Perf($\mathscr{A}$) and perfect chain complexes

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?)

• It's obvious, at least to me. What's the part you don't see? – Fernando Muro Sep 17 '14 at 17:37
• Well, for example why should an object in $X \in \text{Perf}(k)$ be bounded? – Anette Sep 17 '14 at 17:53
• @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. – Tyler Lawson Sep 17 '14 at 18:15
• @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. – Anette Sep 17 '14 at 18:21
• @Anette: because you apply only FINITELY many operations. – Sasha Sep 17 '14 at 19:24