Let $A$ be a left coherent ring, that is, a ring for which the kernel of any homomorphisms between finitely generated projective modules are finitely generated. $T^{\bullet}\in \mathcal{K}^b(A-proj)$ is a tilting comples over $A$ with $B = End_{ \mathfrak{D}^b(A)-mod}(T^{\bullet})$. Then we know that $\mathfrak{D}^b(A-mod)$ and $\mathfrak{D}^b(B-mod)$ are equivalent as triangulated categories.
I have seen in a paper that"There is a $F:\mathfrak{D}^b(B-mod)\rightarrow \mathfrak{D}^b(A-mod) $ an equivalence of triangulated categories such that $F(B)=T^{\bullet}$. $F$ induces an quivalence $F'$ of the homotopy categories from $\mathcal{K}^b(B-proj)$ to $\mathcal{K}^b(A-proj)$".
So I want to ask:
- I know $A$ and $B$ are derived equivalent by definition. But who can tell me how to get the $F$ such that $F(B)=T^{\bullet}$ ? Also how $F$ induces $F'$?
- Can we still get $F$ and $F'$ when $A$ is a general ring or algebra?
