The Stacks project

Lemma 21.52.1. Let $\mathcal{A}$ be a Grothendieck abelian category. Let $S \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ be a set of objects such that

  1. any object of $\mathcal{A}$ is a quotient of a direct sum of elements of $S$, and

  2. for any $E \in S$ the functor $\mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(E, -)$ commutes with direct sums.

Then every compact object of $D(\mathcal{A})$ is a direct summand in $D(\mathcal{A})$ of a finite complex of finite direct sums of elements of $S$.

Proof. Assume $K \in D(\mathcal{A})$ is a compact object. Represent $K$ by a complex $K^\bullet $ and consider the map

\[ K^\bullet \longrightarrow \bigoplus \nolimits _{n \geq 0} \tau _{\geq n} K^\bullet \]

where we have used the canonical truncations, see Homology, Section 12.15. This makes sense as in each degree the direct sum on the right is finite. By assumption this map factors through a finite direct sum. We conclude that $K \to \tau _{\geq n} K$ is zero for at least one $n$, i.e., $K$ is in $D^{-}(R)$.

We may represent $K$ by a bounded above complex $K^\bullet $ each of whose terms is a direct sum of objects from $S$, see Derived Categories, Lemma 13.15.4. Note that we have

\[ K^\bullet = \bigcup \nolimits _{n \leq 0} \sigma _{\geq n}K^\bullet \]

where we have used the stupid truncations, see Homology, Section 12.15. Hence by Derived Categories, Lemmas 13.33.7 and 13.33.9 we see that $1 : K^\bullet \to K^\bullet $ factors through $\sigma _{\geq n}K^\bullet \to K^\bullet $ in $D(R)$. Thus we see that $1 : K^\bullet \to K^\bullet $ factors as

\[ K^\bullet \xrightarrow {\varphi } L^\bullet \xrightarrow {\psi } K^\bullet \]

in $D(\mathcal{A})$ for some complex $L^\bullet $ which is bounded and whose terms are direct sums of elements of $S$. Say $L^ i$ is zero for $i \not\in [a, b]$. Let $c$ be the largest integer $\leq b + 1$ such that $L^ i$ a finite direct sum of elements of $S$ for $i < c$. Claim: if $c < b + 1$, then we can modify $L^\bullet $ to increase $c$. By induction this claim will show we have a factorization of $1_ K$ as

\[ K \xrightarrow {\varphi } L \xrightarrow {\psi } K \]

in $D(\mathcal{A})$ where $L$ can be represented by a finite complex of finite direct sums of elements of $S$. Note that $e = \varphi \circ \psi \in \text{End}_{D(\mathcal{A})}(L)$ is an idempotent. By Derived Categories, Lemma 13.4.14 we see that $L = \mathop{\mathrm{Ker}}(e) \oplus \mathop{\mathrm{Ker}}(1 - e)$. The map $\varphi : K \to L$ induces an isomorphism with $\mathop{\mathrm{Ker}}(1 - e)$ in $D(R)$ and we conclude.

Proof of the claim. Write $L^ c = \bigoplus _{\lambda \in \Lambda } E_\lambda $. Since $L^{c - 1}$ is a finite direct sum of elements of $S$ we can by assumption (2) find a finite subset $\Lambda ' \subset \Lambda $ such that $L^{c - 1} \to L^ c$ factors through $\bigoplus _{\lambda \in \Lambda '} E_\lambda \subset L^ c$. Consider the map of complexes

\[ \pi : L^\bullet \longrightarrow (\bigoplus \nolimits _{\lambda \in \Lambda \setminus \Lambda '} E_\lambda )[-i] \]

given by the projection onto the factors corresponding to $\Lambda \setminus \Lambda '$ in degree $i$. By our assumption on $K$ we see that, after possibly replacing $\Lambda '$ by a larger finite subset, we may assume that $\pi \circ \varphi = 0$ in $D(\mathcal{A})$. Let $(L')^\bullet \subset L^\bullet $ be the kernel of $\pi $. Since $\pi $ is surjective we get a short exact sequence of complexes, which gives a distinguished triangle in $D(\mathcal{A})$ (see Derived Categories, Lemma 13.12.1). Since $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(K, -)$ is homological (see Derived Categories, Lemma 13.4.2) and $\pi \circ \varphi = 0$, we can find a morphism $\varphi ' : K^\bullet \to (L')^\bullet $ in $D(\mathcal{A})$ whose composition with $(L')^\bullet \to L^\bullet $ gives $\varphi $. Setting $\psi '$ equal to the composition of $\psi $ with $(L')^\bullet \to L^\bullet $ we obtain a new factorization. Since $(L')^\bullet $ agrees with $L^\bullet $ except in degree $c$ and since $(L')^ c = \bigoplus _{\lambda \in \Lambda '} E_\lambda $ the claim is proved. $\square$


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