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Let $F: \mathcal{A} \to \mathcal{B}$ be an exact covariant functor of abelian categories and let $$\mathscr{C}: A \to A \to B \to A$$ be an exact couple in $\mathcal{A}$ with corresponding spectral sequence $E^r(\mathscr{C})$.
$$F(\mathscr{C}): F(A) \to F(A) \to F(B) \to F(A)$$ is a exact couple in $\mathcal{B}$ and defines a spectral sequence $E^r(F(\mathscr{C}))$. I am pretty sure there is an isomorphism of spectral sequences that is natural in $\mathscr{C}$: $$E^r(F(\mathscr{C})) \cong F(E^r(\mathscr{C}))$$ Such a isomorphism is equivalent to a natural isomorphism $D(F(\mathscr{C}))\cong F(D(\mathscr{C}))$ for all exact couples where $D(-)$ denotes the derived couple.

Question: Does someone know a reference for such an isomorphism or a place in the literature where such an isomorphism is used (i.e. in this generality, not for particular $F)$ ?

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    $\begingroup$ Of course this is true! By the definition of "exact" and "functor." Such functors commute with taking kernels and cokernels, and this is precisely what one does in forming the derived exact couple. $\endgroup$
    – Dylan Wilson
    Commented Mar 10, 2013 at 3:23
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    $\begingroup$ @Dylan: That's why I asked for a reference. Also note that a formal proof that an exact functor commutes with homology requires some work. So one shouldn't expect that a formal proof for the isomorphism in the question is written down in 1 or 2 lines (as your comment seems to indicate). $\endgroup$
    – Ralph
    Commented Mar 10, 2013 at 6:20
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    $\begingroup$ I think, as Dylan, that a formal proof is not really necessary if you're familiar with the notion of exact functor, so probably you won't find it. I assume you want to put a reference in a paper. I would just check McCleary's book, and if it is not there I'd take its as 'well known'. $\endgroup$
    – Fernando Muro
    Commented Mar 10, 2013 at 8:27
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    $\begingroup$ BTW I you want to deduce something about the target of the spectral sequence you should be careful. If your original sequence is associated with a degreewise finite filtration of the target you're OK, otherwise you must assume some continuity conditions on $F$ to be able to say someting meaningful. $\endgroup$
    – Fernando Muro
    Commented Mar 10, 2013 at 9:09

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