# Natural morphism appearing in Grothendieck spectral sequence

Assume we are in the setting of the Grothendieck spectral sequence (Weibel, 5.8): $G : A \to B, F : B \to C$ are left exact functors such that $G$ sends injective objects to $F$-acyclic objects. Now the edge maps should be some natural maps $(R^p F)(GA) \to R^p(FG)(A)$ and $R^q(FG)(A) \to F((R^q G)(A))$. But how are they defined?

I've managed to write down a definition in the special case of the Leray spectral sequence, there $G : Sh(X) \to Sh(Y)$ is a direct image functor of a map $f : X \to Y$ and $F : Sh(Y) \to Ab$ is a global section functor: The first map is $H^q(Y,f_* A) \to H^q(X,A)$ and the second map is $H^q(X,A) \to H^0(Y,(R^q f_*) A)$. For the first one, you may use the inverse image functor and define $H^q(Y,-) \to H^q(X,f^{-1} -)$ using universal $\delta$-functors, and then compose this with the adjunction morphism $f^{-1} f_* A \to A$. For the second one, you may use an injective resolution $I^*$ of $A$ and use the canonical maps $H^0(Y,Z)/H^0(Y,B) \to H^0(Y,Z/B)$ for a subsheaf $B \subseteq Z$.

But this method does not generalize.

Also, I want to know why these maps are natural with respect to $F$ and $G$. For example in the special case above, if $f' : X' \to Y$ is another map and $g : X \to X'$ is a map over $Y$ and $A'$ is a sheaf on $X'$, why is the diagram [Feel free to edit this!]

$H^q(X',A') \to H^q(X,g^{-1} A')$

$\downarrow ~~~~~~~~~~~~~~~~~~~~~~~~~~ \downarrow$

$H^0(Y,(R^q f'_*) A') \to H^0(Y,(R^q f_*) g^{-1} A')$

commutative? Motivation: This is needed to show that the morphism $Pic(X) \to Pic(X/Y)$ is natural, where $f$ is a morphism of locally ringed spaces.

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The compatibility with natural transformations in $F$ and $G$ is an exercise in looking at the construction. The final commutative diagram is explained in section 12 of Ch. 0 of EGA III; best seen via the answer to the question on the 2nd edge map (for which your question is surely to give intrinsic characterizations of each so they can be identified/computed in specific situations). What's the answer? It's the unique such map from the universal $\delta$-functor ${\rm{R}}^q(FG)$ to the non-exact $\delta$-functor'' $F \circ {\rm{R}}^q(G)$ (exercise: make precise what such a statement means!) –  Boyarsky Jun 20 '10 at 11:55
In the above I should have said at the end that it is the unique such map which is the evident equality for $q=0$. –  Boyarsky Jun 20 '10 at 16:28
Could you please elaborate this? In the section of EGA you mentioned I can't find any answer to my questions. Also, I don't know why univeral $\delta$-functors are also initial (or at least here) for non-exact $\delta$-functors. Otherwise I would not have asked ... –  Martin Brandenburg Jun 21 '10 at 9:27
@Martin: Look at 12.2.5 for the relation of the 2nd edge map with "sheafified pullback" (you may want to provide some justification omitted in EGA), and 12.1.3 for the proof of functoriality in the morphism for sheaf-pullback. Combining these, the commutative square drops out. Concerning "non-exact $\delta$-functor" stuff, I wasn't assuming you knew about it already, which was why I thought pointing you in that direction was a good hint. (I noticed it myself for this application!) I say more: use erasability and look back at the proof of universality of erasable $\delta$-functors. –  Boyarsky Jun 21 '10 at 11:34
@Martin: By the way, for the first edge map, the only "direct description" I ever found for it was in terms of injective/acyclic resolutions (certainly more direct than saying "edge map in spectral sequence", so not a tautology, and still useful for concrete identification of it in some practical situations). Not as nice as for the 2nd edge map. If you notice something better, please mention it here. –  Boyarsky Jun 21 '10 at 11:42

The second edge morphism $\kappa_2: R^p(FG)(X) \to FR^pG(X)$ is induced by universality of $R^*(FG)$ from the identity on $FG(X)$.
The first edge morphism $\kappa_1: R^pF(GX) \to R^p(FG)(X)$ can be described as follows: Assume $G$ has an exact adjoint functor $T$. Choose an injective resolution $X[0] \to J^\bullet$ in $\mathcal{A}$. Apply $G$ and choose an injective resolution $GX[0] \to I^\bullet$ in $\mathcal{B}$. Apply $T$ and use the composition of the natural map $TGX[0] \to X[0] \to J^\bullet$ and the theorem on the natural extending of morphisms between an exact complex and an injective complex to obtain a morphism $TI^\bullet \to J^\bullet$. Adjoint to this is $\phi: I^\bullet \to GJ^\bullet$, which gives us $\kappa_1$ after applying $F$ and taking cohomology. (see also commutative diagram with Yoneda pairing and edge morphism)