Lemma 20.51.2. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $K$ and $E$ be objects of $D(\mathcal{O}_ X)$ with $E$ perfect. The diagram

$\xymatrix{ H^0(X, K \otimes _{\mathcal{O}_ X}^\mathbf {L} E^\vee ) \times H^0(X, E) \ar[r] \ar[d] & H^0(X, K \otimes _{\mathcal{O}_ X}^\mathbf {L} E^\vee \otimes _{\mathcal{O}_ X}^\mathbf {L} E) \ar[d] \\ \mathop{\mathrm{Hom}}\nolimits _ X(E, K) \times H^0(X, E) \ar[r] & H^0(X, K) }$

commutes where the top horizontal arrow is the cup product, the right vertical arrow uses $\epsilon : E^\vee \otimes _{\mathcal{O}_ X}^\mathbf {L} E \to \mathcal{O}_ X$ (Example 20.50.7), the left vertical arrow uses Lemma 20.50.5, and the bottom horizontal arrow is the obvious one.

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

$E = \mathcal{O} \otimes E \xrightarrow {\tilde\xi \otimes 1_ E} K \otimes E^\vee \otimes E \xrightarrow {1_ K \otimes \epsilon } K$

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

\begin{align*} E = \mathcal{O} \otimes R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}, E) & \xrightarrow {\tilde\xi \otimes 1} R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}, K) \otimes R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (E, \mathcal{O}) \otimes R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}, E) \\ & \xrightarrow {\underline{\circ } \otimes 1} R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (E, K) \otimes R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}, E) \\ & \xrightarrow {\underline{\circ }} R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}, K) = K \end{align*}

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. $\square$

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