## 48.8 Right adjoint of pushforward and pullback

Let $f : X \to Y$ be a morphism of quasi-compact and quasi-separated schemes. Let $a$ be the right adjoint of pushforward as in Lemma 48.3.1. For $K, L \in D_\mathit{QCoh}(\mathcal{O}_ Y)$ there is a canonical map

$Lf^*K \otimes ^\mathbf {L}_{\mathcal{O}_ X} a(L) \longrightarrow a(K \otimes _{\mathcal{O}_ Y}^\mathbf {L} L)$

Namely, this map is adjoint to a map

$Rf_*(Lf^*K \otimes ^\mathbf {L}_{\mathcal{O}_ X} a(L)) = K \otimes ^\mathbf {L}_{\mathcal{O}_ Y} Rf_*(a(L)) \longrightarrow K \otimes ^\mathbf {L}_{\mathcal{O}_ Y} L$

(equality by Derived Categories of Schemes, Lemma 36.22.1) for which we use the trace map $Rf_*a(L) \to L$. When $L = \mathcal{O}_ Y$ we obtain a map

48.8.0.1
$$\label{duality-equation-compare-with-pullback} Lf^*K \otimes ^\mathbf {L}_{\mathcal{O}_ X} a(\mathcal{O}_ Y) \longrightarrow a(K)$$

functorial in $K$ and compatible with distinguished triangles.

Lemma 48.8.1. Let $f : X \to Y$ be a morphism of quasi-compact and quasi-separated schemes. The map $Lf^*K \otimes ^\mathbf {L}_{\mathcal{O}_ X} a(L) \to a(K \otimes _{\mathcal{O}_ Y}^\mathbf {L} L)$ defined above for $K, L \in D_\mathit{QCoh}(\mathcal{O}_ Y)$ is an isomorphism if $K$ is perfect. In particular, (48.8.0.1) is an isomorphism if $K$ is perfect.

Proof. Let $K^\vee$ be the “dual” to $K$, see Cohomology, Lemma 20.47.5. For $M \in D_\mathit{QCoh}(\mathcal{O}_ X)$ we have

\begin{align*} \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ Y)}(Rf_*M, K \otimes ^\mathbf {L}_{\mathcal{O}_ Y} L) & = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ Y)}( Rf_*M \otimes ^\mathbf {L}_{\mathcal{O}_ Y} K^\vee , L) \\ & = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}( M \otimes ^\mathbf {L}_{\mathcal{O}_ X} Lf^*K^\vee , a(L)) \\ & = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(M, Lf^*K \otimes ^\mathbf {L}_{\mathcal{O}_ X} a(L)) \end{align*}

Second equality by the definition of $a$ and the projection formula (Cohomology, Lemma 20.50.3) or the more general Derived Categories of Schemes, Lemma 36.22.1. Hence the result by the Yoneda lemma. $\square$

Lemma 48.8.2. Suppose we have a diagram (48.4.0.1) where $f$ and $g$ are tor independent. Let $K \in D_\mathit{QCoh}(\mathcal{O}_ Y)$. The diagram

$\xymatrix{ L(g')^*(Lf^*K \otimes ^\mathbf {L}_{\mathcal{O}_ X} a(\mathcal{O}_ Y)) \ar[r] \ar[d] & L(g')^*a(K) \ar[d] \\ L(f')^*Lg^*K \otimes _{\mathcal{O}_{X'}}^\mathbf {L} a'(\mathcal{O}_{Y'}) \ar[r] & a'(Lg^*K) }$

commutes where the horizontal arrows are the maps (48.8.0.1) for $K$ and $Lg^*K$ and the vertical maps are constructed using Cohomology, Remark 20.28.3 and (48.5.0.1).

Proof. In this proof we will write $f_*$ for $Rf_*$ and $f^*$ for $Lf^*$, etc, and we will write $\otimes$ for $\otimes ^\mathbf {L}_{\mathcal{O}_ X}$, etc. Let us write (48.8.0.1) as the composition

\begin{align*} f^*K \otimes a(\mathcal{O}_ Y) & \to a(f_*(f^*K \otimes a(\mathcal{O}_ Y))) \\ & \leftarrow a(K \otimes f_*a(\mathcal{O}_ K)) \\ & \to a(K \otimes \mathcal{O}_ Y) \\ & \to a(K) \end{align*}

Here the first arrow is the unit $\eta _ f$, the second arrow is $a$ applied to Cohomology, Equation (20.50.2.1) which is an isomorphism by Derived Categories of Schemes, Lemma 36.22.1, the third arrow is $a$ applied to $\text{id}_ K \otimes \text{Tr}_ f$, and the fourth arrow is $a$ applied to the isomorphism $K \otimes \mathcal{O}_ Y = K$. The proof of the lemma consists in showing that each of these maps gives rise to a commutative square as in the statement of the lemma. For $\eta _ f$ and $\text{Tr}_ f$ this is Lemmas 48.7.2 and 48.7.1. For the arrow using Cohomology, Equation (20.50.2.1) this is Cohomology, Remark 20.50.5. For the multiplication map it is clear. This finishes the proof. $\square$

Lemma 48.8.3. Let $f : X \to Y$ be a proper morphism of Noetherian schemes. Let $V \subset Y$ be an open such that $f^{-1}(V) \to V$ is an isomorphism. Then for $K \in D_\mathit{QCoh}^+(\mathcal{O}_ Y)$ the map (48.8.0.1) restricts to an isomorphism over $f^{-1}(V)$.

Proof. By Lemma 48.4.4 the map (48.4.1.1) is an isomorphism for objects of $D_\mathit{QCoh}^+(\mathcal{O}_ Y)$. Hence Lemma 48.8.2 tells us the restriction of (48.8.0.1) for $K$ to $f^{-1}(V)$ is the map (48.8.0.1) for $K|_ V$ and $f^{-1}(V) \to V$. Thus it suffices to show that the map is an isomorphism when $f$ is the identity morphism. This is clear. $\square$

Lemma 48.8.4. Let $f : X \to Y$ and $g : Y \to Z$ be composable morphisms of quasi-compact and quasi-separated schemes and set $h = g \circ f$. Let $a, b, c$ be the adjoints of Lemma 48.3.1 for $f, g, h$. For any $K \in D_\mathit{QCoh}(\mathcal{O}_ Z)$ the diagram

$\xymatrix{ Lf^*(Lg^*K \otimes _{\mathcal{O}_ Y}^\mathbf {L} b(\mathcal{O}_ Z)) \otimes _{\mathcal{O}_ X}^\mathbf {L} a(\mathcal{O}_ Y) \ar@{=}[d] \ar[r] & a(Lg^*K \otimes _{\mathcal{O}_ Y}^\mathbf {L} b(\mathcal{O}_ Z)) \ar[r] & a(b(K)) \ar@{=}[d] \\ Lh^*K \otimes _{\mathcal{O}_ X}^\mathbf {L} Lf^*b(\mathcal{O}_ Z) \otimes _{\mathcal{O}_ X}^\mathbf {L} a(\mathcal{O}_ Y) \ar[r] & Lh^*K \otimes _{\mathcal{O}_ X}^\mathbf {L} c(\mathcal{O}_ Z) \ar[r] & c(K) }$

is commutative where the arrows are (48.8.0.1) and we have used $Lh^* = Lf^* \circ Lg^*$ and $c = a \circ b$.

Proof. In this proof we will write $f_*$ for $Rf_*$ and $f^*$ for $Lf^*$, etc, and we will write $\otimes$ for $\otimes ^\mathbf {L}_{\mathcal{O}_ X}$, etc. The composition of the top arrows is adjoint to a map

$g_*f_*(f^*(g^*K \otimes b(\mathcal{O}_ Z)) \otimes a(\mathcal{O}_ Y)) \to K$

The left hand side is equal to $K \otimes g_*f_*(f^*b(\mathcal{O}_ Z) \otimes a(\mathcal{O}_ Y))$ by Derived Categories of Schemes, Lemma 36.22.1 and inspection of the definitions shows the map comes from the map

$g_*f_*(f^*b(\mathcal{O}_ Z) \otimes a(\mathcal{O}_ Y)) \xleftarrow {g_*\epsilon } g_*(b(\mathcal{O}_ Z) \otimes f_*a(\mathcal{O}_ Y)) \xrightarrow {g_*\alpha } g_*(b(\mathcal{O}_ Z)) \xrightarrow {\beta } \mathcal{O}_ Z$

tensored with $\text{id}_ K$. Here $\epsilon$ is the isomorphism from Derived Categories of Schemes, Lemma 36.22.1 and $\beta$ comes from the counit map $g_*b \to \text{id}$. Similarly, the composition of the lower horizontal arrows is adjoint to $\text{id}_ K$ tensored with the composition

$g_*f_*(f^*b(\mathcal{O}_ Z) \otimes a(\mathcal{O}_ Y)) \xrightarrow {g_*f_*\delta } g_*f_*(ab(\mathcal{O}_ Z)) \xrightarrow {g_*\gamma } g_*(b(\mathcal{O}_ Z)) \xrightarrow {\beta } \mathcal{O}_ Z$

where $\gamma$ comes from the counit map $f_*a \to \text{id}$ and $\delta$ is the map whose adjoint is the composition

$f_*(f^*b(\mathcal{O}_ Z) \otimes a(\mathcal{O}_ Y)) \xleftarrow {\epsilon } b(\mathcal{O}_ Z) \otimes f_*a(\mathcal{O}_ Y) \xrightarrow {\alpha } b(\mathcal{O}_ Z)$

By general properties of adjoint functors, adjoint maps, and counits (see Categories, Section 4.24) we have $\gamma \circ f_*\delta = \alpha \circ \epsilon ^{-1}$ as desired. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).