Deligne’s tensor product of algebra module categories $A$ and $B$ are finite dimensional $\mathbb{k}$-algebras. $\textrm{Mod}_A$ is the category of finite dimensional $A$-modules.
In Proposition 1.46.2. of the note，it is claimed that $\textrm{Mod}_{A\otimes B}$ is the Deligne’s tensor product of $\textrm{Mod}_A$ and $\textrm{Mod}_B$.
For any abelian category $C$ and functor $F:\textrm{Mod}_A\times \textrm{Mod}_B\to C$ which is right exact on both variables, how to construct a right exact functor $\tilde{F}: \textrm{Mod}_{A\otimes B}\to C$?
It’s quite obvious for algebraic closed $\mathbb{k}$ and semisimple $A$ and $B$, because every simple $A\otimes B$-module is a tensor product of simple $A$-module and simple $B$-module. But for a general field and algebra, we can only decompose a finite dimensional module into a direct sum of indecomposable modules. Is there a similar relation for indecomposable modules?
 A: I'm sorry for answering my question by myself, but I think it's harmless to record an answer here.
We should embed $C$ to $\textrm{Mod}_R$ for some algebra R.
Note that $F(A\otimes B)$ is an $R-A\otimes B$ bimodule. The $A\otimes B$ structure is defined by $a\otimes b\mapsto F(-\cdot a,-\cdot b)$. The $A\otimes B$ structure is well defined, since $F$ is bilinear on the morphism. Two module structures are compatible, since $F(-\cdot a,-\cdot b)$ is an $R$-map.
We define $\tilde{F}=F(A\otimes B)\otimes_{A\otimes B}(-)$.
To prove the natural isomorphism $\tilde{F}\circ \boxtimes\cong F$, we can choose free resolutions $M_{\ast}$ and $N_{\ast}$ for $A$-module $M$ and $B$-module $N$ repectively. Note that there is an exact sequence
$$F(M_0,N_1)\oplus F(M_1,N_0)\to F(M_0,N_0)\to F(M,N)\to 0$$
the exactness comes from direct computation or using a simple argument of sepectral sequence. The exact sequence is isomorphic to another sequence
$$F(A,B)\otimes_{A\otimes B}(M_0\otimes N_1\oplus M_1\otimes N_0)\to F(A,B)\otimes_{A\otimes B}(M_0\otimes N_0)\to F(A,B)\otimes_{A\otimes B}(M\otimes N)\to 0$$
since $M_\ast$ and $N_\ast$ are free, $F$ is bilinear and $F(A,B)\otimes_{A\otimes B}(-)$ is right exact. Using comparison lemma and universal property of cokernel, we can prove the naturality.
Finally, we need to prove the image of $\tilde{F}$ lies in $C$. We choose a free resolution $K_\ast$ of an $A\otimes B$-module $K$, then $\tilde{F}(K_1)\to\tilde{F}(K_0)\to \tilde{F}(K)$ is exact. Since $\tilde{F}(K_1)$ and $\tilde{F}(K_0)$ are in $C$ and abelian category $C$ is closed under cokernel, $\tilde{F}(K)$ must in $C$.
