Lemma 75.19.1. Let $S$ be a scheme. Let $X$ be a quasi-compact and quasi-separated algebraic space over $S$. The inclusion functor $D_\mathit{QCoh}(\mathcal{O}_ X) \to D(\mathcal{O}_ X)$ has a right adjoint.
First proof. We will use the induction principle in Lemma 75.9.3 to prove this. If $D(\mathit{QCoh}(\mathcal{O}_ X)) \to D_\mathit{QCoh}(\mathcal{O}_ X)$ is an equivalence, then the lemma is true because the functor $RQ_ X$ of Section 75.11 is a right adjoint to the functor $D(\mathit{QCoh}(\mathcal{O}_ X)) \to D(\mathcal{O}_ X)$. In particular, our lemma is true for affine algebraic spaces, see Lemma 75.11.3. Thus we see that it suffices to show: if $(U \subset X, f : V \to X)$ is an elementary distinguished square with $U$ quasi-compact and $V$ affine and the lemma holds for $U$, $V$, and $U \times _ X V$, then the lemma holds for $X$.
The adjoint exists if and only if for every object $K$ of $D(\mathcal{O}_ X)$ we can find a distinguished triangle
\[ E' \to E \to K \to E'[1] \]
in $D(\mathcal{O}_ X)$ such that $E'$ is in $D_\mathit{QCoh}(\mathcal{O}_ X)$ and such that $\mathop{\mathrm{Hom}}\nolimits (M, K) = 0$ for all $M$ in $D_\mathit{QCoh}(\mathcal{O}_ X)$. See Derived Categories, Lemma 13.40.7. Consider the distinguished triangle
\[ E \to Rj_{U, *}E|_ U \oplus Rj_{V, *}E|_ V \to Rj_{U \times _ X V, *}E|_{U \times _ X V} \to E[1] \]
in $D(\mathcal{O}_ X)$ of Lemma 75.10.2. By Derived Categories, Lemma 13.40.5 it suffices to construct the desired distinguished triangles for $Rj_{U, *}E|_ U$, $Rj_{V, *}E|_ V$, and $Rj_{U \times _ X V, *}E|_{U \times _ X V}$. This reduces us to the statement discussed in the next paragraph.
Let $j : U \to X$ be an étale morphism corresponding with $U$ quasi-compact and quasi-separated and the lemma is true for $U$. Let $L$ be an object of $D(\mathcal{O}_ U)$. Then there exists a distinguished triangle
\[ E' \to Rj_*L \to K \to E'[1] \]
in $D(\mathcal{O}_ X)$ such that $E'$ is in $D_\mathit{QCoh}(\mathcal{O}_ X)$ and such that $\mathop{\mathrm{Hom}}\nolimits (M, K) = 0$ for all $M$ in $D_\mathit{QCoh}(\mathcal{O}_ X)$. To see this we choose a distinguished triangle
\[ L' \to L \to Q \to L'[1] \]
in $D(\mathcal{O}_ U)$ such that $L'$ is in $D_\mathit{QCoh}(\mathcal{O}_ U)$ and such that $\mathop{\mathrm{Hom}}\nolimits (N, Q) = 0$ for all $N$ in $D_\mathit{QCoh}(\mathcal{O}_ U)$. This is possible because the statement in Derived Categories, Lemma 13.40.7 is an if and only if. We obtain a distinguished triangle
\[ Rj_*L' \to Rj_*L \to Rj_*Q \to Rj_*L'[1] \]
in $D(\mathcal{O}_ X)$. Observe that $Rj_*L'$ is in $D_\mathit{QCoh}(\mathcal{O}_ X)$ by Lemma 75.6.1. On the other hand, if $M$ in $D_\mathit{QCoh}(\mathcal{O}_ X)$, then
\[ \mathop{\mathrm{Hom}}\nolimits (M, Rj_*Q) = \mathop{\mathrm{Hom}}\nolimits (Lj^*M, Q) = 0 \]
because $Lj^*M$ is in $D_\mathit{QCoh}(\mathcal{O}_ U)$ by Lemma 75.5.5. This finishes the proof. $\square$
Second proof. The adjoint exists by Derived Categories, Proposition 13.38.2. The hypotheses are satisfied: First, note that $D_\mathit{QCoh}(\mathcal{O}_ X)$ has direct sums and direct sums commute with the inclusion functor (Lemma 75.5.3). On the other hand, $D_\mathit{QCoh}(\mathcal{O}_ X)$ is compactly generated because it has a perfect generator Theorem 75.15.4 and because perfect objects are compact by Proposition 75.16.1. $\square$
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