## 13.34 Compact objects

Here is the definition.

Definition 13.34.1. Let $\mathcal{D}$ be an additive category with arbitrary direct sums. A *compact object* of $\mathcal{D}$ is an object $K$ such that the map

\[ \bigoplus \nolimits _{i \in I} \mathop{\mathrm{Hom}}\nolimits _{\mathcal{D}}(K, E_ i) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _{\mathcal{D}}(K, \bigoplus \nolimits _{i \in I} E_ i) \]

is bijective for any set $I$ and objects $E_ i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$ parametrized by $i \in I$.

This notion turns out to be very useful in algebraic geometry. It is an intrinsic condition on objects that forces the objects to be, well, compact.

Lemma 13.34.2. Let $\mathcal{D}$ be a (pre-)triangulated category with direct sums. Then the compact objects of $\mathcal{D}$ form the objects of a Karoubian, saturated, strictly full, (pre-)triangulated subcategory $\mathcal{D}_ c$ of $\mathcal{D}$.

**Proof.**
Let $(X, Y, Z, f, g, h)$ be a distinguished triangle of $\mathcal{D}$ with $X$ and $Y$ compact. Then it follows from Lemma 13.4.2 and the five lemma (Homology, Lemma 12.5.20) that $Z$ is a compact object too. It is clear that if $X \oplus Y$ is compact, then $X$, $Y$ are compact objects too. Hence $\mathcal{D}_ c$ is a saturated triangulated subcategory. Since $\mathcal{D}$ is Karoubian by Lemma 13.4.13 we conclude that the same is true for $\mathcal{D}_ c$.
$\square$

Lemma 13.34.3. Let $\mathcal{D}$ be a triangulated category with direct sums. Let $E_ i$, $i \in I$ be a family of compact objects of $\mathcal{D}$ such that $\bigoplus E_ i$ generates $\mathcal{D}$. Then every object $X$ of $\mathcal{D}$ can be written as

\[ X = \text{hocolim} X_ n \]

where $X_1$ is a direct sum of shifts of the $E_ i$ and each transition morphism fits into a distinguished triangle $Y_ n \to X_ n \to X_{n + 1} \to Y_ n[1]$ where $Y_ n$ is a direct sum of shifts of the $E_ i$.

**Proof.**
Set $X_1 = \bigoplus _{(i, m, \varphi )} E_ i[m]$ where the direct sum is over all triples $(i, m, \varphi )$ such that $i \in I$, $m \in \mathbf{Z}$ and $\varphi : E_ i[m] \to X$. Then $X_1$ comes equipped with a canonical morphism $X_1 \to X$. Given $X_ n \to X$ we set $Y_ n = \bigoplus _{(i, m, \varphi )} E_ i[m]$ where the direct sum is over all triples $(i, m, \varphi )$ such that $i \in I$, $m \in \mathbf{Z}$, and $\varphi : E_ i[m] \to X_ n$ is a morphism such that $E_ i[m] \to X_ n \to X$ is zero. Choose a distinguished triangle $Y_ n \to X_ n \to X_{n + 1} \to Y_ n[1]$ and let $X_{n + 1} \to X$ be any morphism such that $X_ n \to X_{n + 1} \to X$ is the given one; such a morphism exists by our choice of $Y_ n$. We obtain a morphism $\text{hocolim} X_ n \to X$ by the construction of our maps $X_ n \to X$. Choose a distinguished triangle

\[ C \to \text{hocolim} X_ n \to X \to C[1] \]

Let $E_ i[m] \to C$ be a morphism. Since $E_ i$ is compact, the composition $E_ i[m] \to X \to \text{hocolim} X_ n$ factors through $X_ n$ for some $n$, say by $E_ i[m] \to X_ n$. Then the construction of $Y_ n$ shows that the composition $E_ i[m] \to X_ n \to X_{n + 1}$ is zero. In other words, the composition $E_ i[m] \to C \to \text{hocolim} X_ n$ is zero. This means that our morphism $E_ i[m] \to C$ comes from a morphism $E_ i[m] \to X[-1]$. The construction of $X_1$ then shows that such morphism lifts to $\text{hocolim} X_ n$ and we conclude that our morphism $E_ i[m] \to C$ is zero. The assumption that $\bigoplus E_ i$ generates $\mathcal{D}$ implies that $C$ is zero and the proof is done.
$\square$

Lemma 13.34.4. With assumptions and notation as in Lemma 13.34.3. If $C$ is a compact object and $C \to X_ n$ is a morphism, then there is a factorization $C \to E \to X_ n$ where $E$ is an object of $\langle E_{i_1} \oplus \ldots \oplus E_{i_ t} \rangle $ for some $i_1, \ldots , i_ t \in I$.

**Proof.**
We prove this by induction on $n$. The base case $n = 1$ is clear. If $n > 1$ consider the composition $C \to X_ n \to Y_{n - 1}[1]$. This can be factored through some $E'[1] \to Y_{n - 1}[1]$ where $E'$ is a finite direct sum of shifts of the $E_ i$. Let $I' \subset I$ be the finite set of indices that occur in this direct sum. Thus we obtain

\[ \xymatrix{ E' \ar[r] \ar[d] & C' \ar[r] \ar[d] & C \ar[r] \ar[d] & E'[1] \ar[d] \\ Y_{n - 1} \ar[r] & X_{n - 1} \ar[r] & X_ n \ar[r] & Y_{n - 1}[1] } \]

By induction the morphism $C' \to X_{n - 1}$ factors through $E'' \to X_{n - 1}$ with $E''$ an object of $\langle \bigoplus _{i \in I''} E_ i \rangle $ for some finite subset $I'' \subset I$. Choose a distinguished triangle

\[ E' \to E'' \to E \to E'[1] \]

then $E$ is an object of $\langle \bigoplus _{i \in I' \cup I''} E_ i \rangle $. By construction and the axioms of a triangulated category we can choose morphisms $C \to E$ and a morphism $E \to X_ n$ fitting into morphisms of triangles $(E', C', C) \to (E', E'', E)$ and $(E', E'', E) \to (Y_{n - 1}, X_{n - 1}, X_ n)$. The composition $C \to E \to X_ n$ may not equal the given morphism $C \to X_ n$, but the compositions into $Y_{n - 1}$ are equal. Let $C \to X_{n - 1}$ be a morphism that lifts the difference. By induction assumption we can factor this through a morphism $E''' \to X_{n - 1}$ with $E''$ an object of $\langle \bigoplus _{i \in I'''} E_ i \rangle $ for some finite subset $I' \subset I$. Thus we see that we get a solution on considering $E \oplus E''' \to X_ n$ because $E \oplus E'''$ is an object of $\langle \bigoplus _{i \in I' \cup I'' \cup I'''} E_ i \rangle $.
$\square$

Definition 13.34.5. Let $\mathcal{D}$ be a triangulated category with arbitrary direct sums. We say $\mathcal{D}$ is *compactly generated* if there exists a set $E_ i$, $i \in I$ of compact objects such that $\bigoplus E_ i$ generates $\mathcal{D}$.

The following proposition clarifies the relationship between classical generators and weak generators.

Proposition 13.34.6. Let $\mathcal{D}$ be a triangulated category with direct sums. Let $E$ be a compact object of $\mathcal{D}$. The following are equivalent

$E$ is a classical generator for $\mathcal{D}_ c$ and $\mathcal{D}$ is compactly generated, and

$E$ is a generator for $\mathcal{D}$.

**Proof.**
If $E$ is a classical generator for $\mathcal{D}_ c$, then $\mathcal{D}_ c = \langle E \rangle $. It follows formally from the assumption that $\mathcal{D}$ is compactly generated and Lemma 13.33.3 that $E$ is a generator for $\mathcal{D}$.

The converse is more interesting. Assume that $E$ is a generator for $\mathcal{D}$. Let $X$ be a compact object of $\mathcal{D}$. Apply Lemma 13.34.3 with $I = \{ 1\} $ and $E_1 = E$ to write

\[ X = \text{hocolim} X_ n \]

as in the lemma. Since $X$ is compact we find that $X \to \text{hocolim} X_ n$ factors through $X_ n$ for some $n$ (Lemma 13.31.9). Thus $X$ is a direct summand of $X_ n$. By Lemma 13.34.4 we see that $X$ is an object of $\langle E \rangle $ and the lemma is proven.
$\square$

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