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In the article Grothendieck Ring of pretriangulated categories by Bondal-Larsen-Lunts, two dg-categories $\mathcal A$ and $\mathcal B$ are called quasi-equivalent if there is a chain of dg-categories and quasi-equivalences

$\mathcal A \leftarrow \mathcal C_1 \rightarrow \cdots \leftarrow \mathcal C_n \rightarrow \mathcal B$.

Another possibility, I think, would be to call $\mathcal A$ and $\mathcal B$ quasi-equivalent if they are isomorphic in the homotopy category $\mathrm{Ho}(\mathbf{dgcat})$. Clearly if $\mathcal A$ and $\mathcal B$ are quasi-equivalent as in the definition of Bondal-Larsen-Lunts, they are also isomorphic in $\mathrm{Ho}(\mathbf{dgcat})$. Is the converse true? What is the "better" definition of quasi-equivalent dg-categories?

Thanks in advance!

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  • $\begingroup$ have you had a look at toen's lectures on dg-categories? anyway, I think you're right, if I'm not mistaken a way to define morphisms in a localisation goes exactly via strings $A_1 \leftarrow C_1 \cdots \to B$ as you write. $\endgroup$ Mar 4, 2012 at 17:41
  • $\begingroup$ I'm studying that article, among others. The fact is: if $\mathcal A$ is isomorphic to $\mathcal B$ in the homotopy category, by definition we have diagrams $\mathcal A \leftarrow \mathcal C_1 \cdots \rightarrow \mathcal B$ and $\mathcal B \leftarrow \mathcal D_1 \cdots \rightarrow \mathcal A$, where the arrows "in the wrong direction" are quasi-equivalences or identities, such that the concatenation of the first and the second diagram is equivalent to the identity (and vice-versa). But this does not give immediately a diagram from $\mathcal A$ to $\mathcal B$ with quasi-equivalences only. $\endgroup$ Mar 4, 2012 at 17:59
  • $\begingroup$ it must be a general fact about localising a class of morphisms S that if a roof $A \lefttarrow C \rightarrow B$ represents an iso in Ho then $C \to B$ is in S. anyway don't listen to me, thomas nikolaus below gave you an answer using the model structure. $\endgroup$ Mar 5, 2012 at 0:22

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The converse is true.

You can see this for example as follows: first since DG-categories admit a model structure each morpism $A \to B$ in Ho(dgCat) can be realized as a 3-step span like this $A \leftarrow A' \to B' \leftarrow B$ (here $A'$ is a cofibrant replacement and $B'$ is a fibrant replacement). Then such a morpism is an isomorphism in the homotopy category be the 2-out-of-3 property if and only if the middle morphism $A'\to B'$ is a weak equivalence. Thus each isomorphism in the homotopy category comes from a Zig-Zag of weak equivalences (and this argument is valid in each model category).

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  • $\begingroup$ Yes, I found myself that the converse was indeed true: in Hovey's book on model categories it is proved in Theorem 1.2.10, part (iv) that whenever a map $f: A \rightarrow B$ in a model category is an isomorphism when viewed as a morphism in the homotopy category, then it is a weak equivalence. Could you please explain, in your argument, how to apply the 2-out-of-3 property? Thanks in advance! $\endgroup$ Mar 10, 2012 at 12:20
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    $\begingroup$ On can argue more or less as you say. Given the factorization $A \leftarrow A′ \to B′ \leftarrow B$ this gives a ZigZag in the homotopy category (applying the functor $\dgCat \to Ho(\dgCat)$ where the outer two morphisms are isomorphisms. Thus the morphism in the middle is an isomorphism iff the inner is one, i.e. a weak equuivalence. $\endgroup$ Mar 10, 2012 at 12:55
  • $\begingroup$ All right! I shall approve the answer; thanks! $\endgroup$ Mar 10, 2012 at 14:29

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