Morphisms in the derived category vs. morphisms on cohomology

Let $A$ and $B$ be complexes $R$-modules. Assume that $A^\bullet$ is bounded above and $B^\bullet$ is bounded below. Then there is a convergent spectral sequence $$\prod_{r\in \mathbb{Z}} \operatorname{Ext}^p_R (H^{r-q} (A), H^r (B)) \Longrightarrow \operatorname{Ext}_{\mathcal{D} (R\text{-Mod})}^{p+q} (A,B)$$ —see e.g. the very last result in Verdier's thesis.

If we are interested in complexes of abelian groups ($R = \mathbb{Z}$), then $\operatorname{Ext}^p_\mathbb{Z} = 0$, unless $p = 0,1$, and the above thing is a two-column spectral sequence, which gives us short exact sequences $$0 \to E_2^{1,n-1} \to H^n \to E_2^{0,n} \to 0$$ In particular, we have a short exact sequence $$\tag{*} 0 \to \prod_{i\in \mathbb{Z}} \operatorname{Ext}_\mathbb{Z} (H^i (A), H^{i-1} (B)) \to \operatorname{Hom}_{\mathcal{D} (\text{Ab})} (A,B) \to \prod_{i\in \mathbb{Z}} \operatorname{Hom}_\mathbb{Z} (H^i (A), H^i (B)) \to 0$$ The last morphism here is given by $f \mapsto (H^i (f))_{i\in \mathbb{Z}}$.

1. This looks very natural and I wonder whether there is a direct way to get (*).

2. Does (*) hold if the complexes are not bounded? I am interested in the case where they are both bounded below and not bounded above.

Thank you.

If $\text{Ext}^i_{\mathcal A}(-,-)\equiv 0$ for $i>1$ (that is, if ${\mathcal A}$ is a hereditary category) then any bounded complex is isomorphic to the direct sum of its cohomologies in ${\mathbf D}_{\mathcal A}$, i.e. $X\cong\bigoplus_n \Sigma^{-n}\text{H}^n(X)$ - see e.g. If the cohomology of two objects in the derived category are equal, are the objects isomorphic?.

Therefore, studying $\textbf{D}_{\mathcal A}(X,Y)$ as you do in $(\ast)$ naturally splits in two cases:

1. $X=X^0,Y=Y^0\in{\mathcal A}$, i.e. both are stalk complexes in cohomological degree $0$.

In this case, $(\ast)$ degenerates to $\textbf{D}_{\mathcal A}(X,Y)\cong{\mathcal A}(X,Y)$, i.e. the fact that ${\mathcal A}\to\textbf{D}_{\mathcal A}$ is fully faithful.

2. $X=X^0, Y=\Sigma Y^{-1}$, i.e. $X$ and $Y$ are stalk complexes in cohomological degree $0$ and $-1$, respectively.

In this case $(\ast)$ degenerates to $\textbf{D}_{\mathcal A}(X,Y)\cong\text{Ext}^1_{\mathcal A}(X,Y)$.

• Thank you! Is there a way to do that for $\mathcal{D}^+$ instead of $\mathcal{D}^b$? – Lyosha Jun 12 '17 at 10:09
• Indeed, your decomposition may be deduced without induction, hence the result doesn't need boundedness of complexes. So it solves my problem. – Lyosha Jun 12 '17 at 20:39