Lemma 115.13.1. 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 $D(\mathcal{O}_ X)$.

The proof given here follows the argument given in [Remark 3.4, MS]

**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 20.50.6). 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 20.50.7. 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. $\square$

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