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Everything here is taken from spectral sequences noted by Daniel Murfet (see here).

I provide relevant excerpts through screenshots:

enter image description here

enter image description here

The problem is I don't understand how $B_k(E^{pq}_r)$ and $Z_k(E^{pq})$ are defined. I understand that an inverse image is a pullback along a subobject. Weibel does something similar, but doesn't provide a construction.

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(the beginning part is not necessary since you say you already know how that is defined, but I'm just recalling this for people who might come across your question and not know this)

Suppose you have a map $A\to B$, and a subobject $C\to B$ of $B$.

Then if you take the following pullback : $$\require{AMScd}\begin{CD}A\times_BC @>>>A \\ @VVV @VVV \\ C @>>> B\end{CD}$$

you get that $A\times_B C\to A$ is a subobject (monomorphisms are stable under pullbacks), which you can call "the inverse image of $C$ in $A$" (the morphisms being implicit but of course an essential part of the definition)

Then in your situation, you do the same thing with $C= B_k, Z_k$ for the appropriate $k$. The definition is by induction, and I think there's a small mistake in the screenshot : it should be by induction on $k-r$. So if you want to define $B_k(E_r^{pq})$, you may assume that $B_k(E_{r+1}^{pq}$) is already defined, because $k-(r+1) < k-r$

Now here, you have a quotient map $E_r^{pq}\to E_r^{pq}/B_{r+1}(E_r^{pq})$, and a subobject of that which is $Z_{r+1}(E_r^{pq})/B_{r+1}(E_r^{pq})$ which is identified, via $\alpha_r^{pq}$, with $E_{r+1}^{pq}$.

So now this has subobjects $B_k(E_{r+1}^{pq}), Z_k(E_{r+1}^{pq})$ (which are well-defined, because you're doing induction on $k-r$), and so these are also subobjects of $E_r^{pq}/B_{r+1}(E_r^{pq})$, and so you can take their inverse image along the quotient map.

For $k-r = -1$ (the base case of the induction), you already have a definition of $B_{r+1}(E_r^{pq}), Z_{r+1}(E_r^{pq})$ as cycles and boundaries.

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  • $\begingroup$ I have a question, though: why $Z_{r + 1}(E^{pq}_r)/B_{r + 1}(E^{pq}_r)$ is a subobject of $E^{pq}_r/B_{r + 1}(E^{pq}_r)$? Is this related to the following question? math.stackexchange.com/questions/3656574/… $\endgroup$ – Jxt921 May 3 '20 at 13:17
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    $\begingroup$ Yes, I have just answered that other question : any quotient of a subobject is a subobject of the quotient $\endgroup$ – Maxime Ramzi May 3 '20 at 13:23
  • $\begingroup$ Dear Maxime, your answer to this question was great. Could you, by any chance, check if you have any idea regarding its follow-up: math.stackexchange.com/q/3657800/229776 $\endgroup$ – Jxt921 May 6 '20 at 1:52
  • $\begingroup$ There just really little information about this filtration in the literature. $\endgroup$ – Jxt921 May 6 '20 at 1:53

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