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There are several definitions of a quasi-equivalence $\newcommand{\T}{\mathscr{T}}F : \T \to \T'$ of DG categories in the literature, e.g.

(i) the induced functor $H^0(F) : H^0(\T) \stackrel{\sim}{\to} H^0(\T')$ is an equivalence (Bondal-Kapranov 1990);

(ii) $H^0(F)$ is an equivalence and $H^*(F)$ is fully faithful, i.e. $$ \newcommand{\Hom}{\mathrm{Hom}}\bigoplus_n H^n(\Hom_\T(X,Y)) \stackrel{\sim}{\longrightarrow} \bigoplus_n H^n(\Hom_{\T'}(F(X),F(Y))) $$ is an isomorphism for all $X,Y \in \T$, $n \in \mathbf{Z}$ (Drinfeld 2002);

(iii) $H^0(F)$ is an equivalence and for all $X,Y \in \T$, $n \in \mathbf{Z}$, $$ H^n(\Hom_\T(X,Y)) \stackrel{\sim}{\longrightarrow} H^n(\Hom_{\T'}(F(X),F(Y))) $$ are isomorphisms (To\"en 2004, Keller 2006).

It is obvious that (iii) implies (ii) and (ii) implies (i), but when are the reverse implications true? Is it obvious when $\T$ and $\T'$ are (strongly?) pretriangulated? In particular, I am working with the DG category $\mathbf{L}_{\mathrm{pf}}(X)$ of perfect complexes of sheaves on a smooth proper scheme $X$, and in this case I would hope that all definitions are the same.

I apologize if this question is too elementary for MO.

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(ii) and (iii) are clearly the same, and it is easy to find counterexamples to (i), just take a DG-category with homology concentrated in negative degrees, but not in degree $0$, and compare it to it's 0-dimensional cohomology regarded as a DG-category. However, (i) implies (ii) if the DG-categories are pretriangulated, this is an exercise. – Fernando Muro Aug 26 '13 at 15:14
$H^1(\mathcal{T}(X,Y))=H^0(\mathcal{T}(\Sigma X,Y))$ – Fernando Muro Aug 26 '13 at 15:29
Adding to Fernando's comment, the difference between (ii) and (iii) is purely cultural and semantic: as he says, they are clearly the same. If you are brought up as an algebraic geometer, you generally take direct sums when defining graded algebraic structures. If you are brought up as an algebraic topologist, you do not. The difference can matter (but not here): ask yourself whether or not the Laurent series ring $k((x))$ over a field $k$ is a graded field, where x has degree 2. Spencer Bloch (an algebraic geometer) and I once taught a joint course, and we disagreed about the answer. – Peter May Aug 26 '13 at 15:34
I don't think this is too elementary for MO. It seems like just the sort of question MO was originally designed to answer, i.e. not really general knowledge but the sort of thing a non-expert might need to know. I don't know why there's a vote to close except perhaps that large chunks of what could be an answer are in the comments. – David White Aug 26 '13 at 16:34
Adeel, all DG-functors are pretriangulated, you get $\Sigma F(X)\cong F(\Sigma X)$ for free! – Fernando Muro Aug 26 '13 at 17:41

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