## 20.49 Miscellany

Some results which do not fit anywhere else.

Lemma 20.49.1. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $(K_ n)_{n \in \mathbf{N}}$ be a system of perfect objects of $D(\mathcal{O}_ X)$. Let $K = \text{hocolim} K_ n$ be the derived colimit (Derived Categories, Definition 13.33.1). Then for any object $E$ of $D(\mathcal{O}_ X)$ we have

$R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, E) = R\mathop{\mathrm{lim}}\nolimits E \otimes ^\mathbf {L}_{\mathcal{O}_ X} K_ n^\vee$

where $(K_ n^\vee )$ is the inverse system of dual perfect complexes.

Proof. By Lemma 20.48.5 we have $R\mathop{\mathrm{lim}}\nolimits E \otimes ^\mathbf {L}_{\mathcal{O}_ X} K_ n^\vee = R\mathop{\mathrm{lim}}\nolimits R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K_ n, E)$ which fits into the distinguished triangle

$R\mathop{\mathrm{lim}}\nolimits R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K_ n, E) \to \prod R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K_ n, E) \to \prod R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K_ n, E)$

Because $K$ similarly fits into the distinguished triangle $\bigoplus K_ n \to \bigoplus K_ n \to K$ it suffices to show that $\prod R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K_ n, E) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\bigoplus K_ n, E)$. This is a formal consequence of (20.40.0.1) and the fact that derived tensor product commutes with direct sums. $\square$

Lemma 20.49.2. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $K$ and $E$ be objects of $D(\mathcal{O}_ X)$ with $E$ perfect. The diagram

$\xymatrix{ H^0(X, K \otimes _{\mathcal{O}_ X}^\mathbf {L} E^\vee ) \times H^0(X, E) \ar[r] \ar[d] & H^0(X, K \otimes _{\mathcal{O}_ X}^\mathbf {L} E^\vee \otimes _{\mathcal{O}_ X}^\mathbf {L} E) \ar[d] \\ \mathop{\mathrm{Hom}}\nolimits _ X(E, K) \times H^0(X, E) \ar[r] & H^0(X, K) }$

commutes where the top horizontal arrow is the cup product, the right vertical arrow uses $\epsilon : E^\vee \otimes _{\mathcal{O}_ X}^\mathbf {L} E \to \mathcal{O}_ X$ (Example 20.48.7), the left vertical arrow uses Lemma 20.48.5, and the bottom horizontal arrow is the obvious one.

Proof. We will abbreviate $\otimes = \otimes _{\mathcal{O}_ X}^\mathbf {L}$ and $\mathcal{O} = \mathcal{O}_ X$. We will identify $E$ and $K$ with $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}, E)$ and $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}, K)$ and we will identify $E^\vee$ with $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (E, \mathcal{O})$.

Let $\xi \in H^0(X, K \otimes E^\vee )$ and $\eta \in H^0(X, E)$. Denote $\tilde\xi : \mathcal{O} \to K \otimes E^\vee$ and $\tilde\eta : \mathcal{O} \to E$ the corresponding maps in $D(\mathcal{O})$. By Lemma 20.31.1 the cup product $\xi \cup \eta$ corresponds to $\tilde\xi \otimes \tilde\eta : \mathcal{O} \to K \otimes E^\vee \otimes E$.

We claim the map $\xi ' : E \to K$ corresponding to $\xi$ by Lemma 20.48.5 is the composition

$E = \mathcal{O} \otimes E \xrightarrow {\tilde\xi \otimes 1_ E} K \otimes E^\vee \otimes E \xrightarrow {1_ K \otimes \epsilon } K$

The construction in Lemma 20.48.5 uses the evaluation map (20.40.8.1) which in turn is constructed using the identification of $E$ with $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}, E)$ and the composition $\underline{\circ }$ constructed in Lemma 20.40.5. Hence $\xi '$ is the composition

\begin{align*} E = \mathcal{O} \otimes R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}, E) & \xrightarrow {\tilde\xi \otimes 1} R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}, K) \otimes R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (E, \mathcal{O}) \otimes R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}, E) \\ & \xrightarrow {\underline{\circ } \otimes 1} R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (E, K) \otimes R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}, E) \\ & \xrightarrow {\underline{\circ }} R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}, K) = K \end{align*}

The claim follows immediately from this and the fact that the composition $\underline{\circ }$ constructed in Lemma 20.40.5 is associative (insert future reference here) and the fact that $\epsilon$ is defined as the composition $\underline{\circ } : E^\vee \otimes E \to \mathcal{O}$ in Example 20.48.7.

Using the results from the previous two paragraphs, we find the statement of the lemma is that $(1_ K \otimes \epsilon ) \circ (\tilde\xi \otimes \tilde\eta )$ is equal to $(1_ K \otimes \epsilon ) \circ (\tilde\xi \otimes 1_ E) \circ (1_\mathcal {O} \otimes \tilde\eta )$ which is immediate. $\square$

Lemma 20.49.3. 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 20.40.13 is an isomorphism in the following cases

1. $K$ is perfect,

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

3. $\mathcal{O}_ X$ 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 20.47. Choose a K-flat complex $\mathcal{F}^\bullet$ representing $M$. Apply Lemma 20.44.9 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 20.44.4). 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 20.44.10. 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 20.45. 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. $\square$

Lemma 20.49.4. Let $X$ be a ringed space. Let $K, M$ be objects of $D(\mathcal{O}_ X)$. Let $x \in X$. The canonical map

$R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, M)_ x \longrightarrow R\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X, x}}(K_ x, M_ x)$

is an isomorphism in the following cases

1. $K$ is perfect,

2. $K$ is pseudo-coherent and $M$ is (locally) bounded below.

Proof. Let $Y = \{ x\}$ be the singleton ringed space with structure sheaf given by $\mathcal{O}_{X, x}$. Then apply Lemma 20.49.3 to the flat inclusion morphism $Y \to X$. $\square$

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