Lemma 20.51.3. Let $h : X \to Y$ be a morphism of ringed spaces. Let $K, M$ be objects of $D(\mathcal{O}_ Y)$. The canonical map

\[ Lh^*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, M) \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (Lh^*K, Lh^*M) \]

of Remark 20.42.13 is an isomorphism in the following cases

$K$ is perfect,

$h$ is flat, $K$ is pseudo-coherent, and $M$ is (locally) bounded below,

$\mathcal{O}_ X$ has finite tor dimension over $h^{-1}\mathcal{O}_ Y$, $K$ is pseudo-coherent, and $M$ is (locally) bounded below,

**Proof.**
Proof of (1). The question is local on $Y$, hence we may assume that $K$ is represented by a strictly perfect complex $\mathcal{E}^\bullet $, see Section 20.49. Choose a K-flat complex $\mathcal{F}^\bullet $ representing $M$. Apply Lemma 20.46.9 to see that $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, L)$ is represented by the complex $\mathcal{H}^\bullet = \mathop{\mathcal{H}\! \mathit{om}}\nolimits ^\bullet (\mathcal{E}^\bullet , \mathcal{F}^\bullet )$ with terms $\mathcal{H}^ n = \bigoplus \nolimits _{n = p + q} \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{E}^{-q}, \mathcal{F}^ p)$. By the construction of $Lh^*$ in Section 20.27 we see that $Lh^*K$ is represented by the strictly perfect complex $h^*\mathcal{E}^\bullet $ (Lemma 20.46.4). Similarly, the object $Lh^*M$ is represented by the complex $h^*\mathcal{F}^\bullet $. Finally, the object $Lh^*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, M)$ is represented by $h^*\mathcal{H}^\bullet $ as $\mathcal{H}^\bullet $ is K-flat by Lemma 20.46.10. Thus to finish the proof it suffices to show that $h^*\mathcal{H}^\bullet = \mathop{\mathcal{H}\! \mathit{om}}\nolimits ^\bullet (h^*\mathcal{E}^\bullet , h^*\mathcal{F}^\bullet )$. For this it suffices to note that $h^*\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{E}, \mathcal{F}) = \mathop{\mathcal{H}\! \mathit{om}}\nolimits (h^*\mathcal{E}, \mathcal{F})$ whenever $\mathcal{E}$ is a direct summand of a finite free $\mathcal{O}_ X$-module.

Proof of (2). Since $h$ is flat, we can compute $Lh^*$ by simply using $h^*$ on any complex of $\mathcal{O}_ Y$-modules. In particular we have $H^ i(Lh^*K) = h^*H^ i(K)$ for all $i \in \mathbf{Z}$. Say $H^ i(M) = 0$ for $i < a$. Let $K' \to K$ be a morphism of $D(\mathcal{O}_ Y)$ which defines an isomorphism $H^ i(K') \to H^ i(K)$ for all $i \geq b$. Then the corresponding maps

\[ R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, M) \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K', M) \]

and

\[ R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (Lh^*K, Lh^*M) \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (Lh^*K', Lh^*M) \]

are isomorphisms on cohomology sheaves in degrees $< a - b$ (details omitted). Thus to prove the map in the statement of the lemma induces an isomorphism on cohomology sheaves in degrees $< a - b$ it suffices to prove the result for $K'$ in those degrees. Also, as in the proof of part (1) the question is local on $Y$. Thus we may assume $K$ is represented by a strictly perfect complex, see Section 20.47. This reduces us to case (1).

Proof of (3). The proof is the same as the proof of (2) except one uses that $Lh^*$ has bounded cohomological dimension to get the desired vanishing. We omit the details.
$\square$

## Comments (2)

Comment #7128 by Ivan Serna on

Comment #7286 by Johan on