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I saw this question asked in some form somewhere, but couldn't quite prove the statement that I wanted to, so I am hoping that someone can help. Essentially I want to prove that a spectral sequence is a generalization of a long exact sequence.

Here's the formulation: Let $\mathcal{C}$ and $\mathcal{C'}$ be abelian categories and $F: \mathcal{C} \rightarrow \mathcal{C'}$ an additive left-exact functor. Assume that we know that for any chain complex $A^{\bullet}$ of objects in $\mathcal{C}$ there exist two spectral sequences $$ E_1^{p,q} = R^qF(A^p) \implies R^{p+q}F(A^{\bullet}) $$ and $$ E_2^{p,q} = R^pF(\mathcal{H}^q(A^{\bullet})) \implies R^{p+q}F(A^{\bullet}) $$ Let $0 \rightarrow B' \rightarrow B \rightarrow B'' \rightarrow 0$ be a short exact sequence in $\mathcal{C}$. Prove that from the existence statement above follows that we have a long exact sequence: $$ 0 \rightarrow R^0F(B') \rightarrow R^0F(B) \rightarrow R^0F(B'')\rightarrow R^1F(B') \rightarrow R^1F(B) \rightarrow \cdots $$

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  • $\begingroup$ Ave you seen this thread : math.stackexchange.com/questions/2204494/… ? This is a bit different since this is not in the context of derived functors, but it translates almost immediately : if $0\to B'\to B\to B''\to 0$ then you get a short exact sequence of resolutions $0\to I'\to I\to I''\to 0$. The two spectral sequences arising from this double complex are the two you mentioned in your post. One of them is uninteresting because it is zero. The other gives the long exact sequence. $\endgroup$
    – Roland
    Jan 22, 2019 at 15:52
  • $\begingroup$ @Roland Oh I see. That does help, thanks! $\endgroup$
    – baltazar
    Jan 22, 2019 at 16:04

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