Proposition 101.6.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. The exact functor $f^*$ induces a commutative diagram

$\xymatrix{ D_{\mathcal{M}_\mathcal {X}}(\mathcal{O}_\mathcal {X}) \ar[r] & D(\mathcal{O}_\mathcal {X}) \\ D_{\mathcal{M}_\mathcal {Y}}(\mathcal{O}_\mathcal {Y}) \ar[r] \ar[u]^{f^*} & D(\mathcal{O}_\mathcal {Y}) \ar[u]^{f^*} }$

The composition

$D_{\mathcal{M}_\mathcal {Y}}(\mathcal{O}_\mathcal {Y}) \xrightarrow {f^*} D_{\mathcal{M}_\mathcal {X}}(\mathcal{O}_\mathcal {X}) \xrightarrow {q_\mathcal {X}} D_\mathit{QCoh}(\mathcal{O}_\mathcal {X})$

is left derivable with respect to the localization $D_{\mathcal{M}_\mathcal {Y}}(\mathcal{O}_\mathcal {Y}) \to D_\mathit{QCoh}(\mathcal{O}_\mathcal {Y})$ and we may define $Lf^*_\mathit{QCoh}$ as its left derived functor

$Lf_\mathit{QCoh}^* : D_\mathit{QCoh}(\mathcal{O}_\mathcal {Y}) \longrightarrow D_\mathit{QCoh}(\mathcal{O}_\mathcal {X})$

(see Derived Categories, Definitions 13.14.2 and 13.14.9). If $f$ is quasi-compact and quasi-separated, then $Lf^*_\mathit{QCoh}$ and $Rf_{\mathit{QCoh}, *}$ satisfy the following adjointness:

$\mathop{\mathrm{Hom}}\nolimits _{D_\mathit{QCoh}(\mathcal{O}_\mathcal {X})}(Lf^*_\mathit{QCoh}A, B) = \mathop{\mathrm{Hom}}\nolimits _{D_\mathit{QCoh}(\mathcal{O}_\mathcal {Y})}(A, Rf_{\mathit{QCoh}, *}B)$

for $A \in D_\mathit{QCoh}(\mathcal{O}_\mathcal {Y})$ and $B \in D^{+}_\mathit{QCoh}(\mathcal{O}_\mathcal {X})$.

Proof. To prove the first statement, we have to show that $f^*E$ is an object of $D_{\mathcal{M}_\mathcal {X}}(\mathcal{O}_\mathcal {X})$ for $E$ in $D_{\mathcal{M}_\mathcal {Y}}(\mathcal{O}_\mathcal {Y})$. Since $f^* = f^{-1}$ is exact this follows immediately from the fact that $f^*$ maps $\mathcal{M}_\mathcal {Y}$ into $\mathcal{M}_\mathcal {X}$.

Set $\mathcal{D} = D_{\mathcal{M}_\mathcal {Y}}(\mathcal{O}_\mathcal {Y})$. Let $S$ be the collection of morphisms in $\mathcal{D}$ whose cone is an object of $D_{\mathcal{P}_\mathcal {Y}}(\mathcal{O}_\mathcal {Y})$. Set $\mathcal{D}' = D_\mathit{QCoh}(\mathcal{O}_\mathcal {X})$. Set $F = q_\mathcal {X} \circ f^* : \mathcal{D} \to \mathcal{D}'$. Then $\mathcal{D}, S, \mathcal{D}', F$ are as in Derived Categories, Situation 13.14.1 and Definition 13.14.2. Let us prove that $LF(E)$ is defined for any object $E$ of $\mathcal{D}$. Namely, consider the triangle

$E' \to E \to P \to E'[1]$

constructed in Lemma 101.4.4. Note that $s : E' \to E$ is an element of $S$. We claim that $E'$ computes $LF$. Namely, suppose that $s' : E'' \to E$ is another element of $S$, i.e., fits into a triangle $E'' \to E \to P' \to E''[1]$ with $P'$ in $D_{\mathcal{P}_\mathcal {Y}}(\mathcal{O}_\mathcal {Y})$. By Lemma 101.4.4 (and its proof) we see that $E' \to E$ factors through $E'' \to E$. Thus we see that $E' \to E$ is cofinal in the system $S/E$. Hence it is clear that $E'$ computes $LF$.

To see the final statement, write $B = q_\mathcal {X}(H)$ and $A = q_\mathcal {Y}(E)$. Choose $E' \to E$ as above. We will use on the one hand that $Rf_{\mathit{QCoh}, *}(B) = q_\mathcal {Y}(Rf_*H)$ and on the other that $Lf^*_\mathit{QCoh}(A) = q_\mathcal {X}(f^*E')$.

\begin{align*} \mathop{\mathrm{Hom}}\nolimits _{D_\mathit{QCoh}(\mathcal{O}_\mathcal {X})}(Lf^*_\mathit{QCoh}A, B) & = \mathop{\mathrm{Hom}}\nolimits _{D_\mathit{QCoh}(\mathcal{O}_\mathcal {X})}(q_\mathcal {X}(f^*E'), q_\mathcal {X}(H)) \\ & = \mathop{\mathrm{colim}}\nolimits _{H \to H'} \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_\mathcal {X})}(f^*E', H') \\ & = \mathop{\mathrm{colim}}\nolimits _{H \to H'} \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_\mathcal {Y})}(E', Rf_*H') \\ & = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_\mathcal {Y})}(E', Rf_*H) \\ & = \mathop{\mathrm{Hom}}\nolimits _{D_\mathit{QCoh}(\mathcal{O}_\mathcal {Y})}(A, Rf_{\mathit{QCoh}, *}B) \end{align*}

Here the colimit is over morphisms $s : H \to H'$ in $D^+_{\mathcal{M}_\mathcal {X}}(\mathcal{O}_\mathcal {X})$ whose cone $P(s)$ is an object of $D^+_{\mathcal{P}_\mathcal {X}}(\mathcal{O}_\mathcal {X})$. The first equality we've seen above. The second equality holds by construction of the Verdier quotient. The third equality holds by Cohomology on Sites, Lemma 21.19.1. Since $Rf_*P(s)$ is an object of $D^+_{\mathcal{P}_\mathcal {Y}}(\mathcal{O}_\mathcal {Y})$ by Proposition 101.5.1 we see that $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_\mathcal {Y})}(E', Rf_*P(s)) = 0$. Thus the fourth equality holds. The final equality holds by construction of $E'$. $\square$

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