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86.3 Right adjoint of pushforward

This is the analogue of Duality for Schemes, Section 48.3.

Lemma 86.3.1.reference 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.

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)

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)

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.

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

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)))

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

  1. L and Rf_*L are perfect, or

  2. K is bounded below and L and Rf_*L are pseudo-coherent.

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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