Lemma 15.86.3. Let $R$ be a ring. Let $K, L, M$ be objects of $D(R)$. The map

$K \otimes _ R^\mathbf {L} R\mathop{\mathrm{Hom}}\nolimits _ R(M, L) \longrightarrow R\mathop{\mathrm{Hom}}\nolimits _ R(M, K \otimes _ R^\mathbf {L} L)$

of Lemma 15.68.5 is an isomorphism in the following cases

1. $M$ perfect, or

2. $K$ is perfect, or

3. $M$ is pseudo-coherent, $L \in D^+(R)$, and $K$ has tor amplitude in $[a, \infty ]$.

Proof. Proof in case $M$ is perfect. Note that both sides of the arrow transform distinguished triangles in $M$ into distinguished triangles and commute with direct sums. Hence it suffices to check it holds when $M = R[n]$, see Derived Categories, Remark 13.33.5 and Lemma 15.72.1. In this case the result is obvious.

Proof in case $K$ is perfect. Same argument as in the previous case.

Proof in case (3). We may represent $K$ and $L$ by bounded below complexes of $R$-modules $K^\bullet$ and $L^\bullet$. We may assume that $K^\bullet$ is a K-flat complex consisting of flat $R$-modules, see Lemma 15.63.4. We may represent $M$ by a bounded above complex $M^\bullet$ of finite free $R$-modules, see Definition 15.62.1. Then the object on the LHS is represented by

$\text{Tot}(K^\bullet \otimes _ R \mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , L^\bullet ))$

and the object on the RHS by

$\mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , \text{Tot}(K^\bullet \otimes _ R L^\bullet ))$

This uses Lemma 15.68.2. Both complexes have in degree $n$ the module

$\bigoplus \nolimits _{p + q + r = n} K^ p \otimes \mathop{\mathrm{Hom}}\nolimits _ R(M^{-r}, L^ q) = \bigoplus \nolimits _{p + q + r = n} \mathop{\mathrm{Hom}}\nolimits _ R(M^{-r}, K^ p \otimes _ R L^ q)$

because $M^{-r}$ is finite free (as well these are finite direct sums). The map defined in Lemma 15.68.5 comes from the map of complexes defined in Lemma 15.67.5 which uses the canonical isomorphisms between these modules. $\square$

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