If M is a C-bicomodule, then considering C as a $C$-bicomodule also, is $M \square_C C \cong C$, where $\square_C$ is the cotensor product in $^C\mathscr{M}^C$.
2026-03-26 06:26:25.1774506385
Cotensor and counit?
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Of course not, we have $M \square_C C = M$. The proof is the same as for modules (in fact, the statement holds in great generality in monoidal categories, and statements about monoidal categories can be dualized).