Lemma 56.7.1. Let $f : X' \to X$ be a proper birational morphism of integral Noetherian schemes with $X$ regular. The map $\mathcal{O}_ X \to Rf_*\mathcal{O}_{X'}$ canonically splits in $D(\mathcal{O}_ X)$.

## 56.7 Representability in the regular proper case

Theorem 56.6.3 also holds for regular (for example smooth) proper varieties. This is proven in [BvdB] using their general characterization of “right saturated” $k$-linear triangulated categories. In this section we give a quick and dirty proof of this result using a little bit of duality.

**Proof.**
Set $E = Rf_*\mathcal{O}_{X'}$ in $D(\mathcal{O}_ X)$. Observe that $E$ is in $D^ b_{\textit{Coh}}(\mathcal{O}_ X)$ by Derived Categories of Schemes, Lemma 36.11.3. By Derived Categories of Schemes, Lemma 36.11.8 we find that $E$ is a perfect object of $D(\mathcal{O}_ X)$. Since $\mathcal{O}_{X'}$ is a sheaf of algebras, we have the relative cup product $\mu : E \otimes _{\mathcal{O}_ X}^\mathbf {L} E \to E$ by Cohomology, Remark 20.28.7. Let $\sigma : E \otimes E^\vee \to E^\vee \otimes E$ be the commutativity constraint on the symmetric monoidal category $D(\mathcal{O}_ X)$ (Cohomology, Lemma 20.47.6). Denote $\eta : \mathcal{O}_ X \to E \otimes E^\vee $ and $\epsilon : E^\vee \otimes E \to \mathcal{O}_ X$ the maps constructed in Cohomology, Example 20.47.7. Then we can consider the map

We claim that this map is a one sided inverse to the map in the statement of the lemma. To see this it suffices to show that the composition $\mathcal{O}_ X \to \mathcal{O}_ X$ is the identity map. This we may do in the generic point of $X$ (or on an open subscheme of $X$ over which $f$ is an isomorphism). In this case $E = \mathcal{O}_ X$ and $\mu $ is the usual multiplication map and the result is clear. $\square$

Lemma 56.7.2. Let $X$ be a proper scheme over a field $k$ which is regular. Let $K \in \mathop{\mathrm{Ob}}\nolimits (D_\mathit{QCoh}(\mathcal{O}_ X))$. The following are equivalent

$K \in D^ b_{\textit{Coh}}(\mathcal{O}_ X) = D_{perf}(\mathcal{O}_ X)$, and

$\sum _{i \in \mathbf{Z}} \dim _ k \mathop{\mathrm{Ext}}\nolimits ^ i_ X(E, K) < \infty $ for all perfect $E$ in $D(\mathcal{O}_ X)$.

**Proof.**
The equality in (1) holds by Derived Categories of Schemes, Lemma 36.11.8. The implication (1) $\Rightarrow $ (2) follows from Lemma 56.5.3. The implication (2) $\Rightarrow $ (1) follows from More on Morphisms, Lemma 37.63.6.
$\square$

Lemma 56.7.3. Let $X$ be a proper scheme over a field $k$ which is regular.

Let $F : D_{perf}(\mathcal{O}_ X)^{opp} \to \text{Vect}_ k$ be a $k$-linear cohomological functor such that

\[ \sum \nolimits _{n \in \mathbf{Z}} \dim _ k F(E[n]) < \infty \]for all $E \in D_{perf}(\mathcal{O}_ X)$. Then $F$ is isomorphic to a functor of the form $E \mapsto \mathop{\mathrm{Hom}}\nolimits _ X(E, K)$ for some $K \in D_{perf}(\mathcal{O}_ X)$.

Let $G : D_{perf}(\mathcal{O}_ X) \to \text{Vect}_ k$ be a $k$-linear homological functor such that

\[ \sum \nolimits _{n \in \mathbf{Z}} \dim _ k G(E[n]) < \infty \]for all $E \in D_{perf}(\mathcal{O}_ X)$. Then $G$ is isomorphic to a functor of the form $E \mapsto \mathop{\mathrm{Hom}}\nolimits _ X(K, E)$ for some $K \in D_{perf}(\mathcal{O}_ X)$.

**Proof.**
Proof of (1). The derived category $D_\mathit{QCoh}(\mathcal{O}_ X)$ has direct sums, is compactly generated, and $D_{perf}(\mathcal{O}_ X)$ is the full subcategory of compact objects, see Derived Categories of Schemes, Lemma 36.3.1, Theorem 36.15.3, and Proposition 36.17.1. By Lemma 56.6.2 we may assume $F(E) = \mathop{\mathrm{Hom}}\nolimits _ X(E, K)$ for some $K \in \mathop{\mathrm{Ob}}\nolimits (D_\mathit{QCoh}(\mathcal{O}_ X))$. Then it follows that $K$ is in $D^ b_{\textit{Coh}}(\mathcal{O}_ X)$ by Lemma 56.7.2.

Proof of (2). Consider the contravariant functor $E \mapsto E^\vee $ on $D_{perf}(\mathcal{O}_ X)$, see Cohomology, Lemma 20.47.5. This functor is an exact anti-self-equivalence of $D_{perf}(\mathcal{O}_ X)$. Hence we may apply part (1) to the functor $F(E) = G(E^\vee )$ to find $K \in D_{perf}(\mathcal{O}_ X)$ such that $G(E^\vee ) = \mathop{\mathrm{Hom}}\nolimits _ X(E, K)$. It follows that $G(E) = \mathop{\mathrm{Hom}}\nolimits _ X(E^\vee , K) = \mathop{\mathrm{Hom}}\nolimits _ X(K^\vee , E)$ and we conclude that taking $K^\vee $ works. $\square$

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