# Is the functor $mod(Tw\mathcal{C})\rightarrow mod(\mathcal{C})$ cohomologically full and faithful?

Let $\mathcal{C}$ be a $c$-unital $A_\infty$-category. If $\mathcal{A}$ is a $c$-unital and triangulated $A_\infty$-category, then there is a $c$-unital $A_\infty$- functor $$Tw: fun(\mathcal{C},\mathcal{A})\rightarrow fun(Tw \mathcal{C}, Tw\mathcal{A})\rightarrow fun(Tw \mathcal{C}, \mathcal{A}),$$ See (3.26) in section (3n) of Seidel's book "Fukaya cateogories and Picard-Lefschetz Theory". Here $fun(\mathcal{C},\mathcal{A})$ is the $A_\infty$-category of $c$-unital functors from $\mathcal{C}$ to $\mathcal{A}$, $Tw\mathcal{A}$ means the category of twisted complexes of $\mathcal{A}$ and the latter morphism uses the quasi-equivalence between $Tw\mathcal{A}$ and $\mathcal{A}$.

There is also the restriction functor $$\mathcal{I}^*: fun(Tw \mathcal{C}, \mathcal{A})\rightarrow fun(\mathcal{C},\mathcal{A}).$$ My question is: are the two functors $Tw$ and $\mathcal{I}^*$ in any sense inverse to each other?

My instinct is that if two functors $\mathcal{G}, \mathcal{H}\in fun(Tw\mathcal{C}, \mathcal{A})$ have $\mathcal{I}^*\mathcal{G}\cong \mathcal{I}^*\mathcal{H}$ in $H^0(fun(\mathcal{C},\mathcal{A}))$, then $\mathcal{G}\cong \mathcal{H}$ in $H^0(fun(Tw\mathcal{C},\mathcal{A}))$. But this seems to be too strong.

For example, take $\mathcal{A}=Ch$, the dg category of cochain complexes, I would expect $$\mathcal{I}^*: mod(Tw\mathcal{C})\rightarrow mod(\mathcal{C})$$ is cohomologically full and faithful. But I only see result like Lemma 3.36 in Seidel's book which says $$\tilde{l}: Tw(\mathcal{C})\rightarrow mod(Tw\mathcal{C})\rightarrow mod(\mathcal{C})$$ is cohomologically full and faithful.

Thanks!

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Welcome to MO! It is common (if at all possible) to use at least one of the top-level tags that is those that have a two-letter prefix (they correspond to the math categories on arxiv). Thus, I added the tag ct.category-theory. If you should prefer another one you can always change this, using the link edit just below your question. –  quid Jul 19 '13 at 23:58
@quid Thanks for adding the tag! –  Xin Jin Aug 1 '13 at 6:02