## 94.14 Quasi-coherent sheaves on algebraic stacks

Let $\mathcal{X}$ be an algebraic stack over $S$. By Algebraic Stacks, Lemma 92.16.2 we can find an equivalence $[U/R] \to \mathcal{X}$ where $(U, R, s, t, c)$ is a smooth groupoid in algebraic spaces. Then

\[ \mathit{QCoh}(\mathcal{O}_\mathcal {X}) \cong \mathit{QCoh}(\mathcal{O}_{[U/R]}) \cong \mathit{QCoh}(U, R, s, t, c) \]

where the second equivalence is Proposition 94.13.1. Hence the category of quasi-coherent sheaves on an algebraic stack is equivalent to the category of quasi-coherent modules on a smooth groupoid in algebraic spaces. In particular, by Groupoids in Spaces, Lemma 76.12.5 we see that $\mathit{QCoh}(\mathcal{O}_\mathcal {X})$ is abelian!

There is something slightly disconcerting about our current setup. It is that the fully faithful embedding

\[ \mathit{QCoh}(\mathcal{O}_\mathcal {X}) \longrightarrow \textit{Mod}(\mathcal{O}_\mathcal {X}) \]

is in general **not** exact. However, exactly the same thing happens for schemes: for most schemes $X$ the embedding

\[ \mathit{QCoh}(\mathcal{O}_ X) \cong \mathit{QCoh}((\mathit{Sch}/X)_{fppf}, \mathcal{O}_ X) \longrightarrow \textit{Mod}((\mathit{Sch}/X)_{fppf}, \mathcal{O}_ X) \]

isn't exact, see Descent, Lemma 35.8.13. Parenthetically, the example in the proof of Descent, Lemma 35.8.13 shows that in general the strictly full embedding $\mathit{QCoh}(\mathcal{O}_\mathcal {X}) \to \textit{LQCoh}(\mathcal{O}_\mathcal {X})$ isn't exact either.

We collect all the positive results obtained so far in a single statement.

Lemma 94.14.1. Let $\mathcal{X}$ be an algebraic stack over $S$.

If $[U/R] \to \mathcal{X}$ is a presentation of $\mathcal{X}$ then there is a canonical equivalence $\mathit{QCoh}(\mathcal{O}_\mathcal {X}) \cong \mathit{QCoh}(U, R, s, t, c)$.

The category $\mathit{QCoh}(\mathcal{O}_\mathcal {X})$ is abelian.

The category $\mathit{QCoh}(\mathcal{O}_\mathcal {X})$ has colimits and they agree with colimits in the category $\textit{Mod}(\mathcal{O}_\mathcal {X})$.

Given $\mathcal{F}, \mathcal{G}$ in $\mathit{QCoh}(\mathcal{O}_\mathcal {X})$ the tensor product $\mathcal{F} \otimes _{\mathcal{O}_\mathcal {X}} \mathcal{G}$ in $\textit{Mod}(\mathcal{O}_\mathcal {X})$ is an object of $\mathit{QCoh}(\mathcal{O}_\mathcal {X})$.

Given $\mathcal{F}, \mathcal{G}$ in $\mathit{QCoh}(\mathcal{O}_\mathcal {X})$ with $\mathcal{F}$ locally of finite presentation on $\mathcal{X}_{fppf}$ the sheaf $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_\mathcal {X}}(\mathcal{F}, \mathcal{G})$ in $\textit{Mod}(\mathcal{O}_\mathcal {X})$ is an object of $\mathit{QCoh}(\mathcal{O}_\mathcal {X})$.

**Proof.**
Properties (3), (4), and (5) were proven in Lemma 94.11.9. Part (1) is Proposition 94.13.1. Part (2) follows from Groupoids in Spaces, Lemma 76.12.5 as discussed above.
$\square$

Proposition 94.14.2. Let $\mathcal{X}$ be an algebraic stack over $S$.

The category $\mathit{QCoh}(\mathcal{O}_\mathcal {X})$ is a Grothendieck abelian category. Consequently, $\mathit{QCoh}(\mathcal{O}_\mathcal {X})$ has enough injectives and all limits.

The inclusion functor $\mathit{QCoh}(\mathcal{O}_\mathcal {X}) \to \textit{Mod}(\mathcal{O}_\mathcal {X})$ has a right adjoint^{1}

\[ Q : \textit{Mod}(\mathcal{O}_\mathcal {X}) \longrightarrow \mathit{QCoh}(\mathcal{O}_\mathcal {X}) \]

such that for every quasi-coherent sheaf $\mathcal{F}$ the adjunction mapping $Q(\mathcal{F}) \to \mathcal{F}$ is an isomorphism.

**Proof.**
This proof is a repeat of the proof in the case of schemes, see Properties, Proposition 28.23.4 and the case of algebraic spaces, see Properties of Spaces, Proposition 64.32.2. We advise the reader to read either of those proofs first.

Part (1) means $\mathit{QCoh}(\mathcal{O}_\mathcal {X})$ (a) has all colimits, (b) filtered colimits are exact, and (c) has a generator, see Injectives, Section 19.10. By Lemma 94.14.1 colimits in $\mathit{QCoh}(\mathcal{O}_ X)$ exist and agree with colimits in $\textit{Mod}(\mathcal{O}_ X)$. By Modules on Sites, Lemma 18.14.2 filtered colimits are exact. Hence (a) and (b) hold.

Choose a presentation $\mathcal{X} = [U/R]$ so that $(U, R, s, t, c)$ is a smooth groupoid in algebraic spaces and in particular $s$ and $t$ are flat morphisms of algebraic spaces. By Lemma 94.14.1 above we have $\mathit{QCoh}(\mathcal{O}_\mathcal {X}) = \mathit{QCoh}(U, R, s, t, c)$. By Groupoids in Spaces, Lemma 76.13.2 there exists a set $T$ and a family $(\mathcal{F}_ t)_{t \in T}$ of quasi-coherent sheaves on $\mathcal{X}$ such that every quasi-coherent sheaf on $\mathcal{X}$ is the directed colimit of its subsheaves which are isomorphic to one of the $\mathcal{F}_ t$. Thus $\bigoplus _ t \mathcal{F}_ t$ is a generator of $\mathit{QCoh}(\mathcal{O}_ X)$ and we conclude that (c) holds. The assertions on limits and injectives hold in any Grothendieck abelian category, see Injectives, Theorem 19.11.7 and Lemma 19.13.2.

Proof of (2). To construct $Q$ we use the following general procedure. Given an object $\mathcal{F}$ of $\textit{Mod}(\mathcal{O}_\mathcal {X})$ we consider the functor

\[ \mathit{QCoh}(\mathcal{O}_\mathcal {X})^{opp} \longrightarrow \textit{Sets}, \quad \mathcal{G} \longmapsto \mathop{\mathrm{Hom}}\nolimits _\mathcal {X}(\mathcal{G}, \mathcal{F}) \]

This functor transforms colimits into limits, hence is representable, see Injectives, Lemma 19.13.1. Thus there exists a quasi-coherent sheaf $Q(\mathcal{F})$ and a functorial isomorphism $\mathop{\mathrm{Hom}}\nolimits _\mathcal {X}(\mathcal{G}, \mathcal{F}) = \mathop{\mathrm{Hom}}\nolimits _\mathcal {X}(\mathcal{G}, Q(\mathcal{F}))$ for $\mathcal{G}$ in $\mathit{QCoh}(\mathcal{O}_\mathcal {X})$. By the Yoneda lemma (Categories, Lemma 4.3.5) the construction $\mathcal{F} \leadsto Q(\mathcal{F})$ is functorial in $\mathcal{F}$. By construction $Q$ is a right adjoint to the inclusion functor. The fact that $Q(\mathcal{F}) \to \mathcal{F}$ is an isomorphism when $\mathcal{F}$ is quasi-coherent is a formal consequence of the fact that the inclusion functor $\mathit{QCoh}(\mathcal{O}_\mathcal {X}) \to \textit{Mod}(\mathcal{O}_\mathcal {X})$ is fully faithful.
$\square$

## Comments (0)