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

of Lemma 20.42.9 is an isomorphism.

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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