It is well known that a bigraded exact couple of objects of an abelian category yields a spectral sequence (cf. https://ncatlab.org/nlab/show/exact+couple#SpectralSequencesFromExactCouples). My question is:under which conditions does this spectral sequence degenerate at $E_1$ (actually, I am more interested in necessary conditions)? This condition appears to be equivalent to the image of the morphism $f_1:E_1\to D_1$ lying inside all levels of the filtration of $D_1$ by $g_1^i(D_1)$ (here I ignore the upper indices, and $g_1^i$ denotes the $i$th iterate of $g_1:D_1\to D_1$). Is this true; are there any references for this fact?

Actually, I would like to conclude that $f_1=0$; does this follow from the degeneration at $E_1$? I am more interested in the bounded case; so an answer to the first part of my question is sufficient for my purposes; yet are there any other conditions ensuring that $f_1=0$ whenever the spectral sequence degenerates at $E_1$?


Degenerating at $E_1$ is, as you describe, equivalent to the image of the map $f_1: D_1 \to E_1$ being contained in the image of $g_1^i$ for all $i$. Roughly, this is because the definition of the $d_r$-differential on a class $x$ is the equivalence class of any element $h_1 g_1^{1-r} f_1(x)$, where $h_1$ is the map $D_1 \to E_1$. If the spectral sequence degenerates, then $E_1 = E_r$ for all $r$ and so we don't need to worry about equivalence classes: then this asks that $g_1^{1-r} f_1(x)$ is in the kernel of $h_1$, hence the image of $g_1$. Then we repeat.

I do not know a place in the literature where this is explicitly stated.

It is not necessarily the case that $f_1 = 0$ if the spectral sequence degenerates. For example, if the map $h_1$ is zero so that $0 \to E_1 \xrightarrow{f_1} D_1 \to D_1 \to 0$ is an exact sequence, the spectral sequence still degenerates at the $E_1$-page. This is not uncommon behavior for degenerating spectral sequences in cohomology, rather than homology.


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