Lemma 20.50.4. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $K, L, M \in D(\mathcal{O}_ X)$. If $K$ is perfect, then the map

$R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (L, M) \otimes _{\mathcal{O}_ X}^\mathbf {L} K \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, L), M)$

of Lemma 20.42.9 is an isomorphism.

Proof. Since the map is globally defined and since formation of the right and left hand side commute with localization (see Lemma 20.42.3), to prove this we may work locally on $X$. Thus we may assume $K$ is represented by a strictly perfect complex $\mathcal{E}^\bullet$.

If $K_1 \to K_2 \to K_3$ is a distinguished triangle in $D(\mathcal{O}_ X)$, then we get distinguished triangles

$R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (L, M) \otimes _{\mathcal{O}_ X}^\mathbf {L} K_1 \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (L, M) \otimes _{\mathcal{O}_ X}^\mathbf {L} K_2 \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (L, M) \otimes _{\mathcal{O}_ X}^\mathbf {L} K_3$

and

$R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K_1, L), M) \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K_2, L), M) R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K_3, L), M)$

See Section 20.26 and Lemma 20.42.4. The arrow of Lemma 20.42.9 is functorial in $K$ hence we get a morphism between these distinguished triangles. Thus, if the result holds for $K_1$ and $K_3$, then the result holds for $K_2$ by Derived Categories, Lemma 13.4.3.

Combining the remarks above with the distinguished triangles

$\sigma _{\geq n}\mathcal{E}^\bullet \to \mathcal{E}^\bullet \to \sigma _{\leq n - 1}\mathcal{E}^\bullet$

of stupid trunctions, we reduce to the case where $K$ consists of a direct summand of a finite free $\mathcal{O}_ X$-module placed in some degree. By an obvious compatibility of the problem with direct sums (similar to what was said above) and shifts this reduces us to the case where $K = \mathcal{O}_ X^{\oplus n}$ for some integer $n$. This case is clear. $\square$

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