The Stacks project

Proof. By Lemma 20.50.5 we have $R\mathop{\mathrm{lim}}\nolimits E \otimes ^\mathbf {L}_{\mathcal{O}_ X} K_ n^\vee = R\mathop{\mathrm{lim}}\nolimits R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K_ n, E)$ which fits into the distinguished triangle

Because $K$ similarly fits into the distinguished triangle $\bigoplus K_ n \to \bigoplus K_ n \to K$ it suffices to show that $\prod R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K_ n, E) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\bigoplus K_ n, E)$. This is a formal consequence of (20.42.0.1) and the fact that derived tensor product commutes with direct sums. $\square$

Proof. We will abbreviate $\otimes = \otimes _{\mathcal{O}_ X}^\mathbf {L}$ and $\mathcal{O} = \mathcal{O}_ X$. We will identify $E$ and $K$ with $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}, E)$ and $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}, K)$ and we will identify $E^\vee $ with $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (E, \mathcal{O})$.

Let $\xi \in H^0(X, K \otimes E^\vee )$ and $\eta \in H^0(X, E)$. Denote $\tilde\xi : \mathcal{O} \to K \otimes E^\vee $ and $\tilde\eta : \mathcal{O} \to E$ the corresponding maps in $D(\mathcal{O})$. By Lemma 20.31.1 the cup product $\xi \cup \eta $ corresponds to $\tilde\xi \otimes \tilde\eta : \mathcal{O} \to K \otimes E^\vee \otimes E$.

We claim the map $\xi ' : E \to K$ corresponding to $\xi $ by Lemma 20.50.5 is the composition

The construction in Lemma 20.50.5 uses the evaluation map (20.42.8.1) which in turn is constructed using the identification of $E$ with $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}, E)$ and the composition $\underline{\circ }$ constructed in Lemma 20.42.5. Hence $\xi '$ is the composition

The claim follows immediately from this and the fact that the composition $\underline{\circ }$ constructed in Lemma 20.42.5 is associative (insert future reference here) and the fact that $\epsilon $ is defined as the composition $\underline{\circ } : E^\vee \otimes E \to \mathcal{O}$ in Example 20.50.7.

Using the results from the previous two paragraphs, we find the statement of the lemma is that $(1_ K \otimes \epsilon ) \circ (\tilde\xi \otimes \tilde\eta )$ is equal to $(1_ K \otimes \epsilon ) \circ (\tilde\xi \otimes 1_ E) \circ (1_\mathcal {O} \otimes \tilde\eta )$ which is immediate. $\square$

Proof. Proof of (1). The question is local on $Y$, hence we may assume that $K$ is represented by a strictly perfect complex $\mathcal{E}^\bullet $, see Section 20.49. Choose a K-flat complex $\mathcal{F}^\bullet $ representing $M$. Apply Lemma 20.46.9 to see that $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, L)$ is represented by the complex $\mathcal{H}^\bullet = \mathop{\mathcal{H}\! \mathit{om}}\nolimits ^\bullet (\mathcal{E}^\bullet , \mathcal{F}^\bullet )$ with terms $\mathcal{H}^ n = \bigoplus \nolimits _{n = p + q} \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{E}^{-q}, \mathcal{F}^ p)$. By the construction of $Lh^*$ in Section 20.27 we see that $Lh^*K$ is represented by the strictly perfect complex $h^*\mathcal{E}^\bullet $ (Lemma 20.46.4). Similarly, the object $Lh^*M$ is represented by the complex $h^*\mathcal{F}^\bullet $. Finally, the object $Lh^*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, M)$ is represented by $h^*\mathcal{H}^\bullet $ as $\mathcal{H}^\bullet $ is K-flat by Lemma 20.46.10. Thus to finish the proof it suffices to show that $h^*\mathcal{H}^\bullet = \mathop{\mathcal{H}\! \mathit{om}}\nolimits ^\bullet (h^*\mathcal{E}^\bullet , h^*\mathcal{F}^\bullet )$. For this it suffices to note that $h^*\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{E}, \mathcal{F}) = \mathop{\mathcal{H}\! \mathit{om}}\nolimits (h^*\mathcal{E}, \mathcal{F})$ whenever $\mathcal{E}$ is a direct summand of a finite free $\mathcal{O}_ X$-module.

Proof of (2). Since $h$ is flat, we can compute $Lh^*$ by simply using $h^*$ on any complex of $\mathcal{O}_ Y$-modules. In particular we have $H^ i(Lh^*K) = h^*H^ i(K)$ for all $i \in \mathbf{Z}$. Say $H^ i(M) = 0$ for $i < a$. Let $K' \to K$ be a morphism of $D(\mathcal{O}_ Y)$ which defines an isomorphism $H^ i(K') \to H^ i(K)$ for all $i \geq b$. Then the corresponding maps

and

are isomorphisms on cohomology sheaves in degrees $< a - b$ (details omitted). Thus to prove the map in the statement of the lemma induces an isomorphism on cohomology sheaves in degrees $< a - b$ it suffices to prove the result for $K'$ in those degrees. Also, as in the proof of part (1) the question is local on $Y$. Thus we may assume $K$ is represented by a strictly perfect complex, see Section 20.47. This reduces us to case (1).

Proof of (3). The proof is the same as the proof of (2) except one uses that $Lh^*$ has bounded cohomological dimension to get the desired vanishing. We omit the details. $\square$

Proof. Let $Y = \{ x\} $ be the singleton ringed space with structure sheaf given by $\mathcal{O}_{X, x}$. Then apply Lemma 20.51.3 to the flat inclusion morphism $Y \to X$. $\square$