Let $\mathcal{C}$ and $\mathcal{D}$ be dg categories over a field $k$ of characteristic zero. Then one can form their tensor product $\mathcal{C} \otimes \mathcal{D}$: the objects of the tensor product are pairs of objects in $\mathcal{C}$ and $\mathcal{D}$, and the morphisms are the obvious ones. (see section 2.3: http://www.mi-ras.ru/~akuznet/dgcat/Keller%20On%20differential%20graded%20categories.pdf)
My question: how does tensor product behave with respect to passing to the cohomology category? Since we are over a field of characteristic zero, I have always believed that $H^*(\mathcal{A}) \otimes H^*(\mathcal{B}) = H^*(\mathcal{A} \otimes \mathcal{B})$. However, it seems from a paper I am reading that this is not automatically the case.
Could someone explain what goes wrong? The Künneth formula seems to imply that the morphisms spaces coincide... maybe the issue has to do with naturality of multiplication?
If the above equality is indeed false, does one at least have a faithful functor $H^*(\mathcal{A}) \otimes H^*(\mathcal{B}) \to H^*(\mathcal{A} \otimes \mathcal{B})$?
EDIT: Actually, I think my confusion was caused by a notational clash. The paper I was looking at was actually only considering the zero-graded part of the cohomology category: it's indeed not true that $H^0(\mathcal{A}) \otimes H^0(\mathcal{B}) \to H^0(\mathcal{A} \otimes \mathcal{B})$ is an equivalence (however, it is faithful) if we consider $\mathbb{Z}$-graded dg categories. However, I believe it is true in general (by Künneth) that $H^*(\mathcal{A}) \otimes H^*(\mathcal{B}) \to H^*(\mathcal{A} \otimes \mathcal{B})$ is an equivalence.
(Not sure whether I should delete the question -- I will leave it open for now in case it's somehow ever useful).
