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$\require{AMScd}$The following is taken from these notes by Daniel Murfet.

Let $ \cdots \subseteq F^{p + 1}(C) \subseteq F^p(C) \subseteq F^{p - 1}(C) \subseteq \cdots$ be a filtration of a complex $C$ in an abelian category.

enter image description here

There is either a mistake or I don't understand something. I think that $\ddot{A^{pq}_r} \subseteq \ddot{A^{pq}_{r + 1}}$. Indeed, $A^{pq}_r$ defined by the following pullback

$$\begin{CD} A^{pq}_r @>>> F^p(C^{p + q}) \\ @VVV @VVV \\ F^{p + r}(C^{p + q + 1}) @>>> F^p(C^{p + q + 1}) \end{CD}$$

where the bottom morphism is a subobject inclusion and the left morphism is a differential of $F^p(C)$. Concretely, $A^{pq}_r$ is a pullback of $d^{p,p+q}$ along the subobject inclusion $F^{p + r}(C^{p + q + 1}) \subseteq F^p(C^{p + q + 1})$. Then $\ddot{A^{pq}_r}$ is the image of the composition in the following diagram

$$\begin{CD} A^{p - r + 1, q + r - 2}_{r - 1} @>>> F^{p - r + 1}(C^{p + q - 1}) \\ @VVV @VVV \\ F^p(C^{p + q}) @>>> F^{p - r + 1}(C^{p + q}) \end{CD}$$

and $\ddot{A^{pq}_{r + 1}}$ is the image of the composition in the following diagram

$$\begin{CD} A^{p - r, q + r - 1}_r @>>> F^{p - r}(C^{p + q - 1}) \\ @VVV @VVV \\ F^p(C^{p + q}) @>>> F^{p - r}(C^{p + q}) \end{CD}$$

To have a map from from one image to another, we need a map between their domains and codomains. But the pullback universal property only gives a morphism from $A^{p - r + 1, q + r - 2}$ to $A^{p - r, q + r - 1}$. Similarly, for the following screenshot

enter image description here

I only see how to construct a map from $A^{p + r, q - r + 1} \to A^{pq}_r$, for similar reasons.

So, my question is: is there is a mistake? If yes, can the proof be salvaged? If not, what am I missing?

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    $\begingroup$ What follows "I think that" is unclear because of a probable typo (more generally, your "dot dot" on top of $A$ is not always well written). Are you saying you think there is a mistake in the order of inclusion of the "A dot dot"'s ? $\endgroup$ May 10 '20 at 7:57
  • $\begingroup$ @MaximeRamzi Exactly. In a similar vein, I think a domain and a codomain of a morphism $A^{pq}_r \to A^{p + r, q - r + 1}_r$ Murfet defines are reversed. $\endgroup$
    – Jxt921
    May 10 '20 at 8:07
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    $\begingroup$ I haven't read all of it yet, but have you tried forgetting you're in a general abelian category, pretending you're in $R-\mathbf{Mod}$ and reading all images, intersections, pre-images etc. as the usual set-theoretic constructions and seeing what happens ? If you do this, you should either be able to confirm your doubts, or see where you're mistaken, and what's actually happening. Then you can translate back to general abelian categories $\endgroup$ May 10 '20 at 8:20
  • $\begingroup$ @MaximeRamzi Yeah, I've tried and I think this confirms my suspicion, though I may be making a similar mistake in the category of modules. Besides, does this mean the proof is completely wrong, or it can be salvaged? I don't know. $\endgroup$
    – Jxt921
    May 10 '20 at 18:14
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We have $$\ddot{A^{p,q}_r}=\partial \newcommand\of[1]{\left({#1}\right)} \of{ F^{p-r+1}C^{p+q-1}\cap \partial^{-1} \of{ F^{p+1}C^{p+q} } }. $$ Note that $\partial^{-1}(F^{p+1}C^{p+q})$ is independent of $r$, and as $r$ increases, the filtration index decreases, and therefore the groups get larger, so you appear to be correct that the order of inclusions should be $\ddot{A}^{p,q}_r\subseteq \ddot{A}^{p,q}_{r+1}$.

This doesn't really affect the proof in any way (at least the visible part), because the excerpted discussion doesn't use the ordering at all. However, I will say that this ordering should be the correct one, since we want $Z^{p,q}_{r+1}\subseteq Z^{p,q}_r$ and $B^{p,q}_{r+1}\supseteq B^{p,q}_r$ so that $E_{r+1}^{p,q}$ is a subquotient of $E_r^{p,q}$.

$r$-cocycles should get smaller and $r$-coboundaries should get larger so that we have a sensible notion of convergence.

I imagine this was a typo, it's really easy to get indices and ordering confused here.

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  • $\begingroup$ Thanks for you answer. But what do you think about morphism $A^{pq}_r \to A^{p + r, q - r + 1}$? Does the order of the domain and the codomain reversed too? But one needs this morphism to construct a morphism $d^{pq}_r\colon E^{pq}_r \to E^{p + r, q - r + 1}_r$ (a differential of a spectral sequence) $\endgroup$
    – Jxt921
    May 10 '20 at 19:47
  • $\begingroup$ I hope this is not too much to ask, but for that, could you, please, see Lemma 2 in the link for that? $\endgroup$
    – Jxt921
    May 10 '20 at 19:49

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