## 48.3 Right adjoint of pushforward

References for this section and the following are , [LN], , and .

Let $f : X \to Y$ be a morphism of schemes. In this section we consider the right adjoint to the functor $Rf_* : D_\mathit{QCoh}(\mathcal{O}_ X) \to D_\mathit{QCoh}(\mathcal{O}_ Y)$. In the literature, if this functor exists, then it is sometimes denoted $f^{\times }$. This notation is not universally accepted and we refrain from using it. We will not use the notation $f^!$ for such a functor, as this would clash (for general morphisms $f$) with the notation in [RD].

Lemma 48.3.1. Let $f : X \to Y$ be a morphism between quasi-separated and quasi-compact schemes. 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 Schemes, Lemma 36.3.1. The category $D_\mathit{QCoh}(\mathcal{O}_ X)$ is compactly generated by Derived Categories of Schemes, Theorem 36.15.3. Since $X$ and $Y$ are quasi-compact and quasi-separated, so is $f$, see Schemes, Lemmas 26.21.13 and 26.21.14. Hence the functor $Rf_*$ commutes with direct sums, see Derived Categories of Schemes, Lemma 36.4.5. This finishes the proof. $\square$

Example 48.3.2. Let $A \to B$ be a ring map. Let $Y = \mathop{\mathrm{Spec}}(A)$ and $X = \mathop{\mathrm{Spec}}(B)$ and $f : X \to Y$ the morphism corresponding to $A \to B$. Then $Rf_* : D_\mathit{QCoh}(\mathcal{O}_ X) \to D_\mathit{QCoh}(\mathcal{O}_ Y)$ corresponds to restriction $D(B) \to D(A)$ via the equivalences $D(B) \to D_\mathit{QCoh}(\mathcal{O}_ X)$ and $D(A) \to D_\mathit{QCoh}(\mathcal{O}_ Y)$. Hence the right adjoint corresponds to the functor $K \longmapsto R\mathop{\mathrm{Hom}}\nolimits (B, K)$ of Dualizing Complexes, Section 47.13.

Example 48.3.3. If $f : X \to Y$ is a separated finite type morphism of Noetherian schemes, then the right adjoint of $Rf_* : D_\mathit{QCoh}(\mathcal{O}_ X) \to D_\mathit{QCoh}(\mathcal{O}_ Y)$ does not map $D_{\textit{Coh}}(\mathcal{O}_ Y)$ into $D_{\textit{Coh}}(\mathcal{O}_ X)$. Namely, let $k$ be a field and consider the morphism $f : \mathbf{A}^1_ k \to \mathop{\mathrm{Spec}}(k)$. By Example 48.3.2 this corresponds to the question of whether $R\mathop{\mathrm{Hom}}\nolimits (B, -)$ maps $D_{\textit{Coh}}(A)$ into $D_{\textit{Coh}}(B)$ where $A = k$ and $B = k[x]$. This is not true because

$R\mathop{\mathrm{Hom}}\nolimits (k[x], k) = \left(\prod \nolimits _{n \geq 0} k\right)[0]$

which is not a finite $k[x]$-module. Hence $a(\mathcal{O}_ Y)$ does not have coherent cohomology sheaves.

Example 48.3.4. If $f : X \to Y$ is a proper or even finite morphism of Noetherian schemes, then the right adjoint of $Rf_* : D_\mathit{QCoh}(\mathcal{O}_ X) \to D_\mathit{QCoh}(\mathcal{O}_ Y)$ does not map $D_\mathit{QCoh}^-(\mathcal{O}_ Y)$ into $D_\mathit{QCoh}^-(\mathcal{O}_ X)$. Namely, let $k$ be a field, let $k[\epsilon ]$ be the dual numbers over $k$, let $X = \mathop{\mathrm{Spec}}(k)$, and let $Y = \mathop{\mathrm{Spec}}(k[\epsilon ])$. Then $\mathop{\mathrm{Ext}}\nolimits ^ i_{k[\epsilon ]}(k, k)$ is nonzero for all $i \geq 0$. Hence $a(\mathcal{O}_ Y)$ is not bounded above by Example 48.3.2.

Lemma 48.3.5. Let $f : X \to Y$ be a morphism of quasi-compact and quasi-separated schemes. Let $a : D_\mathit{QCoh}(\mathcal{O}_ Y) \to D_\mathit{QCoh}(\mathcal{O}_ X)$ be the right adjoint to $Rf_*$ of Lemma 48.3.1. 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 Schemes, Lemma 36.4.1 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 $f : X \to Y$ be a morphism of quasi-separated and quasi-compact schemes. 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

48.3.5.1
$$\label{duality-equation-sheafy-trace} Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(L, a(K)) \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*L, K)$$

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, Remark 20.39.11 and the second arrow is the counit $Rf_*a(K) \to K$ of the adjunction.

Lemma 48.3.6. Let $f : X \to Y$ be a morphism of quasi-compact and quasi-separated schemes. 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 (48.3.5.1)

$Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(L, a(K)) \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*L, K)$

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 Schemes, Section 36.21.

Proof. The statement makes sense as $DQ_ Y$ exists by Derived Categories of Schemes, Lemma 36.21.1. 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 (48.3.5.1) 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, Lemma 20.28.1. The second equality by Cohomology, Lemma 20.39.2. The third equality by construction of $a$. The fourth equality by Derived Categories of Schemes, Lemma 36.22.1 (this is the important step). The fifth by Cohomology, Lemma 20.39.2. $\square$

Example 48.3.7. The statement of Lemma 48.3.6 is not true without applying the “coherator” $DQ_ Y$. Indeed, suppose $Y = \mathop{\mathrm{Spec}}(R)$ and $X = \mathbf{A}^1_ R$. Take $L = \mathcal{O}_ X$ and $K = \mathcal{O}_ Y$. The left hand side of the arrow is in $D_\mathit{QCoh}(\mathcal{O}_ Y)$ but the right hand side of the arrow is isomorphic to $\prod _{n \geq 0} \mathcal{O}_ Y$ which is not quasi-coherent.

Remark 48.3.8. In the situation of Lemma 48.3.6 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 Schemes, Lemma 36.21.2. 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 (48.3.5.1) is an isomorphism. Combining this with Derived Categories of Schemes, Lemma 36.10.8 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 48.3.5.

Example 48.3.9. Let $f : X \to Y$ be a proper morphism of Noetherian schemes, $L \in D^-_{\textit{Coh}}(X)$ 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 Schemes, Lemmas 36.10.3 and 36.11.3 and the discussion in Remark 48.3.8 applies.

Lemma 48.3.10. Let $f : X \to Y$ be a morphism of quasi-separated and quasi-compact schemes. For all $L \in D_\mathit{QCoh}(\mathcal{O}_ X)$ and $K \in D_\mathit{QCoh}(\mathcal{O}_ Y)$ (48.3.5.1) 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 the construction in Cohomology, Section 20.41 we have

$R\mathop{\mathrm{Hom}}\nolimits _ X(L, a(K)) = R\Gamma (X, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(L, a(K))) = R\Gamma (Y, Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(L, a(K)))$

and

$R\mathop{\mathrm{Hom}}\nolimits _ Y(Rf_*L, K) = R\Gamma (Y, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*L, K))$

Thus the lemma is a consequence of Lemma 48.3.6. 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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