The Stacks project

Proof. Observe that $\textit{Coh}(\mathcal{O}_ X) \subset \mathit{QCoh}(\mathcal{O}_ X)$ is a Serre subcategory, see Homology, Definition 12.10.1 and Lemma 12.10.2 and Cohomology of Schemes, Lemmas 30.9.2 and 30.9.3. On the other hand, if $\mathcal{G} \to \mathcal{F}$ is a surjection from a quasi-coherent $\mathcal{O}_ X$-module to a coherent $\mathcal{O}_ X$-module, then there exists a coherent submodule $\mathcal{G}' \subset \mathcal{G}$ which surjects onto $\mathcal{F}$. Namely, we can write $\mathcal{G}$ as the filtered union of its coherent submodules by Properties, Lemma 28.22.3 and then one of these will do the job. Thus the lemma follows from Derived Categories, Lemma 13.17.4. $\square$

Proof. Consider the commutative diagram

By Lemma 36.11.1 the left vertical arrow is fully faithful. By Proposition 36.8.3 the bottom arrow is an equivalence. By construction the right vertical arrow is fully faithful. We conclude that the top horizontal arrow is fully faithful. If $K$ is an object of $D^-_{\textit{Coh}}(\mathcal{O}_ X)$ then the object $K'$ of $D^-(\mathit{QCoh}(\mathcal{O}_ X))$ which corresponds to it by Proposition 36.8.3 will have coherent cohomology sheaves. Hence $K'$ is in the essential image of the left vertical arrow by Lemma 36.11.1 and we find that the top horizontal arrow is essentially surjective. This finishes the proof for the bounded above case. The bounded case follows immediately from the bounded above case. $\square$

Proof. Consider the spectral sequence

see Derived Categories, Lemma 13.21.3. By assumption and Cohomology of Schemes, Lemma 30.26.10 the sheaves $R^ pf_*H^ q(E)$ are coherent. Hence $R^{p + q}f_*E$ is coherent, i.e., $Rf_*E \in D_{\textit{Coh}}(\mathcal{O}_ S)$. Boundedness from below is trivial. Boundedness from above follows from Cohomology of Schemes, Lemma 30.4.5 or from Lemma 36.4.1. $\square$

Proof. The proof is the same as the proof of Lemma 36.11.3. You can also deduce it from Lemma 36.11.3 by considering what the exact functor $Rf_*$ does to the distinguished triangles $\tau _{\leq a}E \to E \to \tau _{\geq a + 1}E \to \tau _{\leq a}E[1]$. $\square$

Proof. It suffices to prove this when $X$ is the spectrum of a Noetherian ring $A$. By Lemma 36.10.3 we see that $K$ is pseudo-coherent. Then we can use Lemma 36.10.8 to translate the problem into the following algebra problem: for $L \in D^+_{\textit{Coh}}(A)$ and $K$ in $D^-_{\textit{Coh}}(A)$, then $R\mathop{\mathrm{Hom}}\nolimits _ A(K, L)$ is in $D^+_{\textit{Coh}}(A)$. Since $L$ is bounded below and $K$ is bounded below there is a convergent spectral sequence

and there are convergent spectral sequences

See Injectives, Remarks 19.13.9 and 19.13.11. This finishes the proof as the modules $\mathop{\mathrm{Ext}}\nolimits ^ p_ A(M, N)$ are finite for finite $A$-modules $M$, $N$ by Algebra, Lemma 10.71.9. $\square$

Proof. Since $X$ is quasi-compact, each of these statements can be checked over the members of any open covering of $X$. Thus we may assume $E$ is represented by a bounded complex $\mathcal{E}^\bullet $ of finite free modules, see Cohomology, Lemma 20.49.3. In this case each of the statements is clear as both $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(E, L)$ and $E \otimes _{\mathcal{O}_ X}^\mathbf {L} L$ can be computed on the level of complexes using $\mathcal{E}^\bullet $, see Cohomology, Lemmas 20.46.9 and 20.26.9. Some details omitted. $\square$

Proof. Recall that

see Cohomology, Lemma 20.42.1 and Cohomology, Section 20.13. Thus the result follows from Lemmas 36.11.5 and 36.11.4. $\square$

Proof. Let $K$ be an object of $D^ b_{\textit{Coh}}(\mathcal{O}_ X)$. To check that $K$ is perfect, we may work affine locally on $X$ (see Cohomology, Section 20.49). Then $K$ is perfect by Lemma 36.10.7 and More on Algebra, Lemma 15.74.14. The converse is Lemma 36.11.6. $\square$