On the definition of a derived $A_\infty$-category

Question

Let $\mathcal{A}$ be an $A_\infty$-category. The derived $A_\infty$-category is defined to be the 0th cohomology category of the category of twisted complexes of $\mathcal{A}$.

I have troubles understanding this definition fully, and try to see the analogy to the definition of usual derived categories (as built of from the homotopy category).

Here is what I see: We want a triangulated category, eventually, which $\mathcal{A}$ is not necessarily. So, we embed into a larger category, namely $\operatorname{Tw}\mathcal{A}$. Now, the twisted complexes play the role of ordinary complexes. But why are we particularly interested in $H^0$? How does this tie back to the classical setting?

Edit: I believe a shift allows to recover also higher cohomologies, for if $X,Y \in \operatorname{Tw}\mathcal{A}$, then $\hom(X,Y[m]) = H^0(\hom(X, Y[m]) = H^m(\hom(X,Y))$. So, it seems no loss of generality to take $H^0$.

Answer 1

The category $H^0(\operatorname{Tw}\mathcal{A})$ is rather analogous to what is sometimes called the perfect derived category, that is for a ring $R$, the category of bounded complexes of projective $R$-modules $K^b(\operatorname{proj} R)$.

More or less by construction $\operatorname{Tw}\mathcal{A}$ is what you obtain from $\mathcal{A}$ when you close under shifts and cones. If you do this in the case where $\mathcal{A}=\operatorname{proj} R$ is the category of projective modules over a ring $R$ (viewed as an $A_\infty$-category with trivial differential and higher multiplications), then $H^0(\operatorname{Tw}\operatorname{proj} R)\cong K^b(\operatorname{proj} R)$, the $H^0$ corresponding to factoring out by homotopy. For a ring $R$, the perfect derived category coincides with the usual bounded derived category $D^b(\operatorname{mod} R)$ if and only if $R$ is of finite global dimension.