# Two basic questions on derived categories

Let $\mathcal{A}, \mathcal{B}$ be two abelian categories with sufficiently many injective objects (in my case these are categories of sheaves of vector spaces on a manifold). Let $f_*\colon \mathcal{A}\to\mathcal{B}$ be a left exact functor which commutes with direct sums (in my case it is push-forward on sheaves). Let $Rf_*\colon D^+\mathcal{A}\to D^+\mathcal{B}$ be the derived functor between the derived categories. Let $F^\bullet$ be a bounded complex of objects of $\mathcal{A}$.

Question 1. Assume that $R^qf_*(F^p)=0$ for every $q\ne i$ where $i$ is a fixed number, and for every $p$. Is it true that $Rf_*F^\bullet$ is isomorphic in $D^+\mathcal{B}$ to the complex $$\dots\to R^if_*(F^k)\to R^if_*(F^{k+1})\to R^if_*(F^{k+2})\to\dots \,?$$ (For $i=0$ this is well known to be true. I believe that it should be true in general. A reference would also be helpful.)

Question 2. Assume now that for a fixed integer $a$ one has $$\begin{eqnarray*} R^qf_*(F^a)=0 \mbox{ for every } q;\\ R^qf_*(F^p)=0 \mbox{ for every } q\ne 0, p>a;\\ R^qf_*(F^p)=0 \mbox{ for every } q\ne 1, p<a. \end{eqnarray*}$$ Is it true that $Rf_*(F^\bullet)$ is isomorphic to the complex (with an appropriate cohomological shift) $$\to R^1f_*F^{a-2}\to R^1f_*F^{a-1} \overset{\Delta}{\to}R^0f_*F^{a+1}\to R^0f_*F^{a+1}\to \dots$$ where $\Delta$ is the only non-trivial differential of second order in the well known spectral sequence $E^{p.q}_r$ which converges to $R^{p+q}f_*(F^\bullet)$ and whose first term is equal to $E_1^{p,q}=R^qf_*(F^p)$? (The other arrows in the complex are induced in the obvious way by arrows in the complex $F^\bullet$.)

Remark. The second question is motivated by the fact that the two complexes have the same cohomology, as it follows from the spectral sequence considerations.

• I'm assuming that in question 1 it should be $$\to R^if_*(F^{k+1}) \to R^if_*(F^{k+2}) \to?$$ – Simon Rose Oct 12 '14 at 10:50
• If by "any q" in question 1 you mean "every q", this is true by the same (simple) argument as the i=0 case. You should understand the argument yourself. – anon Oct 12 '14 at 13:09
• @anon: Corrected to "every q". Thanks. – MKO Oct 12 '14 at 14:44

First of all, the functor $Rf_\ast$ is a red herring. Take an injective resolution $I^{\bullet,\bullet}$ of $F^\bullet$ and apply $f_\ast$; then you just have two questions regarding double complexes. E.g. in the first case you're asking about double complexes whose columns have cohomology concentrated in a single degree.
Let $C^\bullet$ be a cochain complex. Recall that there are wise truncation functors $\tau_{\geq i}$, $\tau_{\leq i}$ with canonical maps $C^\bullet \to \tau_{\geq i}C^\bullet$ and $\tau_{\leq i}C^\bullet \to C^\bullet$, which are quasi-isomorphisms if $C^\bullet$ has no cohomology below (resp. above) degree $i$.
If $A^{\bullet,\bullet}$ is a double complex, let $\tau^V_{\geq i}$ be the double complex obtained by applying $\tau_{\geq i}$ to each column separately (V is for "vertical"). Then there is a zig-zag $$A^{\bullet,\bullet} \to \tau_{\geq i}^V A^{\bullet,\bullet} \leftarrow \tau_{\leq i}^V\tau^V_{\geq i} A^{\bullet,\bullet}.$$ If all columns have cohomology concentrated in degree $i$ then both these maps are quasi-isomorphisms of double complexes. This answers your first question affirmatively, since $\tau_{\leq i}^V\tau^V_{\geq i} A^{\bullet,\bullet}$ can be identified with the cochain complex you wrote down. (Remark: I slightly disagree with anon's comment that the case $i>0$ is identical to the case $i=0$, since in the latter case we don't need to apply $\tau_{\geq i}$. So in that case we don't even need a zig-zag, there is just a quasi-isomorphism staring us in the face.)
The second requires a bit more fiddling, I think, but maybe I'm missing an easy argument. But the following should work. First do a truncation argument similar to the one above to reduce to the case when $A^{p,q} = 0$ unless $p\leq a$ and $q=1$, or $p \geq a$ and $q=0$; moreover, $A^{a,0} \to A^{a,1}$ is an isomorphism. Now let $B^{\bullet,\bullet}$ be the double complex with $B^{p,q} = A^{p,q}$ for $p \neq a$, but $B^{a,0} = B^{a,1} = A^{a-1,1}$. The differentials in $B$ are given by letting $B^{a-1,1} \to B^{a,1}$ and $B^{a,0} \to B^{a,1}$ be the obvious isomorphisms and $B^{a,0} \to B^{a+1,0}$ the map given by the $E_2$ differential in the spectral sequence for $A^{\bullet,\bullet}$. There is a natural map $B^{\bullet,\bullet} \to A^{\bullet,\bullet}$, and also a natural map from $\mathrm{Tot}(B^{\bullet,\bullet})$ to the cochain complex which you wrote down, both of which are quasi-isomorphisms.