## 19.14 The Gabriel-Popescu theorem

In this section we discuss the main theorem of [GP]. The method of proof follows a write-up by Jacob Lurie and another by Akhil Mathew who in turn follow the presentation by Kuhn in [Kuhn]. See also [Takeuchi].

Let $\mathcal{A}$ be a Grothendieck abelian category and let $U$ be a generator for $\mathcal{A}$, see Definition 19.10.1. Let $R = \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(U, U)$. Consider the functor $G : \mathcal{A} \to \text{Mod}_ R$ given by

$G(A) = \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(U, A)$

endowed with its canonical right $R$-module structure.

Lemma 19.14.1. The functor $G$ above has a left adjoint $F : \text{Mod}_ R \to \mathcal{A}$.

Proof. We will give two proofs of this lemma.

The first proof will use the adjoint functor theorem, see Categories, Theorem 4.25.3. Observe that that $G : \mathcal{A} \to \text{Mod}_ R$ is left exact and sends products to products. Hence $G$ commutes with limits. To check the set theoretical condition in the theorem, suppose that $M$ is an object of $\text{Mod}_ R$. Choose a suitably large cardinal $\kappa$ and denote $E$ a set of objects of $\mathcal{A}$ such that every object $A$ with $|A| \leq \kappa$ is isomorphic to an element of $E$. This is possible by Lemma 19.11.4. Set $I = \coprod _{A \in E} \mathop{\mathrm{Hom}}\nolimits _ R(M, G(A))$. We think of an element $i \in I$ as a pair $(A_ i, f_ i)$. Finally, let $A$ be an arbitrary object of $\mathcal{A}$ and $f : M \to G(A)$ arbitrary. We are going to think of elements of $\mathop{\mathrm{Im}}(f) \subset G(A) = \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(U, A)$ as maps $u : U \to A$. Set

$A' = \mathop{\mathrm{Im}}(\bigoplus \nolimits _{u \in \mathop{\mathrm{Im}}(f)} U \xrightarrow {u} A)$

Since $G$ is left exact, we see that $G(A') \subset G(A)$ contains $\mathop{\mathrm{Im}}(f)$ and we get $f' : M \to G(A')$ factoring $f$. On the other hand, the object $A'$ is the quotient of a direct sum of at most $|M|$ copies of $U$. Hence if $\kappa = |\bigoplus _{|M|} U|$, then we see that $(A', f')$ is isomorphic to an element $(A_ i, f_ i)$ of $E$ and we conclude that $f$ factors as $M \xrightarrow {f_ i} G(A_ i) \to G(A)$ as desired.

The second proof will give a construction of $F$ which will show that “$F(M) = M \otimes _ R U$” in some sense. Namely, for any $R$-module $M$ we can choose a resolution

$\bigoplus \nolimits _{j \in J} R \to \bigoplus \nolimits _{i \in I} R \to M \to 0$

Then we define $F(M)$ by the corresponding exact sequence

$\bigoplus \nolimits _{j \in J} U \to \bigoplus \nolimits _{i \in I} U \to F(M) \to 0$

This construction is independent of the choice of the resolution and is functorial; we omit the details. For any $A$ in $\mathcal{A}$ we obtain an exact sequence

$0 \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(F(M), A) \to \prod \nolimits _{i \in I} G(A) \to \prod \nolimits _{j \in J} G(A)$

which is isomorphic to the sequence

$0 \to \mathop{\mathrm{Hom}}\nolimits _ R(M, G(A)) \to \mathop{\mathrm{Hom}}\nolimits _ R(\bigoplus \nolimits _{i \in I} R, G(A)) \to \mathop{\mathrm{Hom}}\nolimits _ R(\bigoplus \nolimits _{j \in J} R, G(A))$

which shows that $F$ is the left adjoint to $G$. $\square$

Lemma 19.14.2. Let $f : M \to G(A)$ be an injective map in $\text{Mod}_ R$. Then the adjoint map $f' : F(M) \to A$ is injective too.

Proof. Choose a map $R^{\oplus n} \to M$ and consider the corresponding map $U^{\oplus n} \to F(M)$. Consider a map $v : U \to U^{\oplus n}$ such that the composition $U \to U^{\oplus n} \to F(M) \to A$ is $0$. Then this arrow $v : U \to U^{\oplus n}$ is an element $v$ of $R^{\oplus n}$ mapping to zero in $G(A)$. Since $f$ is injective, we conclude that $v$ maps to zero in $M$ which means that $U \to U^{\oplus n} \to F(M)$ is zero by construction of $F(M)$ in the proof of Lemma 19.14.1. Since $U$ is a generator we conclude that

$\mathop{\mathrm{Ker}}(U^{\oplus n} \to F(M) \to A) = \mathop{\mathrm{Ker}}(U^{\oplus n} \to F(M))$

To finish the proof we choose a surjection $\bigoplus _{i \in I} R \to M$ and we consider the corresponding surjection

$\pi : \bigoplus \nolimits _{i \in I} U \longrightarrow F(M)$

To prove $f'$ is injective it suffices to show that $\mathop{\mathrm{Ker}}(\pi ) = \mathop{\mathrm{Ker}}(f' \circ \pi )$ as subobjects of $\bigoplus _{i \in I} U$. However, now we can write $\bigoplus _{i \in I} U$ as the filtered colimit of its subobjects $\bigoplus _{i \in I'} U$ where $I' \subset I$ ranges over the finite subsets. Since filtered colimits are exact by AB5 for $\mathcal{A}$, we see that

$\mathop{\mathrm{Ker}}(\pi ) = \mathop{\mathrm{colim}}\nolimits _{I' \subset I\text{ finite}} \left(\bigoplus \nolimits _{i \in I'} U\right) \bigcap \mathop{\mathrm{Ker}}(\pi )$

and

$\mathop{\mathrm{Ker}}(f' \circ \pi ) = \mathop{\mathrm{colim}}\nolimits _{I' \subset I\text{ finite}} \left(\bigoplus \nolimits _{i \in I'} U\right) \bigcap \mathop{\mathrm{Ker}}(f' \circ \pi )$

and we get equality because the same is true for each $I'$ by the first displayed equality above. $\square$

Theorem 19.14.3. Let $\mathcal{A}$ be a Grothendieck abelian category. Then there exists a (noncommutative) ring $R$ and functors $G : \mathcal{A} \to \text{Mod}_ R$ and $F : \text{Mod}_ R \to \mathcal{A}$ such that

1. $F$ is the left adjoint to $G$,

2. $G$ is fully faithful, and

3. $F$ is exact.

Moreover, the functors are the ones constructed above.

Proof. We first prove $G$ is fully faithful, or equivalently that $F \circ G \to \text{id}$ is an isomorphism, see Categories, Lemma 4.24.4. First, given an object $A$ the map $F(G(A)) \to A$ is surjective, because every map of $U \to A$ factors through $F(G(A))$ by construction. On the other hand, the map $F(G(A)) \to A$ is the adjoint of the map $\text{id} : G(A) \to G(A)$ and hence injective by Lemma 19.14.2.

The functor $F$ is right exact as it is a left adjoint. Since $\text{Mod}_ R$ has enough projectives, to show that $F$ is exact, it is enough to show that the first left derived functor $L_1F$ is zero. To prove $L_1F(M) = 0$ for some $R$-module $M$ choose an exact sequence $0 \to K \to P \to M \to 0$ of $R$-modules with $P$ free. It suffices to show $F(K) \to F(P)$ is injective. Now we can write this sequence as a filtered colimit of sequences $0 \to K_ i \to P_ i \to M_ i \to 0$ with $P_ i$ a finite free $R$-module: just write $P$ in this manner and set $K_ i = K \cap P_ i$ and $M_ i = \mathop{\mathrm{Im}}(P_ i \to M)$. Because $F$ is a left adjoint it commutes with colimits and because $\mathcal{A}$ is a Grothendieck abelian category, we find that $F(K) \to F(P)$ is injective if each $F(K_ i) \to F(P_ i)$ is injective. Thus it suffices to check $F(K) \to F(P)$ is injective when $K \subset P = R^{\oplus n}$. Thus $F(K) \to U^{\oplus n}$ is injective by an application of Lemma 19.14.2. $\square$

Lemma 19.14.4. Let $\mathcal{A}$ be a Grothendieck abelian category. Let $R$, $F$, $G$ be as in the Gabriel-Popescu theorem (Theorem 19.14.3). Then we obtain derived functors

$RG : D(\mathcal{A}) \to D(\text{Mod}_ R) \quad \text{and}\quad F : D(\text{Mod}_ R) \to D(\mathcal{A})$

such that $F$ is left adjoint to $RG$, $RG$ is fully faithful, and $F \circ RG = \text{id}$.

Proof. The existence and adjointness of the functors follows from Theorems 19.14.3 and 19.12.6 and Derived Categories, Lemmas 13.31.6, 13.16.9, and 13.30.3. The statement $F \circ RG = \text{id}$ follows because we can compute $RG$ on an object of $D(\mathcal{A})$ by applying $G$ to a suitable representative complex $I^\bullet$ (for example a K-injective one) and then $F(G(I^\bullet )) = I^\bullet$ because $F \circ G = \text{id}$. Fully faithfulness of $RG$ follows from this by Categories, Lemma 4.24.4. $\square$

Comment #6837 by Thiago Landim on

I believe there is a typo in the last exact sequence of the proof of the Lemma 19.14.1. It should be $\operatorname{Hom}_R(M, G(A))$ instead of $\operatorname{Hom}_R(M, A)$.

Comment #6838 by Thiago Landim on

I also believe in the second line of the proof of Lemma 19.14.2, it should be "Consider a map $v:U \rightarrow U^{\oplus n}$ such that the composition $U\rightarrow U^{\oplus n} \rightarrow F(M)\rightarrow A$ is 0."

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0F5R. Beware of the difference between the letter 'O' and the digit '0'.