Let $\mathbb{G}$ be a compact quantum group with function algebra $(C(\mathbb{G}), \Delta)$ (in the sense of Woronowicz). Let $X \in M(B_0(H) \otimes C(\mathbb{G}))$ be a (possibly infinite-dimensional) representation of the quantum group $\mathbb{G}$, and let $K$ be an $X$-invariant subspace of $H$, i.e. if $p\in B(H)$ is the projection on the closed subspace $K$, then $(p\otimes 1)X (p\otimes 1) = X(p\otimes 1).$

Where one encounters an invariant subspace $K$, one hopes to define a subrepresentation $X_K \in M(B_0(K)\otimes C(\mathbb{G}))$. It looks like there is an obvious way to do this:

Consider the surjective strict completely positive map $$\Psi: B_0(H) \to B_0(K): x \mapsto pxp^*$$ The strict completely positive map $$\Psi \otimes \iota: B_0(H) \otimes C(\mathbb{G}) \to B_0(K) \otimes C(\mathbb{G})$$ extends uniquely to a bounded linear map $$\Psi \otimes \iota: M(B_0(H) \otimes C(\mathbb{G}) \to M(B_0(K)\otimes C(\mathbb{G}))$$ which is strictly continuous on bounded subsets and we define $$X_K:= (\Psi \otimes \iota)(X)$$ as our candidate for a subrepresentation. Everything works out nicely, for example, it is easily verified that $$(\iota \otimes \Delta)(X_K) = (X_K)_{12}(X_K)_{13}.$$ However, what is not clear to me is why $X_K$ must be invertible (I consider representations to be invertible elements in the multiplier algebra by definition, and representations do not need to be unitary). Does invertibility of $X \in M(B_0(H)\otimes C(\mathbb{G}))$ imply invertibility of $X_K \in M(B_0(K) \otimes C(\mathbb{G}))?$ I have a feeling that the answer might be negative because compressing with a projection can make things non-invertible.