Construction of the spectral sequence of Katz/Oda

Question

In their famous paper "On the differentiation of De Rham cohomology classes..." Katz and Oda construct the spectral sequence for de Rham cohomology for the situation of a smooth morphism

$\pi: X \rightarrow S$

of smooth $k$-schemes ($k$ a field), where $S$ is assumed affine.

There is a step which is not clear to me: in Lemma 8 of this paper they say that the equality

$(*) \qquad \mathbb R^0\Gamma_X =\Gamma_S\circ \mathbb R^0\pi_*$

yields a spectral sequence of composition

$(**) \qquad E_2^{a,b}=R^a \Gamma_S\circ \mathbb R^b\pi_* \Rightarrow \mathbb R^{a+b}\Gamma_X$.

They seem to consider $(*)$ as an equality of functors on the category of complexes of abelian sheaves on $X$ - Arguments involving quasi-coherence and affineness of $S$ only appear in the next step.

My question is a very simple one which I nevertheless can't figure out:

Why do they have $(**)$?

It is a spectral sequence of composition, hence only exists when one knows that $\mathbb R^0 \pi_* $ sends injective objects in the category of complexes of sheaves on $X$ to $\Gamma_S$-acyclic ones. I don't see why this is true: imagine a two-term complex $I^0\rightarrow I^1$ of injective abelian sheaves on $X$, then $\mathbb R^0 \pi_*$ of it is $ker(\pi_*I^0 \rightarrow \pi_*I^1)$. This has no reason to be acyclic for $\Gamma_S$, hasn't it?

A similar problem arises a bit later in Lemma 9 when they consider the equality

$(+) \qquad \mathbb R^0\Gamma_S = H^0\circ \Gamma_S$

which I see as equality of functors on the category of complexes of abelian sheaves on $S$.

Can anybody give a hint how to cleanly resolve these two problems?

Answer 1

Let me try to say the same thing as Will but in a different way. Let $f \colon A \to B$ be left exact between abelian categories and assume there are enough injectives. Then one can distinguish between derived functors $$\newcommand{\R}{\mathrm R} \R^i f \colon A \to B,$$ as well as hyper-derived functors $$\newcommand{\RR}{\mathbb R} \RR^i f \colon \mathrm{Kom}^+(A) \to B,$$ and the total derived functor $$ \newcommand{\RRR}{\mathbf R} \RRR f \colon D^+(A) \to D^+(B).$$ Now suppose we also have $g \colon B \to C$ with the same hypotheses and that $f$ maps injective objects of $A$ to $g$-acyclic objects of $B$. Then there is a natural isomorphism $$ \RRR(g \circ f) \cong \RRR g \circ \RRR f.$$ This isomorphism specializes not only to the usual spectral sequence $$ E_2^{p,q} = \R^p g( \R^q f(X)) \implies \R^{p+q}(g \circ f)(X),$$ for $X$ any object of $A$, but also to the spectral sequence $$ E_2^{p,q} = \R^p g( \RR^q f(X^\bullet)) \implies \RR^{p+q}(g \circ f)(X^\bullet),$$ for $X^\bullet$ any object of $\mathrm{Kom}^+(A)$, in a similar way. In particular one needs no extra hypotheses to get the latter spectral sequence; even though the domain of $\RR^i f$ is $\mathrm{Kom}^+(A)$, one should not assume that complexes of injectives are sent to acyclics.

Answer 2

You only need injective objects in the underlying abelian category to be mapped to acyclics, not in the category of complexes, to get a Grothendieck spectral sequence. This is because if you take a complex $A$, and take an injective resolution over that complex (to compute hypercohomology), then after applying $R \pi_*$ to this complex, you get an acyclic resolution of $R \pi_* A$. Then you can use that to compute hypercohomology of $\Gamma_s$, which you do by just applying $R^0 \Gamma_s$ to each object, which is the same as applying $R^0 \Gamma$ to the original injective resolution, which is just computing hypercohomology of $R^0 \Gamma$.