Unitary in adjointable operators associated with equivariant Hilbert module Consider the following fragment from the article "Tannaka–Krein duality for compact quantum
homogeneous spaces. I. General theory" by De Commer and Yamashita:

How exactly is $\mathcal{E}\otimes_{\alpha_{\mathbb{X}}} (C(\mathbb{X})\otimes C(\mathbb{G}))$ here a right $C(\mathbb{X})\otimes C(\mathbb{G})$ module? I am not sure what the notation $\otimes_{\alpha_{\mathbb{X}}}$ means here. Presumably, it refers to the inner tensor product of Hilbert modules, but I'm not sure how exactly this works because then we would need a $*$-morphism into the adjointable operators.
Thanks in advance for your help!
 A: The answer is rather easy; I think what is confusing is the sheer number of objects flying around.
What is $C(\mathbb X)$?  This is Definition 1.4 in the paper: this is a $C^*$-algebra with an ergodic action of $\mathbb G$.  So there is a coaction $\alpha_{\mathbb X}:C(\mathbb X) \rightarrow C(\mathbb X) \otimes C(\mathbb G)$, which is a unital $*$-homomorphism.
What is the interior tensor product?  In general, we have $E=E_A$ (a Hilbert $C^*$-module over $A$) and $F=F_B$ and a non-degenerate $*$-homomorphism $\phi:A\rightarrow\mathcal L(F)$.  We then form $E\odot F$ and complete for the $B$-valued form
$$ ( e_1\otimes f_1 | e_2\otimes f_2 ) = (f_1 | \phi((e_1|e_2)) f_2). $$
In this setup, $\mathcal E$ is by definition a Hilbert $C^*$-module over $C(\mathbb X)$ and we consider the algebra $C(\mathbb X)\otimes C(\mathbb G)$ as a module over itself, and in particular identify the "compact" operators on $C(\mathbb X)\otimes C(\mathbb G)$ with the algebra itself.  Thus $\alpha_{\mathbb X}$ may be regarded as a non-degenerate (because it is unital!) $*$-homomorphism $C(\mathbb X) \rightarrow \mathcal L(C(\mathbb X)\otimes C(\mathbb G))$.  This allows us to form the interior-tensor product
$$ \mathcal E \otimes_{\alpha_{\mathbb X}} (C(\mathbb X)\otimes C(\mathbb G)). $$
(Then writing out the definition of $X_{\mathcal E}$ will show that it is an isometry, and hence well-defined.  That it is unitary then follows by the density condititon in Definition 3.1.)
