The Stacks project

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

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


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

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. $\square$

Comments (3)

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 instead of .

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 such that the composition is 0."

PS.: Isn't this method of proof due to Takeuchi

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