)$ and $(\sigma, v)$ are exterior equivalent, there is a unitary element $w\in B\otimes K^0$ such that $$\sigma ={{\rm{Ad}}}(w)\circ \rho, \quad
v=(w\otimes 1^0 )(\rho\otimes {{\rm{id}}}_{K^0})(w)u({{\rm{id}}}_B \otimes\Delta_K^0 )(w)^* .$$ Since $\rho=(\pi\otimes\lambda^0 )^{-1}\circ\rho_{\pi, \lambda^0}\circ\pi$ and $\sigma=(\pi\otimes\lambda^0 )^{-1}\circ\sigma_{\pi. \lambda^0 }\circ\pi$, $$(\pi\otimes\lambda^0 )^{-1}\circ\sigma_{\pi, \lambda^0}\circ\pi
={{\rm{Ad}}}(w)\circ(\pi\otimes\lambda^0 )^{-1}\circ\rho_{\pi, \lambda^0}\circ\pi .$$ Hence $$\sigma_{\pi, \lambda^0 }={{\rm{Ad}}}((\pi\otimes\lambda^0 )(w))\circ\rho_{\pi, \lambda^0 } .$$ Furthermore, $$\begin{aligned}
v_{\pi, \lambda^0}
=(\pi\otimes\lambda^0 \otimes\lambda^0 )(v) 
&=((\pi\otimes\lambda^0 )(w)\otimes 1^0 )(\pi\otimes\lambda^0 \otimes\lambda^0 )((\rho\otimes{{\rm{id}}}_{K^0})(w)) \\
& \times u_{\pi, \lambda^0 }(\pi\otimes\lambda^0 \otimes\lambda^0 )(({{\rm{id}}}_B \otimes\Delta_K^0 )(w^* )) .\end{aligned}$$ Since $(\lambda^0 \otimes\lambda^0 )\circ\Delta_K^0 =\Delta_H^0 \circ\lambda^0 $, $$(\pi\otimes\lambda^0 \otimes\lambda^0 )(({{\rm{id}}}_B \otimes\Delta_K^0 )(w^* ))
=({{\rm{id}}}_A 