Lemma 48.2.4. Let $X$ be a Noetherian scheme. Let $K, L, M \in D_\mathit{QCoh}(\mathcal{O}_ X)$. 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 Cohomology, Lemma 20.42.9 is an isomorphism in the following two cases

$K \in D^-_{\textit{Coh}}(\mathcal{O}_ X)$, $L \in D^+_{\textit{Coh}}(\mathcal{O}_ X)$, and $M$ affine locally has finite injective dimension (see proof), or

$K$ and $L$ are in $D_{\textit{Coh}}(\mathcal{O}_ X)$, the object $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (L, M)$ has finite tor dimension, and $L$ and $M$ affine locally have finite injective dimension (in particular $L$ and $M$ are bounded).

**Proof.**
Proof of (1). We say $M$ has affine locally finite injective dimension if $X$ has an open covering by affines $U = \mathop{\mathrm{Spec}}(A)$ such that the object of $D(A)$ corresponding to $M|_ U$ (Derived Categories of Schemes, Lemma 36.3.5) has finite injective dimension^{1}. To prove the lemma we may replace $X$ by $U$, i.e., we may assume $X = \mathop{\mathrm{Spec}}(A)$ for some Noetherian ring $A$. Observe that $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, L)$ is in $D^+_{\textit{Coh}}(\mathcal{O}_ X)$ by Derived Categories of Schemes, Lemma 36.11.5. Moreover, the formation of the left and right hand side of the arrow commutes with the functor $D(A) \to D_\mathit{QCoh}(\mathcal{O}_ X)$ by Lemma 48.2.3 and Derived Categories of Schemes, Lemma 36.10.8 (to be sure this uses the assumptions on $K$, $L$, $M$ and what we just proved about $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, L)$). Then finally the arrow is an isomorphism by More on Algebra, Lemmas 15.98.1 part (2).

Proof of (2). We argue as above. A small change is that here we get $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, L)$ in $D_{\textit{Coh}}(\mathcal{O}_ X)$ because affine locally (which is allowable by Lemma 48.2.3) we may appeal to Dualizing Complexes, Lemma 47.15.2. Then we finally conclude by More on Algebra, Lemma 15.98.2.
$\square$

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