## 20.27 Derived pullback

Let $f : (X, \mathcal{O}_ X) \to (Y, \mathcal{O}_ Y)$ be a morphism of ringed spaces. We can use K-flat resolutions to define a derived pullback functor

$Lf^* : D(\mathcal{O}_ Y) \to D(\mathcal{O}_ X)$

Namely, for every complex of $\mathcal{O}_ Y$-modules $\mathcal{G}^\bullet$ we can choose a K-flat resolution $\mathcal{K}^\bullet \to \mathcal{G}^\bullet$ and set $Lf^*\mathcal{G}^\bullet = f^*\mathcal{K}^\bullet$. You can use Lemmas 20.26.8, 20.26.12, and 20.26.13 to see that this is well defined. However, to cross all the t's and dot all the i's it is perhaps more convenient to use some general theory.

Lemma 20.27.1. The construction above is independent of choices and defines an exact functor of triangulated categories $Lf^* : D(\mathcal{O}_ Y) \to D(\mathcal{O}_ X)$.

Proof. To see this we use the general theory developed in Derived Categories, Section 13.14. Set $\mathcal{D} = K(\mathcal{O}_ Y)$ and $\mathcal{D}' = D(\mathcal{O}_ X)$. Let us write $F : \mathcal{D} \to \mathcal{D}'$ the exact functor of triangulated categories defined by the rule $F(\mathcal{G}^\bullet ) = f^*\mathcal{G}^\bullet$. We let $S$ be the set of quasi-isomorphisms in $\mathcal{D} = K(\mathcal{O}_ Y)$. This gives a situation as in Derived Categories, Situation 13.14.1 so that Derived Categories, Definition 13.14.2 applies. We claim that $LF$ is everywhere defined. This follows from Derived Categories, Lemma 13.14.15 with $\mathcal{P} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$ the collection of $K$-flat complexes: (1) follows from Lemma 20.26.12 and to see (2) we have to show that for a quasi-isomorphism $\mathcal{K}_1^\bullet \to \mathcal{K}_2^\bullet$ between K-flat complexes of $\mathcal{O}_ Y$-modules the map $f^*\mathcal{K}_1^\bullet \to f^*\mathcal{K}_2^\bullet$ is a quasi-isomorphism. To see this write this as

$f^{-1}\mathcal{K}_1^\bullet \otimes _{f^{-1}\mathcal{O}_ Y} \mathcal{O}_ X \longrightarrow f^{-1}\mathcal{K}_2^\bullet \otimes _{f^{-1}\mathcal{O}_ Y} \mathcal{O}_ X$

The functor $f^{-1}$ is exact, hence the map $f^{-1}\mathcal{K}_1^\bullet \to f^{-1}\mathcal{K}_2^\bullet$ is a quasi-isomorphism. By Lemma 20.26.8 applied to the morphism $(X, f^{-1}\mathcal{O}_ Y) \to (Y, \mathcal{O}_ Y)$ the complexes $f^{-1}\mathcal{K}_1^\bullet$ and $f^{-1}\mathcal{K}_2^\bullet$ are K-flat complexes of $f^{-1}\mathcal{O}_ Y$-modules. Hence Lemma 20.26.13 guarantees that the displayed map is a quasi-isomorphism. Thus we obtain a derived functor

$LF : D(\mathcal{O}_ Y) = S^{-1}\mathcal{D} \longrightarrow \mathcal{D}' = D(\mathcal{O}_ X)$

see Derived Categories, Equation (13.14.9.1). Finally, Derived Categories, Lemma 13.14.15 also guarantees that $LF(\mathcal{K}^\bullet ) = F(\mathcal{K}^\bullet ) = f^*\mathcal{K}^\bullet$ when $\mathcal{K}^\bullet$ is K-flat, i.e., $Lf^* = LF$ is indeed computed in the way described above. $\square$

Lemma 20.27.2. Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of ringed spaces. Then $Lf^* \circ Lg^* = L(g \circ f)^*$ as functors $D(\mathcal{O}_ Z) \to D(\mathcal{O}_ X)$.

Proof. Let $E$ be an object of $D(\mathcal{O}_ Z)$. By construction $Lg^*E$ is computed by choosing a K-flat complex $\mathcal{K}^\bullet$ representing $E$ on $Z$ and setting $Lg^*E = g^*\mathcal{K}^\bullet$. By Lemma 20.26.8 we see that $g^*\mathcal{K}^\bullet$ is K-flat on $Y$. Then $Lf^*Lg^*E$ is given by $f^*g^*\mathcal{K}^\bullet = (g \circ f)^*\mathcal{K}^\bullet$ which also represents $L(g \circ f)^*E$. $\square$

Lemma 20.27.3. Let $f : (X, \mathcal{O}_ X) \to (Y, \mathcal{O}_ Y)$ be a morphism of ringed spaces. There is a canonical bifunctorial isomorphism

$Lf^*( \mathcal{F}^\bullet \otimes _{\mathcal{O}_ Y}^{\mathbf{L}} \mathcal{G}^\bullet ) = Lf^*\mathcal{F}^\bullet \otimes _{\mathcal{O}_ X}^{\mathbf{L}} Lf^*\mathcal{G}^\bullet$

for $\mathcal{F}^\bullet , \mathcal{G}^\bullet \in \mathop{\mathrm{Ob}}\nolimits (D(\mathcal{O}_ Y))$.

Proof. We may assume that $\mathcal{F}^\bullet$ and $\mathcal{G}^\bullet$ are K-flat complexes. In this case $\mathcal{F}^\bullet \otimes _{\mathcal{O}_ Y}^{\mathbf{L}} \mathcal{G}^\bullet$ is just the total complex associated to the double complex $\mathcal{F}^\bullet \otimes _{\mathcal{O}_ Y} \mathcal{G}^\bullet$. By Lemma 20.26.5 $\text{Tot}(\mathcal{F}^\bullet \otimes _{\mathcal{O}_ Y} \mathcal{G}^\bullet )$ is K-flat also. Hence the isomorphism of the lemma comes from the isomorphism

$\text{Tot}(f^*\mathcal{F}^\bullet \otimes _{\mathcal{O}_ X} f^*\mathcal{G}^\bullet ) \longrightarrow f^*\text{Tot}(\mathcal{F}^\bullet \otimes _{\mathcal{O}_ Y} \mathcal{G}^\bullet )$

whose constituents are the isomorphisms $f^*\mathcal{F}^ p \otimes _{\mathcal{O}_ X} f^*\mathcal{G}^ q \to f^*(\mathcal{F}^ p \otimes _{\mathcal{O}_ Y} \mathcal{G}^ q)$ of Modules, Lemma 17.16.4. $\square$

Lemma 20.27.4. Let $f : (X, \mathcal{O}_ X) \to (Y, \mathcal{O}_ Y)$ be a morphism of ringed spaces. There is a canonical bifunctorial isomorphism

$\mathcal{F}^\bullet \otimes _{\mathcal{O}_ X}^{\mathbf{L}} Lf^*\mathcal{G}^\bullet = \mathcal{F}^\bullet \otimes _{f^{-1}\mathcal{O}_ Y}^{\mathbf{L}} f^{-1}\mathcal{G}^\bullet$

for $\mathcal{F}^\bullet$ in $D(\mathcal{O}_ X)$ and $\mathcal{G}^\bullet$ in $D(\mathcal{O}_ Y)$.

Proof. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module and let $\mathcal{G}$ be an $\mathcal{O}_ Y$-module. Then $\mathcal{F} \otimes _{\mathcal{O}_ X} f^*\mathcal{G} = \mathcal{F} \otimes _{f^{-1}\mathcal{O}_ Y} f^{-1}\mathcal{G}$ because $f^*\mathcal{G} = \mathcal{O}_ X \otimes _{f^{-1}\mathcal{O}_ Y} f^{-1}\mathcal{G}$. The lemma follows from this and the definitions. $\square$

Lemma 20.27.5. Let $f : (X, \mathcal{O}_ X) \to (Y, \mathcal{O}_ Y)$ be a morphism of ringed spaces. Let $\mathcal{K}^\bullet$ and $\mathcal{M}^\bullet$ be complexes of $\mathcal{O}_ Y$-modules. The diagram

$\xymatrix{ Lf^*(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ Y}^\mathbf {L} \mathcal{M}^\bullet ) \ar[r] \ar[d] & Lf^*\text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ Y} \mathcal{M}^\bullet ) \ar[d] \\ Lf^*\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X}^\mathbf {L} Lf^*\mathcal{M}^\bullet \ar[d] & f^*\text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ Y} \mathcal{M}^\bullet ) \ar[d] \\ f^*\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X}^\mathbf {L} f^*\mathcal{M}^\bullet \ar[r] & \text{Tot}(f^*\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} f^*\mathcal{M}^\bullet ) }$

commutes.

Proof. We will use the existence of K-flat resolutions as in Lemma 20.26.8. If we choose such resolutions $\mathcal{P}^\bullet \to \mathcal{K}^\bullet$ and $\mathcal{Q}^\bullet \to \mathcal{M}^\bullet$, then we see that

$\xymatrix{ Lf^*\text{Tot}(\mathcal{P}^\bullet \otimes _{\mathcal{O}_ Y} \mathcal{Q}^\bullet ) \ar[r] \ar[d] & Lf^*\text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ Y} \mathcal{M}^\bullet ) \ar[d] \\ f^*\text{Tot}(\mathcal{P}^\bullet \otimes _{\mathcal{O}_ Y} \mathcal{Q}^\bullet ) \ar[d] \ar[r] & f^*\text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ Y} \mathcal{M}^\bullet ) \ar[d] \\ \text{Tot}(f^*\mathcal{P}^\bullet \otimes _{\mathcal{O}_ X} f^*\mathcal{Q}^\bullet ) \ar[r] & \text{Tot}(f^*\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} f^*\mathcal{M}^\bullet ) }$

commutes. However, now the left hand side of the diagram is the left hand side of the diagram by our choice of $\mathcal{P}^\bullet$ and $\mathcal{Q}^\bullet$ and Lemma 20.26.5. $\square$

Comment #7936 by Karl Schwede on

For Lemma 079U, presumably $F, G$ should be in $D(O_Y)$?

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