## 20.28 Cohomology of unbounded complexes

Let $(X, \mathcal{O}_ X)$ be a ringed space. The category $\textit{Mod}(\mathcal{O}_ X)$ is a Grothendieck abelian category: it has all colimits, filtered colimits are exact, and it has a generator, namely

$\bigoplus \nolimits _{U \subset X\text{ open}} j_{U!}\mathcal{O}_ U,$

see Modules, Section 17.3 and Lemmas 17.17.5 and 17.17.6. By Injectives, Theorem 19.12.6 for every complex $\mathcal{F}^\bullet$ of $\mathcal{O}_ X$-modules there exists an injective quasi-isomorphism $\mathcal{F}^\bullet \to \mathcal{I}^\bullet$ to a K-injective complex of $\mathcal{O}_ X$-modules all of whose terms are injective $\mathcal{O}_ X$-modules and moreover this embedding can be chosen functorial in the complex $\mathcal{F}^\bullet$. It follows from Derived Categories, Lemma 13.31.7 that

1. any exact functor $F : K(\textit{Mod}(\mathcal{O}_ X)) \to \mathcal{D}$ into a trianguated category $\mathcal{D}$ has a right derived functor $RF : D(\mathcal{O}_ X) \to \mathcal{D}$,

2. for any additive functor $F : \textit{Mod}(\mathcal{O}_ X) \to \mathcal{A}$ into an abelian category $\mathcal{A}$ we consider the exact functor $F : K(\textit{Mod}(\mathcal{O}_ X)) \to D(\mathcal{A})$ induced by $F$ and we obtain a right derived functor $RF : D(\mathcal{O}_ X) \to K(\mathcal{A})$.

By construction we have $RF(\mathcal{F}^\bullet ) = F(\mathcal{I}^\bullet )$ where $\mathcal{F}^\bullet \to \mathcal{I}^\bullet$ is as above.

Here are some examples of the above:

1. The functor $\Gamma (X, -) : \textit{Mod}(\mathcal{O}_ X) \to \text{Mod}_{\Gamma (X, \mathcal{O}_ X)}$ gives rise to

$R\Gamma (X, -) : D(\mathcal{O}_ X) \to D(\Gamma (X, \mathcal{O}_ X))$

We shall use the notation $H^ i(X, K) = H^ i(R\Gamma (X, K))$ for cohomology.

2. For an open $U \subset X$ we consider the functor $\Gamma (U, -) : \textit{Mod}(\mathcal{O}_ X) \to \text{Mod}_{\Gamma (U, \mathcal{O}_ X)}$. This gives rise to

$R\Gamma (U, -) : D(\mathcal{O}_ X) \to D(\Gamma (U, \mathcal{O}_ X))$

We shall use the notation $H^ i(U, K) = H^ i(R\Gamma (U, K))$ for cohomology.

3. For a morphism of ringed spaces $f : (X, \mathcal{O}_ X) \to (Y, \mathcal{O}_ Y)$ we consider the functor $f_* : \textit{Mod}(\mathcal{O}_ X) \to \textit{Mod}(\mathcal{O}_ Y)$ which gives rise to the total direct image

$Rf_* : D(\mathcal{O}_ X) \longrightarrow D(\mathcal{O}_ Y)$

on unbounded derived categories.

Lemma 20.28.1. Let $f : (X, \mathcal{O}_ X) \to (Y, \mathcal{O}_ Y)$ be a morphism of ringed spaces. The functor $Rf_*$ defined above and the functor $Lf^*$ defined in Lemma 20.27.1 are adjoint:

$\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(Lf^*\mathcal{G}^\bullet , \mathcal{F}^\bullet ) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ Y)}(\mathcal{G}^\bullet , Rf_*\mathcal{F}^\bullet )$

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

Proof. This follows formally from the fact that $Rf_*$ and $Lf^*$ exist, see Derived Categories, Lemma 13.30.3. $\square$

Lemma 20.28.2. Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of ringed spaces. Then $Rg_* \circ Rf_* = R(g \circ f)_*$ as functors $D(\mathcal{O}_ X) \to D(\mathcal{O}_ Z)$.

Proof. By Lemma 20.28.1 we see that $Rg_* \circ Rf_*$ is adjoint to $Lf^* \circ Lg^*$. We have $Lf^* \circ Lg^* = L(g \circ f)^*$ by Lemma 20.27.2 and hence by uniqueness of adjoint functors we have $Rg_* \circ Rf_* = R(g \circ f)_*$. $\square$

Remark 20.28.3. The construction of unbounded derived functor $Lf^*$ and $Rf_*$ allows one to construct the base change map in full generality. Namely, suppose that

$\xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ S' \ar[r]^ g & S }$

is a commutative diagram of ringed spaces. Let $K$ be an object of $D(\mathcal{O}_ X)$. Then there exists a canonical base change map

$Lg^*Rf_*K \longrightarrow R(f')_*L(g')^*K$

in $D(\mathcal{O}_{S'})$. Namely, this map is adjoint to a map $L(f')^*Lg^*Rf_*K \to L(g')^*K$ Since $L(f')^*Lg^* = L(g')^*Lf^*$ we see this is the same as a map $L(g')^*Lf^*Rf_*K \to L(g')^*K$ which we can take to be $L(g')^*$ of the adjunction map $Lf^*Rf_*K \to K$.

Remark 20.28.4. Consider a commutative diagram

$\xymatrix{ X' \ar[r]_ k \ar[d]_{f'} & X \ar[d]^ f \\ Y' \ar[r]^ l \ar[d]_{g'} & Y \ar[d]^ g \\ Z' \ar[r]^ m & Z }$

of ringed spaces. Then the base change maps of Remark 20.28.3 for the two squares compose to give the base change map for the outer rectangle. More precisely, the composition

\begin{align*} Lm^* \circ R(g \circ f)_* & = Lm^* \circ Rg_* \circ Rf_* \\ & \to Rg'_* \circ Ll^* \circ Rf_* \\ & \to Rg'_* \circ Rf'_* \circ Lk^* \\ & = R(g' \circ f')_* \circ Lk^* \end{align*}

is the base change map for the rectangle. We omit the verification.

Remark 20.28.5. Consider a commutative diagram

$\xymatrix{ X'' \ar[r]_{g'} \ar[d]_{f''} & X' \ar[r]_ g \ar[d]_{f'} & X \ar[d]^ f \\ Y'' \ar[r]^{h'} & Y' \ar[r]^ h & Y }$

of ringed spaces. Then the base change maps of Remark 20.28.3 for the two squares compose to give the base change map for the outer rectangle. More precisely, the composition

\begin{align*} L(h \circ h')^* \circ Rf_* & = L(h')^* \circ Lh_* \circ Rf_* \\ & \to L(h')^* \circ Rf'_* \circ Lg^* \\ & \to Rf''_* \circ L(g')^* \circ Lg^* \\ & = Rf”_* \circ L(g \circ g')^* \end{align*}

is the base change map for the rectangle. We omit the verification.

Lemma 20.28.6. Let $f : (X, \mathcal{O}_ X) \to (Y, \mathcal{O}_ Y)$ be a morphism of ringed spaces. Let $\mathcal{K}^\bullet$ be a complex of $\mathcal{O}_ X$-modules. The diagram

$\xymatrix{ Lf^*f_*\mathcal{K}^\bullet \ar[r] \ar[d] & f^*f_*\mathcal{K}^\bullet \ar[d] \\ Lf^*Rf_*\mathcal{K}^\bullet \ar[r] & \mathcal{K}^\bullet }$

coming from $Lf^* \to f^*$ on complexes, $f_* \to Rf_*$ on complexes, and adjunction $Lf^* \circ Rf_* \to \text{id}$ commutes in $D(\mathcal{O}_ X)$.

Proof. We will use the existence of K-flat resolutions and K-injective resolutions, see Lemma 20.26.8 and the discussion above. Choose a quasi-isomorphism $\mathcal{K}^\bullet \to \mathcal{I}^\bullet$ where $\mathcal{I}^\bullet$ is K-injective as a complex of $\mathcal{O}_ X$-modules. Choose a quasi-isomorphism $\mathcal{Q}^\bullet \to f_*\mathcal{I}^\bullet$ where $\mathcal{Q}^\bullet$ is K-flat as a complex of $\mathcal{O}_ Y$-modules. We can choose a K-flat complex of $\mathcal{O}_ Y$-modules $\mathcal{P}^\bullet$ and a diagram of morphisms of complexes

$\xymatrix{ \mathcal{P}^\bullet \ar[r] \ar[d] & f_*\mathcal{K}^\bullet \ar[d] \\ \mathcal{Q}^\bullet \ar[r] & f_*\mathcal{I}^\bullet }$

commutative up to homotopy where the top horizontal arrow is a quasi-isomorphism. Namely, we can first choose such a diagram for some complex $\mathcal{P}^\bullet$ because the quasi-isomorphisms form a multiplicative system in the homotopy category of complexes and then we can replace $\mathcal{P}^\bullet$ by a K-flat complex. Taking pullbacks we obtain a diagram of morphisms of complexes

$\xymatrix{ f^*\mathcal{P}^\bullet \ar[r] \ar[d] & f^*f_*\mathcal{K}^\bullet \ar[d] \ar[r] & \mathcal{K}^\bullet \ar[d] \\ f^*\mathcal{Q}^\bullet \ar[r] & f^*f_*\mathcal{I}^\bullet \ar[r] & \mathcal{I}^\bullet }$

commutative up to homotopy. The outer rectangle witnesses the truth of the statement in the lemma. $\square$

Remark 20.28.7. Let $f : (X, \mathcal{O}_ X) \to (Y, \mathcal{O}_ Y)$ be a morphism of ringed spaces. The adjointness of $Lf^*$ and $Rf_*$ allows us to construct a relative cup product

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

in $D(\mathcal{O}_ Y)$ for all $K, L$ in $D(\mathcal{O}_ X)$. Namely, this map is adjoint to a map $Lf^*(Rf_*K \otimes _{\mathcal{O}_ Y}^\mathbf {L} Rf_*L) \to K \otimes _{\mathcal{O}_ X}^\mathbf {L} L$ for which we can take the composition of the isomorphism $Lf^*(Rf_*K \otimes _{\mathcal{O}_ Y}^\mathbf {L} Rf_*L) = Lf^*Rf_*K \otimes _{\mathcal{O}_ X}^\mathbf {L} Lf^*Rf_*L$ (Lemma 20.27.3) with the map $Lf^*Rf_*K \otimes _{\mathcal{O}_ X}^\mathbf {L} Lf^*Rf_*L \to K \otimes _{\mathcal{O}_ X}^\mathbf {L} L$ coming from the counit $Lf^* \circ Rf_* \to \text{id}$.

Comment #7063 by Nick on

The line "and similarly for any left exact functor, see Derived Categories, Lemma 13.31.7" after the second displayed equation seems out of place. (At least I don't see what it is saying and how the referenced Lemma has anything to say about left exact functors)

Comment #7064 by Nick on

I was confusing left and right... P079lease ignore the previous comment.

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