# Is the derived category of $A$-dg-modules as a dg-category coincide with the ordinary definition of derived category?

Let $A$ be a unital dg-algebra over a base field $k$. We consider the category of (unbounded) right $A$-dg-modules with morphisms closed degree $0$ maps. We denote this category by dg-mod-$A$. We could consider homotopic morphisms and get the homotopy category [dg-mod-$A$]. Moreover we could invert quasi-isomorphisms in [dg-mod-$A$] and obtain its derived category. Let us denote this category by $D(A)$.

On the other hand, we could consider the dg-category DG-MOD-$A$: its objects are right $A$-dg-modules and its morphisms are chain complexes of maps between dg-modules.

We notice that there is a general construction of the derived category of a dg-category $\mathcal{C}$: We first consider the dg-category of right modules over $\mathcal{C}$, which is the dg-category of contravariant dg-functors from $\mathcal{C}$ to Ch$(k)$. Let us denote this dg-category by DGM-$\mathcal{C}$. We also consider the full dg-subcategory Acycl-$\mathcal{C}$, which consists of dg-functors $\mathcal{F}:\mathcal{C}\to \text{Ch}(k)$ such that $\mathcal{F}(c)$ is acyclic for any object $c\in \mathcal{C}$. Then the derived category of $\mathcal{C}$ is given by the Verdier quotient $$[\text{DGM}-\mathcal{C}]/[\text{Acycl}-\mathcal{C}]$$ where $[\cdot]$ is the homotopy category of dg-categories. We denote the derived category of $\mathcal{C}$ by $D(\mathcal{C})$. We could show that the Yoneda functor induces a fully faithful functor $[\mathcal{C}]\to D(\mathcal{C})$.

Now go back to the dg-algebra $A$ and the dg-category DG-MOD-$A$.

My question is: is the derived category $D(A)$ equivalent to the derived category $D(\text{DG-MOD}-A)$? Why?

• If one interprets what you wrote literally, then D(DG-MOD-A) is formed from the dg-category of right modules over the dg-category DG-MOD-A, which itself consists of right modules over A. Do you really mean to take dg-modules over dg-modules here? Such a category would be clearly much bigger than ordinary dg-modules over A, which form D(A). Presumable one would want to take a suitable subcategory of D(DG-MOD-A). – Dmitri Pavlov Jan 25 '18 at 4:49

I think it is not good point of view to think that $A \mapsto H^0(Mod-A)[Qis^{-1}]$ is canonical construction for dg-algebras and $\mathcal{C} \mapsto H^0(Mod-\mathcal{C})/H^0(Acyc-\mathcal{C})$ is canonical construction for dg-categories. In fact, both constructions can be done for algebras and for categories in the same way. Third very useful model for $D(\mathcal{A})$ is $H^0(SF-\mathcal{A})$ - homotopy category of all semi-free dg-modules-$\mathcal{A}$ which just strictly full triangulated subcategory of $H^0(Mod-\mathcal{A})$. So I see there are 2 questions:

1. Let $A$ is dg-algebra, why $D(A)$ and $D(Mod-A)$ canonically triangulated equivalent?
2. Let $\mathcal{C}$ is dg-category, why $H^0(Mod-\mathcal{C})[Qis^{-1}]$ and $H^0(Mod-\mathcal{C})/H^0(Acyc-\mathcal{C})$ canonically triangulated equivalent?

Both questions are quite classical.

Let $F : A \to Mod-A$ canonical embedding of $A$ (which we consider as category with one object $*$ and $End(*) = A$) to $Mod-A$ as rank 1 free module-$A$.

First question is composition of following facts

1. Let $SF-\mathcal{A} \subset Mod-\mathcal{A}$ full dg-category which consists all semi-free dg-modules. Then induced localization functor $H^0(SF-\mathcal{A}) \to D(A)$ is quasi-equivalence. (Every dg-module-$\mathcal{A}$ have semi-free resolution). Keller Deriving DG-categories 3.1

2. Let $G : \mathcal{B} \to \mathcal{C}$ be full embedding of dg-categories. And let $G^* : SF-\mathcal{B} \to SF-\mathcal{C}$ be the extension dg-functor. Then the induced functor $H^0(G^*) : D(\mathcal{B}) \to D(\mathcal{C})$ is fully faithful. If, in addition, the category $H^0(\mathcal{C})$ is classically generated by $Ob \mathcal{B}$ then $H^0(G^*)$ is an equivalence. Proposition 1.15 in Luntz Orlov, "UNIQUENESS OF ENHANCEMENT FOR TRIANGULATED CATEGORIES"

3. $F$ is full embedding of classical generator of $SF-\mathcal{A}$. It is quite obvious, you can construct any semi-free modules from free using cone operation. Argument of such type must be also written in Keller Deriving DG-categories

Your second question is more or less also implies from theorem 3 in Keller Deriving DG-categories.