M\ot A$. The stability condition in the case of the usual stable anti Yetter-Drinfeld modules ensures that $\tau^{n+1}_n=\Id.$ To obtain the same conclusion in our general case leads us to the requirement that the single cyclic map $\tau_0$ be the identity. In summary, from the above considerations, we guess that whereas $YD= \mathcal{Z}( {{_H\mathcal{M}}}_{fd})$ (center of a monoidal category), $AYD=\mathcal{Z}({{{^{**}_H\mathcal{M}}}}_{fd})$ (center of a bimodule category) where ${{{^{**}_H\mathcal{M}}}}_{fd}$ is simply ${{_H\mathcal{M}}}_{fd}$ with the left action modified by ${(-)^{**}}$. This guess turns out to be correct. Note that both $\tau_n$ and $$AYD({\mathcal{C}})=\mathcal{Z}_{{\mathcal{C}}}({^{**}{\mathcal{C}}})$$ make sense for any rigid category.

### Covariant cohomology theory.

If instead we consider $C^n=\Hom_H(k, M\ot C^{\ot n+1})$ and try to define $\tau$ directly, we need to *slide the first copy of $C$ past $M$ and then to the back.* Again let us consider the *finite* dimensional left modules over a Hopf algebra $H$. Using the standard adjunction properties of rigidity, for any $V, W\in {{_H\mathcal{M}}}$, we have:

$$\Hom_{H}(1, V\ot W)\simeq \Hom_{H}( 1\ot {^* W}, V)\simeq \Hom_{H}( {^* W}\ot 1, V)\simeq \Hom_H(1, {^{**}W}\ot V).$$

Thus for an $M\in {{_H\mathcal{M}}}$ and a coalgebra object $C$ (though the coalgebra structure plays no role at this stage) in ${{_H\mathcal{M}}}$ we obtain

$$\label{adj2} \Hom_{H}(1, 