Lemma 86.3.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism between quasi-separated and quasi-compact algebraic spaces over $S$. The functor $Rf_* : D_\mathit{QCoh}(X) \to D_\mathit{QCoh}(Y)$ has a right adjoint.
86.3 Right adjoint of pushforward
This is the analogue of Duality for Schemes, Section 48.3.
Proof. We will prove a right adjoint exists by verifying the hypotheses of Derived Categories, Proposition 13.38.2. First off, the category $D_\mathit{QCoh}(\mathcal{O}_ X)$ has direct sums, see Derived Categories of Spaces, Lemma 75.5.3. The category $D_\mathit{QCoh}(\mathcal{O}_ X)$ is compactly generated by Derived Categories of Spaces, Theorem 75.15.4. Since $X$ and $Y$ are quasi-compact and quasi-separated, so is $f$, see Morphisms of Spaces, Lemmas 67.4.10 and 67.8.9. Hence the functor $Rf_*$ commutes with direct sums, see Derived Categories of Spaces, Lemma 75.6.2. This finishes the proof. $\square$
Lemma 86.3.2. Notation and assumptions as in Lemma 86.3.1. Let $a : D_\mathit{QCoh}(\mathcal{O}_ Y) \to D_\mathit{QCoh}(\mathcal{O}_ X)$ be the right adjoint to $Rf_*$. Then $a$ maps $D^+_\mathit{QCoh}(\mathcal{O}_ Y)$ into $D^+_\mathit{QCoh}(\mathcal{O}_ X)$. In fact, there exists an integer $N$ such that $H^ i(K) = 0$ for $i \leq c$ implies $H^ i(a(K)) = 0$ for $i \leq c - N$.
Proof. By Derived Categories of Spaces, Lemma 75.6.1 the functor $Rf_*$ has finite cohomological dimension. In other words, there exist an integer $N$ such that $H^ i(Rf_*L) = 0$ for $i \geq N + c$ if $H^ i(L) = 0$ for $i \geq c$. Say $K \in D^+_\mathit{QCoh}(\mathcal{O}_ Y)$ has $H^ i(K) = 0$ for $i \leq c$. Then
\[ \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(\tau _{\leq c - N}a(K), a(K)) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ Y)}(Rf_*\tau _{\leq c - N}a(K), K) = 0 \]
by what we said above. Clearly, this implies that $H^ i(a(K)) = 0$ for $i \leq c - N$. $\square$
Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of quasi-separated and quasi-compact algebraic spaces over $S$. Let $a$ denote the right adjoint to $Rf_* : D_\mathit{QCoh}(\mathcal{O}_ X) \to D_\mathit{QCoh}(\mathcal{O}_ Y)$. For every $K \in D_\mathit{QCoh}(\mathcal{O}_ Y)$ and $L \in D_\mathit{QCoh}(\mathcal{O}_ X)$ we obtain a canonical map
86.3.2.1
\begin{equation} \label{spaces-duality-equation-sheafy-trace} Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(L, a(K)) \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*L, K) \end{equation}
Namely, this map is constructed as the composition
\[ Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(L, a(K)) \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*L, Rf_*a(K)) \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*L, K) \]
where the first arrow is Cohomology on Sites, Remark 21.35.10 and the second arrow is the counit $Rf_*a(K) \to K$ of the adjunction.
Lemma 86.3.3. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of quasi-compact and quasi-separated algebraic spaces over $S$. Let $a$ be the right adjoint to $Rf_* : D_\mathit{QCoh}(\mathcal{O}_ X) \to D_\mathit{QCoh}(\mathcal{O}_ Y)$. Let $L \in D_\mathit{QCoh}(\mathcal{O}_ X)$ and $K \in D_\mathit{QCoh}(\mathcal{O}_ Y)$. Then the map (86.3.2.1) becomes an isomorphism after applying the functor $DQ_ Y : D(\mathcal{O}_ Y) \to D_\mathit{QCoh}(\mathcal{O}_ Y)$ discussed in Derived Categories of Spaces, Section 75.19.
\[ Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(L, a(K)) \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*L, K) \]
Proof. The statement makes sense as $DQ_ Y$ exists by Derived Categories of Spaces, Lemma 75.19.1. Since $DQ_ Y$ is the right adjoint to the inclusion functor $D_\mathit{QCoh}(\mathcal{O}_ Y) \to D(\mathcal{O}_ Y)$ to prove the lemma we have to show that for any $M \in D_\mathit{QCoh}(\mathcal{O}_ Y)$ the map (86.3.2.1) induces an bijection
\[ \mathop{\mathrm{Hom}}\nolimits _ Y(M, Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(L, a(K))) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _ Y(M, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*L, K)) \]
To see this we use the following string of equalities
\begin{align*} \mathop{\mathrm{Hom}}\nolimits _ Y(M, Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(L, a(K))) & = \mathop{\mathrm{Hom}}\nolimits _ X(Lf^*M, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(L, a(K))) \\ & = \mathop{\mathrm{Hom}}\nolimits _ X(Lf^*M \otimes _{\mathcal{O}_ X}^\mathbf {L} L, a(K)) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y(Rf_*(Lf^*M \otimes _{\mathcal{O}_ X}^\mathbf {L} L), K) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y(M \otimes _{\mathcal{O}_ Y}^\mathbf {L} Rf_*L, K) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y(M, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*L, K)) \end{align*}
The first equality holds by Cohomology on Sites, Lemma 21.19.1. The second equality by Cohomology on Sites, Lemma 21.35.2. The third equality by construction of $a$. The fourth equality by Derived Categories of Spaces, Lemma 75.20.1 (this is the important step). The fifth by Cohomology on Sites, Lemma 21.35.2. $\square$
Example 86.3.4. The statement of Lemma 86.3.3 is not true without applying the “coherator” $DQ_ Y$. See Duality for Schemes, Example 48.3.7.
Remark 86.3.5. In the situation of Lemma 86.3.3 we have by Derived Categories of Spaces, Lemma 75.19.2. Thus if $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(L, a(K)) \in D_\mathit{QCoh}(\mathcal{O}_ X)$, then we can “erase” the $DQ_ Y$ on the left hand side of the arrow. On the other hand, if we know that $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*L, K) \in D_\mathit{QCoh}(\mathcal{O}_ Y)$, then we can “erase” the $DQ_ Y$ from the right hand side of the arrow. If both are true then we see that (86.3.2.1) is an isomorphism. Combining this with Derived Categories of Spaces, Lemma 75.13.10 we see that $Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(L, a(K)) \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*L, K)$ is an isomorphism if
$L$ and $Rf_*L$ are perfect, or
$K$ is bounded below and $L$ and $Rf_*L$ are pseudo-coherent.
\[ DQ_ Y(Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(L, a(K))) = Rf_* DQ_ X(R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(L, a(K))) \]
For (2) we use that $a(K)$ is bounded below if $K$ is bounded below, see Lemma 86.3.2.
Example 86.3.6. Let $S$ be a scheme. Let $f : X \to Y$ be a proper morphism of Noetherian algebraic spaces over $S$, $L \in D^-_{\textit{Coh}}(X)$ and $K \in D^+_{\mathit{QCoh}}(\mathcal{O}_ Y)$. Then the map $Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(L, a(K)) \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*L, K)$ is an isomorphism. Namely, the complexes $L$ and $Rf_*L$ are pseudo-coherent by Derived Categories of Spaces, Lemmas 75.13.7 and 75.8.1 and the discussion in Remark 86.3.5 applies.
Lemma 86.3.7. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of quasi-separated and quasi-compact algebraic spaces over $S$. For all $L \in D_\mathit{QCoh}(\mathcal{O}_ X)$ and $K \in D_\mathit{QCoh}(\mathcal{O}_ Y)$ (86.3.2.1) induces an isomorphism $R\mathop{\mathrm{Hom}}\nolimits _ X(L, a(K)) \to R\mathop{\mathrm{Hom}}\nolimits _ Y(Rf_*L, K)$ of global derived homs.
Proof. By construction (Cohomology on Sites, Section 21.36) the complexes
\[ R\mathop{\mathrm{Hom}}\nolimits _ X(L, a(K)) = R\Gamma (X, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(L, a(K))) = R\Gamma (Y, Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(L, a(K))) \]
and
\[ R\mathop{\mathrm{Hom}}\nolimits _ Y(Rf_*L, K) = R\Gamma (Y, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(Rf_*L, a(K))) \]
Thus the lemma is a consequence of Lemma 86.3.3. Namely, a map $E \to E'$ in $D(\mathcal{O}_ Y)$ which induces an isomorphism $DQ_ Y(E) \to DQ_ Y(E')$ induces a quasi-isomorphism $R\Gamma (Y, E) \to R\Gamma (Y, E')$. Indeed we have $H^ i(Y, E) = \mathop{\mathrm{Ext}}\nolimits ^ i_ Y(\mathcal{O}_ Y, E) = \mathop{\mathrm{Hom}}\nolimits (\mathcal{O}_ Y[-i], E) = \mathop{\mathrm{Hom}}\nolimits (\mathcal{O}_ Y[-i], DQ_ Y(E))$ because $\mathcal{O}_ Y[-i]$ is in $D_\mathit{QCoh}(\mathcal{O}_ Y)$ and $DQ_ Y$ is the right adjoint to the inclusion functor $D_\mathit{QCoh}(\mathcal{O}_ Y) \to D(\mathcal{O}_ Y)$. $\square$
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