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Suppose $\mathscr{A}$ is a dg category and $ \mathcal{D} (\mathscr{A})$ its associated derived category. For an object in $ a \in \mathscr{A}$ there is associated (right )dg $ \mathscr{A}$ module, $ \hat{a} := \mathscr{A} (?, a)$. One can define the subcategory $ \text{Perf}(\mathscr{A})$ of $ \mathcal{D} ( \mathscr{A})$ as the smallest thick subcategory ( i.e. strict, triangulated and closed under taking direct summands) containing the set of objects $\{ \hat{a}| a \in \mathscr{A} \}$.

Suppose we take our dg category is $ \mathscr{A}:= k$ i.e. the category with one object whose endomorphism space is some commutative ring $k$. Then $\mathcal{D}( k)$ is just the derived category category of chain complexes over $k$.

Question: How does one see that an object in Perf($k$), as defined above, is a perfect chain complex in the usual sense of homological algebra, i.e. quasi-isomorphic to a bounded complex of finitely generated projectives without using the fact that these are just the compact objects? ( or is this even possible?)

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  • $\begingroup$ It's obvious, at least to me. What's the part you don't see? $\endgroup$ – Fernando Muro Sep 17 '14 at 17:37
  • $\begingroup$ Well, for example why should an object in $ X \in \text{Perf}(k)$ be bounded? $\endgroup$ – Anette Sep 17 '14 at 17:53
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    $\begingroup$ @Anette An equivalent definition of Perf(k) is that it consists of all objects one can build by starting with $\{\hat k\}$ and applying finitely many operations of the form (take an isomorphic object, take a direct summand, take a shift, take the mapping cone of a map). This is because the resulting set of objects form a thick subcategory, and it's necessarily the smallest. One can show by induction (on the number of operations needed to construct an object) that every such object is represented by a complex of finitely generated projectives and conversely. $\endgroup$ – Tyler Lawson Sep 17 '14 at 18:15
  • $\begingroup$ @TylerLawson, thanks for your comment! I think I almost understand, but I don't see why the object you end up with is necessarily bounded or rather quasi-isormophic to something bounded. $\endgroup$ – Anette Sep 17 '14 at 18:21
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    $\begingroup$ @Anette: because you apply only FINITELY many operations. $\endgroup$ – Sasha Sep 17 '14 at 19:24

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