gluing of DG-categories

Question

In the article "Categorical resolution of irrational singularities" of Kuzentsov and Lunts is described how one can glue DG-categories. For this let $\mathcal{D}_1$ and $\mathcal{D}_2$ be two small DG-categories. Now consider a bimodule $\phi\in(\mathcal{D}^{op}_2\otimes \mathcal{D}_1)-dgm$. Then the gluing of these two categories along $\phi$ is denoted by $\mathcal{D}=\mathcal{D}_1\times_{\phi}\mathcal{D}_2$ is defined as follows: Objects are triples $M=(M_1,M_2,\mu)$ with $M_i\in\mathcal{D}_i$ and $\mu\in\phi(M_2,M_1)$. Morphism complex is defined via: $\mathrm{Hom}^k_{\mathcal{D}}(M,N)=\mathrm{Hom}^k_{\mathcal{D}_1}(M_1,N_1)\oplus\mathrm{Hom}^k_{\mathcal{D}_2}(M_2,N_2)\oplus \phi^{k-1}(N_2,M_1)$. Now suppose we consider $\phi^{op}\in(\mathcal{D}^{op}_2\otimes\mathcal{D}_1)^{op}-dgm=(\mathcal{D}^{op}_1\otimes\mathcal{D}_2)-dgm$. Now gluing $\mathcal{D}'=\mathcal{D}_2\times_{\phi^{op}}\mathcal{D}_1$. My question is now: Is it possible to get a DG-functor $F:\mathcal{D}\rightarrow \mathcal{D}'$ in the following way: send $M=(M_1,M_2,\mu)$ to $M'=(M_2,M_1,\mu^{op})$ and anagously for the Morphism complex??? The second question is: If we have a DG-functor $F:\mathcal{D}\rightarrow \mathcal{D}'$ this clearly induces a functor between the homotopy-category $F':H^0(\mathcal{D})\rightarrow H^0(\mathcal{D}')$. Does $\mathrm{Hom}_{H^0}(M,N)=0$ imply $\mathrm{Hom}(F'(M),F'(N))=0$?

Answer 1

As mentioned in the comments, we need to fix isomorphisms $\newcommand{\D}{\mathscr{D}}\D_1 \xrightarrow{\sim} \D_1^{op}$ and $\D_2 \xrightarrow{\sim} \D_2^{op}$ for the question to make sense. They key point is that in this case, there is a canonical isomorphism $\D_1^{op} \otimes \D_2 \xrightarrow{\sim} \D_2^{op} \otimes \D_1$, inducing equivalences between the categories of $\D_1$-$\D_2$-bimodules and $\D_2$-$\D_1$-bimodules.

Hence one can consider the gluings $\D = \D_1 \times_\varphi \D_2$ and $\D' = \D_2 \times_\psi \D_1$, where $\psi$ is the $\D_2$-$\D_1$-bimodule corresponding to $\varphi$.

Indeed one has a functor $\D \to \D'$ as you described, where $\mu^{op}$ means the element of $\psi^0(M_1, M_2)$ corresponding to $\mu \in \varphi^0(M_2, M_1)$ under the isomorphisms induced by duality. Further this functor $\D \to \D'$ is clearly an equivalence, swapping the first two components and acting by duality on the third.