The Stacks project

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

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 20.50.5. 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 20.54.3) 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 (20.54.2.1) 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 (20.54.2.1) this is Cohomology, Remark 20.54.5. 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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0B6N. Beware of the difference between the letter 'O' and the digit '0'.