Right derived functors

Question

I am reading the book “Fourier-Mukai Transforms in Algebraic Geometry” by Daniel Huybrechts and I am trying to understand the proof of Corollary 2.68. This Corollary is as follows:

Suppose $F:K^+(\mathcal{A})\to K^+(\mathcal{B})$ is an exact functor which admits a right derived functor $RF:D^+(\mathcal{A})\to D^+(\mathcal{B})$ and assume that $\mathcal{A}$ has enough injectives.

i) Suppose $\mathcal{C}\subset \mathcal{B}$ is a thick subcategory with $R^iF(A)\in \mathcal{C}$ for all $A\in \mathcal{A}$ and that there exists $n\in \mathbb{Z}$ with $R^iF(A)=0$ for $i<n$ and all $A\in \mathcal{A}$. Then, $RF$ takes values in $D^b_{\mathcal{C}}(\mathcal{B})$, i.e. $$RF:D^+(\mathcal{A})\to D^+_{\mathcal{C}}(\mathcal{B})$$

ii) If $RF(A)\in D^b(\mathcal{B})$ for any object $A\in \mathcal{A}$, then $RF(A^{\bullet})\in D^b(\mathcal{B})$ for any complex $A^{\bullet}\in D^b(\mathcal{A})$, i.e. $RF$ induces an exact functor $$RF:D^b(\mathcal{A})\to D^b(\mathcal{B})$$

In its proof, it is written that both assertions immediately follow from the spectral sequence $E_2^{p,q}=R^pF(H^q(A^{\bullet}))\Longrightarrow R^{p+q}(A^{\bullet})$. My question is that why is this true?

Answer 1

For i) : we have to show that for any $n$ and any $A^\bullet\in\mathcal{A}$ then $R^nF(A^\bullet)\in\mathcal{C}$. But from the spectral sequence, we know that $R^nF(A^\bullet)$ is an extension of subquotients of $R^pF(H^q(A^\bullet))$ (with $p+q=n$). But by assumption, the objects $R^pF(H^q(A^\bullet))$ are all in $\mathcal{C}$. Because $\mathcal{C}$ is thick, so are their subquotient and their extensions, and thus we indeed have $R^nF(A^\bullet)\in\mathcal{C}$.

For ii), we have to prove that if $RF(A^\bullet)$ is bounded. Since $A^\bullet$ is bounded, we can say that its cohomology is concentrated in degree $[m,M]$. For each $q\in [m,M]$, $RF(H^q(A))\in D^b(\mathcal{B})$ by assumption so $R^pF(H^q(A^\bullet))=0$ for all $p\not\in [0,M_q]$. Take $M'$ the max of all the $M_q$. From the spectral sequence, it follows that $R^nF(A^\bullet)=0$ for all $n\not\in [m,M+M']$. Hence $RF(A^\bullet)$ is bounded.