2021-10-05

Compact objects

An important part of establishing Morita equivalence between $R$ and $M_n(R)$ was that fact that the natural map $\bigoplus_{i \in I} \operatorname{Hom}(U, N_i) \to \operatorname{Hom}(U, \bigoplus_{i \in I} N_i)$ was an isomorphism.

This is equivalent to saying we can factor any $f: U \to \bigoplus_{i \in I} N_i$ as

where $|I_f| < \infty$ .

Objects with the property are very useful. They go be different names depending on the context. We will be focused on triangulated categories.

Definition. Let $\mathcal T$ be a triangulated category. An object $X$ is called compact if the natural map $\bigoplus_{i \in I} \operatorname{Hom}(X,Y_i) \to \operatorname{Hom}(X, \bigoplus_{i \in I} Y_i)$ is isomorphism for any $Y_i, i \in I$ .

The full subcategory of compact objects in $\mathcal T$ is denoted by $\mathcal T^c$ .

Lemma. $\mathcal T^c$ is triangulated and closed under retracts.

Proof. (Expand to view)

We have $\operatorname{Hom}(X[n], \bigoplus Y_i) \cong \operatorname{Hom}(X, \bigoplus Y_i[n])$ so $X[n] \in \mathcal T^c$ for any $n \in \mathbb{Z}$ . Next, take a map $X \to X^\prime$ of compact objects and complete it to a triangle $X \to X^\prime \to C \to X[1]$ We get a commutative diagram

with exact rows and where two out of the three vertical maps are isomorphisms. Thus, the third is also. Finally, if

X^\prime

is a retract of

X

, the map

\bigoplus_{i \in I} \operatorname{Hom}(X^\prime,Y_i) \to \operatorname{Hom}(X^\prime, \bigoplus_{i \in I} Y_i)

is a retract of an isomorphism. Hence it is an isomorphism. ■

Definition. We say $\mathcal T$ is compactly generated if it has small coproducts and there exists a set of compact objects $\mathcal C$ such that $\operatorname{Hom}(C,Y) = 0 \ \forall C \in \mathcal C$ implies $Y = 0$ . The set $\mathcal C$ is called a set of compact generators.

Lemma. A map $X \to Y$ in $\mathcal T$ with $\operatorname{Hom}(C,X) \to \operatorname{Hom}(C,Y)$ an isomorphism for all $C$ in a set of compact generators is an isomorphism.

Proof. (Expand to view)

The map $X \to Y$ is an isomorphism if and only if $\operatorname{cone}(X \to Y) = 0$ . Since we have a set of compact generators, we only need to check that $\operatorname{Hom}(C,\operatorname{cone}(X \to Y)) = 0$ for all $C$ . But this is equivalent to the map $\operatorname{Hom}(C,X) \to \operatorname{Hom}(C,Y)$ being an isomorphism for all $C$ . ■

Example. $D(R)$ is compactly generated. Indeed, $\operatorname{Hom}_{D(R)}(R[n],Y) \cong H^{-n}Y$ If all of these are $0$ , then $Y$ is acyclic and hence $0$ in $D(R)$ .

Definition. Given a sequence of composable maps $X_0 \overset{f_0}{\to} X_1 \overset{f_1}{\to} X_2 \overset{f_2}{\to} \cdots$ the homotopy colimit is the cone over the morphism $\bigoplus X_i \to \bigoplus X_i$ induced by the maps $X_i \overset{1,-f_i}{\to} X_i \oplus X_{i+1}$

Lemma. If $C$ is compact, then the natural map $\operatorname{colim} \operatorname{Hom}(C,X_\bullet) \to \operatorname{Hom}(C,\operatorname{hocolim} X_\bullet)$ is an isomorphism.

Proof. (Expand to view)

We have a commutative diagram

where the left two vertical maps are isomorphisms and the rows are exact. Since the map

\bigoplus \operatorname{Hom}(C,X_i) \to \bigoplus \operatorname{Hom}(C,X_i)

is injective we see that we have a short exact sequence

0 \to \operatorname{Hom}(C, \bigoplus X_i) \to \operatorname{Hom}(C, \bigoplus X_i) \to \operatorname{Hom}(C,\operatorname{hocolim} X_\bullet) \to 0

also. Thus, the map

\operatorname{colim}\operatorname{Hom}(C,X_\bullet) \to \operatorname{Hom}(C,\operatorname{hocolim} X_\bullet)

is also an isomorphism. ■

Proposition. If $\mathcal C$ is a compact set of generators, then the smallest subcategory of $\mathcal T$ that

is closed under $\bigoplus$ ’s
is triangulated and
contains $\mathcal C$

is $\mathcal T$ itself.

Proof. (Expand to view)

Let $U_0 := \bigcup_{C \in \mathcal C} \operatorname{Hom}(C,X)$ Set $X_0 := \bigoplus_{(C,f) \in U_0} C$ This comes with a natural evaluation map $X_0 \to X.$ We work not by induction having constructed $\nu_i : X_i \to X$ . We let $U_i := \bigcup_{C \in \mathcal C} \lbrace f : C \to X_i \mid \nu_i f = 0 \rbrace$ Let $K_i := \bigoplus_{(C,f) \in U_i} C$ and set $X_{i+1} = \operatorname{cone}(K_i \to X_i)$ . The composition $K_i \to X_i \to X$ is $0$ so there exists a $X_{i+1} \to X$

We let

X = \operatorname{hocolim} X_\bullet

. Using the above lemma, it suffices to check that

\operatorname{Hom}(C,\operatorname{hocolim}X_\bullet) \to \operatorname{Hom}(C,X)

is any isomorphism for any

C \in \mathcal C

. From the other lemma, this is the the same as

\operatorname{colim}\operatorname{Hom}(C,X_\bullet) \to \operatorname{Hom}(C,X)

being an isomorphism. Any element of the colimit is represented by some

f: C \to X_i

. If

\nu f : C \to X

0

, then

f \in U_i

be definition and, in fact,

f = 0

in the colimit. Given any

g: C \to X

, then

g \in U_0

. So there is some

C \to X_0

whose composition with

X_0 \to X

g

. Thus, we have an isomorphism. ■

If we have a compact generating set $\mathcal C$ , we can also use it characterize all compact objects in $\mathcal T$ . The following is due to Neeman.

Proposition. Let $\mathcal C$ be a set of compact generators for $\mathcal T$ . The smallest triangulated subcategory of $\mathcal T$

containing $\mathcal C$ and
closed under retracts

is $\mathcal T^c$ .

Proof. (Expand to view)

Pick some $X \in \mathcal T^c$ . From the proof the proposition, we know that $X \cong \operatorname{hocolim} X_\bullet$ where each $X_i$ is obtained as a cone over a map $K_{i-1} \to X_{i-1}$ with $K_{i-1}$ a sum of elements of $\mathcal C$ . We also know that $1_X \in \operatorname{Hom}(X,X) \cong \operatorname{colim}\operatorname{Hom}(X,X_i)$ Thus, there exists $X \to X_i$ such that

X

is a retract of

X_i

. The composition

X \to X_i \to K_{i-1}[1]

factors through a finite summand of

K_{i-1}

. Denote that by

K_{i-1}^\prime

and let

X^\prime_{i-1}[1]

be the cone over

X \to K_{i-1}^\prime[1]

. Now repeat the procedure to produce

From repeated applications of the octahedral axiom, the cone of the map

X^\prime_l \to X

is sequence of cones over the cones of

X^\prime_{j-1} \to X^\prime_j

. Each of these cones is a finite sum of objects from

\mathcal C

by construction. Thus the cone over

X_l \to X

for any

l

lies in the smallest triangulated subcategory containing the objects of

\mathcal C

. From the commutative square

we see that

X^\prime_{-1} \to X

0

X \to X_i \to X

is the identity. Thus,

X

and

X^\prime_{-1}

are summands over the cone of

X_{-1}^\prime \to X

. ■

Definition. A complex $P$ in $D(R)$ is called perfect if it is quasi-isomorphic to a bounded complex of finitely generated projectives. The full subcategory of perfect objects is denoted by $\operatorname{perf} R$ .

Proposition. We have $D(R)^c = \operatorname{perf} R$

Proof. (Expand to view)

It is easy to see that $\operatorname{perf} R \subseteq D(R)^c$ since compact objects are closed under finite direct sums, summands, and cones. But $\operatorname{perf} R$ - contains $R[n]$ for any $n \in \mathbb{Z}$ (a set of compact generators) - is triangulated and - is closed under retracts. Thus, the previous proposition says $D(R)^c \subseteq \operatorname{perf} R$ ■

Finally it useful to introduce some notation for generation.

Definition. Given a subcategory $\mathcal S$ of a triangulated category $\mathcal T$ , we let $\langle \mathcal S \rangle$ denote the smallest triangulated subcategory of $\mathcal T$ containing $\mathcal S$ .

We also let $\overline{\langle \mathcal S \rangle}$ denote the smallest triangulated, closed under retracts, containing $\mathcal S$ .