Below screenshots are from these notes by Daniel Murfet.

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In an answer to this question, it has been clarfied that $B_k(E^{pq}_r)$ and $Z_k(E^{pq}_r)$ are constructed inductively as follows. Suppose that objects $B_k(E^{pq}_r)$ and $Z_k(E^{pq})$ are given along with monomorphisms $m^{pq}_{k,r+1}\colon B_k(E^{pq}_{r + 1}) \to Z_k(E^{pq}_{r + 1})$ and $\gamma^{pq}_{k,r+1}\colon Z_k(E^{pq}_{r + 1})\to E^{pq}_{r + 1}$ and an isomorphism $\beta^{pq}_{k,r + 1}\colon Z_k(E^{pq}_{r + 1})/B_k(E^{pq}_{r + 1})\to E^{pq}_{k + 1}$. Let $i^{pq}_r\colon B^{r + 1}(E^{pq}_r)\to E^{pq}_r$ be an image of $d^{p - r, q + r - 1}_r$ and $h^{pq}_r\colon E^{pq}_r\to E^{pq}_r/B_{r + 1}(E^{pq}_r)$ its cokernel. Then we can form pullbacks $B_k(E^{pq}_{r + 1}) \xleftarrow{s^{pq}_{k,r}} B_k(E^{pq}_r) \xrightarrow{t^{pq}_{k,r}} E^{pq}_r$ and $Z_k(E^{pq}_{r + 1}) \xleftarrow{b^{pq}_{k,r}} Z_k(E^{pq}_r) \xrightarrow{c^{pq}_{k,r}} E^{pq}_r$ of $B_k(E^{pq}_{r + 1}) \xrightarrow{\beta^{pq}_{r + 1}\circ \alpha^{pq}_r\circ \gamma^{pq}_{k,r + 1}\circ m^{pq}_{k,r+1}} E^{pq}_r/B_{r + 1}(E^{pq}_r) \xleftarrow{h^{pq}_r} E^{pq}_r$ and $Z_k(E^{pq}_{r + 1}) \xrightarrow{\beta^{pq}_{r + 1}\circ \alpha^{pq}_r\circ \gamma^{pq}_{k,r + 1}} E^{pq}_r/B_{r + 1}(E^{pq}_r) \xleftarrow{h^{pq}_r} E^{pq}_r$, respectively. Then, since $\beta^{pq}_{r + 1}\circ \alpha^{pq}_r\circ \gamma^{pq}_{k,r + 1}\circ m^{pq}_{k,r + 1}\circ s^{pq}_{k,r} = h^{pq}_r\circ t^{pq}_{k,r}$, there is, by the universal property of a pullback, a unique morphism $u^{pq}_{k,r+1}\colon B_k(E^{pq}_r)\to Z_k(E^{pq}_r)$ such that $b^{pq}_{k,r}\circ u^{pq}_{k,r+1} = m^{pq}_{k,r + 1}\circ s^{pq}_{k,r}$ and $c^{pq}_{k,r}\circ u^{pq}_{k,r+1} = t^{pq}_{k,r}$. Now, my questions:

$(1)$ Murfet then claims that said square is a pullback (in my notation, this means that $Z_k(E^{pq}_r) \xleftarrow{u^{pq}_{k,r+1}} B_k(E^{pq}_r) \xrightarrow{s^{pq}_{k,r}} B_k(E^{pq}_{r + 1})$ is a pullback of $Z_k(E^{pq}_r) \xrightarrow{b^{pq}_{k,r}} Z_k(E^{pq}_{r + 1}) \xleftarrow{m^{pq}_{k,r + 1}} B_k(E^{pq}_{r + 1})$. I can't show it. Indeed, suppose that $f\colon T\to Z_k(E^{pq}_r)$ and $g\colon T\to B_k(E^{pq}_{r + 1})$ are morphisms such that $b^{pq}_{k,r}\circ f = m^{pq}_{k,r+1}\circ g$. Then $h^{pq}_r\circ c^{pq}_{k,r} = \beta^{pq}_{k,r+1}\circ \alpha^{pq}_r\circ \gamma^{pq}_{k,r+1}\circ b^{pq}_{k,r}\circ f = \beta^{pq}_{k,r+1}\circ \alpha^{pq}_r\circ \gamma^{pq}_{k,r+1}\circ m^{pq}_{k,r+1}\circ g$. Seems like we're in the right direction, but the compositions belong to different pullbacks, so we can't invoke the universal property of a pullback to obtain a morphism $T \to B_k(E^{pq}_r)$.

$(2)$ Is this

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lemma correct?

$(3)$ Assuming said square is a pullback and said lemma holds, how does it follow from this that the induced epimorphism $Z_k(E^{pq}_r)/B_k(E^{pq}_r) \to Z_k(E^{pq}_{r + 1})/B_k(E^{pq}_{r + 1})$ is in fact isomorphism? If $\mathrm{coker}(u^{pq}_{k,r+1})\colon Z_k(E^{pq}_r)\to Z_k(E^{pq}_r)/B_k(E^{pq}_r)$ and $\mathrm{coker}(m^{pq}_{k,r+1})\colon Z_k(E^{pq}_{r + 1})\to Z_k(E^{pq}_{r + 1})/B_k(E^{pq}_{R + 1})$ are cokernels, $v^{pq}_{k,r+1} \colon Z_k(E^{pq}_r)/B_k(E^{pq}_r) \to Z_k(E^{pq}_{r + 1})/B_k(E^{pq}_{r + 1})$ is the unique morphism such that $v^{pq}_{k,r+1}\circ \mathrm{coker}(u^{pq}_{k,r+1}) = \mathrm{coker}(m^{pq}_{k,r+1})\circ b^{pq}_{k,r}$.

$(4)$ Why $B_k(E^{pq}_r) \subseteq B_{k + 1}(E^{pq}_r)$ and $Z_{k + 1}(E^{pq}_r) \subseteq Z_k(E^{pq}_r)$? I assume that we need to construct recursively a monomorphisms $B_k(E^{pq}_r) \to B_{k + 1}(E^{pq}_r)$ and $Z_{k + 1}(E^{pq}_r) \to Z_k(E^{pq}_r)$ assuming these exist for $r + 1$ using the fact that $B_k(E^{pq}_{r + 1}) \xleftarrow{s^{pq}_{k,r}} B_k(E^{pq}_r) \xrightarrow{t^{pq}_{k,r}} E^{pq}_{k,r}$ is a pullback of $B_k(E^{pq}_{r + 1}) \xrightarrow{\beta^{pq}_{r + 1}\circ \alpha^{pq}_r\circ \gamma^{pq}_{k,r + 1}\circ m^{pq}_{k,r+1}} E^{pq}_r/B_{r + 1}(E^{pq}_r) \xleftarrow{h^{pq}_r} E^{pq}_r$. Indeed, for monomorphisms $n^{pq}_{k,r+1}\colon B_k(E^{pq}_{r+1})\to B_{k+1}(E^{pq}_{r+1})$ are given, then we may wish to use the pullback property to obtain morphisms $n^{pq}_{k,r}\colon B_k(E^{pq}_r)\to B_{k+1}(E^{pq}_r)$ such that $n^{pq}_{k,r+1}\circ s^{pq}_{k,r} = s^{pq}_{k+1,r}\circ n^{pq}_{k,r}$. But for this we need to have $\gamma^{pq}_{k+1,r+1}\circ m^{pq}_{k+1,r+1}\circ n^{pq}_{k,r+1} = \gamma^{pq}_{k,r+1}\circ m^{pq}_{k,r+1}$. This should be a recursion trick, but I get stuck in details.


$\require{AMScd}$(as an aside : the way you describe pullbacks is not really optimal - you should use the CD command, by first recquiring AMScd, and then using \begin and \end{CD} for commutative diagrams - check out the code of my answer to see how this is done)

Also, before answering, note that all "local" questions like that on abelian categories can be answered easily be looking in the specific case of $R$-modules ($R$ a nonnecessarily commutative ring), by Freyd's embedding theorem. By local, I mean things involving finite limits, finite colimits, finitely many objects etc.

It's important to be careful about infinite (co)limits because those aren't "local" : they depend on the whole abelian category, and not only on the smallest abelian subcategory containing your objects (to which you can apply Freyd's embedding theorem).

But here, everything is finite, so Freyd's embedding theorem gives good heuristics, which are in fact more than heuristics. Of course it feels better not to use it, so I'll answer without using it - but know that it's an option.

Finally, the last thing I'll say before writing up actual answers is about equations and diagrams : it seems from what you've written that you use equations between morphisms to understand what's happening. But it's important to note that diagrammatic reasoning and writing has been developped precisely for these kinds of things, because it's easier to understand what's happening with diagrams than with equations. So I don't know if you're using equations because you're limited by MSE, or you also do it that way in your drafts, but you should really try to use diagrams because it makes a lot of things easier.

(1) Recall that the defining pullback for $B_k(E_r^{pq})$ is the following :

$$\begin{CD} B_k(E_r^{pq}) @>>> E_r^{pq} \\ @VVV @VVV \\ B_k(E_{r+1}^{pq}) @>>> E_r^{pq}/B_{r+1}(E_r^{pq})\end{CD}$$

where $E_r^{pq}/B_{r+1}(E_r^{pq})\supset Z_{r+1}(E_r^{pq})/B_{r+1}(E_r^{pq}) \cong E_{r+1}^{pq}$

But $B_k(E_r^{pq})\to E_r^{pq}$ factors through $Z_k(E_r^{pq})$ (note that since the latter is a subobject of $E_r^{pq}$, this is a property, not an additional structure, so it's not important to record the map $B_k\to Z_k$ that we get)

Moreover, both maps $B_k(E_{r+1}^{pq})\to E_{r+1}^{pq}$ and $Z_k(E_r^{pq})\to E_{r+1}^{pq}$ (the latter is obtained because $Z_k(E_r^{pq})\to E_r^{pq}/B_{r+1}(E_r^{pq})$ factors through $E_{r+1}^{pq}$) in this diagram actually factor through $Z_k(E_{r+1}^{pq})$ (again, that's a property because it's a subobject, so it's not necessary to record the map), and so we get a commutative diagram :

$$\begin{CD} B_k(E_r^{pq}) @>>> Z_k(E_r^{pq})\\ @VVV @VVV \\ B_k(E_{r+1}^{pq}) @>>> Z_k(E_{r+1}^{pq})\end{CD}$$

and the claim is that this is a pullback diagram. But if we have maps $T\to Z_k(E_r^{pq}), T\to B_k(E_{r+1}^{pq})$ that coincide in $Z_k(E^{pq}_{r+1})$, then we get (by composing with the obvious inclusions) maps $T\to E_r^{pq}$ and $T\to B_k(E_{r+1}^{pq})$ that coincide in $E_{r+1}^{pq}$, thus we get a map $T\to B_k(E_r^{pq})$ such that all the good things commute.

But those good things factor through $Z_k(E^{pq}_r)$ and since $Z_k(E^{pq}_r)\to E_r^{pq}$ is a monomorphism, it means that they commute with values in that as well.

You should check that this at least makes sense for you in a classical setting, e.g. with sets and actual subsets rather than monomorphisms, and then see why it actually only relies on monomorphisms. Essentially I'm saying :

If we have a monomorphism $A\to B$, and maps $C\to A, C\to D, D\to A$, such that the composites $C\to A\to B$ and $C\to D\to A \to B$ agree, then $C\to A$ and $C\to D\to A$ agree.

which is pretty much obvious.

(2) Well if the square can be completed to a pullback, then certainly $A'\to A$ is a monomorphism, in fact it's the inverse image of $B'\to B$, and $B'\to B$ is the kernel of $B\to B/B'$, so certainly the property sounds reasonable.

To make it precise, let's look at the following diagram :

$$\begin{CD} A' @>>> A \\ & @VVV \\ B'@>>> B \\ @VVV @VVV \\ 0 @>>> B/B' \end{CD}$$

The bottom square is a pullback. So if you take a pullback in the top square, as follows :

$$\begin{CD} A\times_B B' @>>> A \\ @VVV @VVV \\ B'@>>> B \\ @VVV @VVV \\ 0 @>>> B/B' \end{CD}$$

by the pullback lemma, it follows that

$$\begin{CD} A\times_B B' @>>> A \\ @VVV @VVV \\ 0 @>>> B/B' \end{CD}$$

is also a pullback, i.e. $A\times_B B'\to A$ is the kernel of $A\to B\to B/B'$. The lemma easily follows, as pullbacks are entirely determined up to isomorphism.

(3) From the lemma and the pullback, it follows that $B_k(E_r^{pq})\to Z_k(E^{pq}_r)$ is the kernel of $Z_k(E_r^{pq})\to Z_k(E_{r+1}^{pq}) \to Z_k(E_{r+1}^{pq})/B_k(E_{r+1}^{pq})$, that is, the natural map $Z_k(E_r^{pq})/B_k(E^{pq}_r) \to Z_k(E_{r+1}^{pq})/B_k(E_{r+1}^{pq})$ is a monomorphism.

Since you already know it's an epimorphism, and abelian categories are balanced (an epimorphism which is a monomorphism is an isomorphism) it follows that it's an isomorphism.

(4) It indeed follows by induction. I'll only do it for $B$, for $Z$ it's the same story.

Once again, it's good to use induction on $k-r$ rather than $k$. Indeed, note that for $k-r = 0$ (the base case again), $B_r(E_r^{pq}) = 0$, so it's always included in $B_{r+2}(E_r^{pq})$.

Recall that $B_k$ is defined via the following pullback :

$$\begin{CD} B_k(E_r^{pq}) @>>> E_r^{pq} \\ @VVV @VVV \\ B_k(E_{r+1}^{pq}) @>>> E_r^{pq}/B_{r+1}(E_r^{pq})\end{CD}$$.

Independently of $k$, this is defined by the pullback along this map : $E_r^{pq}\to E_r^{pq}/B_{r+1}(E_r^{pq})$.

Then you can prove the following (easy) lemma, valid in any category with pullbacks :

Let $f:A\to B$ be any map; then pullback along $f$ induces a map $Sub(B) \to Sub(A)$. This map preserves the order on both sides.

where $Sub(X)$ denotes the poset of subobjects of $X$. In fact, this lemma is a decategorication of the fact that pullback along $f$ induces a functor between the categories of subobjects : a map $D\to C$ of subobjects of $B$ induces a map $A\times_B C\to A\times_B C$ of subobjects of $A$.

In our situation, we assumed (by induction on $k-r$) that, as subobjects of $E_r^{pq}/B_{r+1}(E_r^{pq})$ (in fact, as objects of $E_{r+1}^{pq}\subset E_r^{pq}/B_{r+1}(E_r^{pq})$, but of course it changes nothing), we have $B_k(E_{r+1}^{pq}) \leq B_{k+1}(E_{r+1}^{pq})$, and so the same holds of their pullbacks: $B_k(E_r^{pq}) \leq B_{k+1}(E_r^{pq})$.

This makes us go one step up in $k-r$, so it allows us to get our induction going.

  • $\begingroup$ Nice answer, Maxime, thank you. $\endgroup$
    – Jxt921
    May 6 '20 at 20:29
  • $\begingroup$ Dear Maxime, I've found a missing detail I can't myself fill. When one proves that $B_k(E^{pq}_r) \subseteq B_{k + 1}(E^{pq}_r)$ and $Z_{k + 1}(E^{pq}_r) \subseteq Z_k(E^{pq}_r)$ one needs a base step: that is, that $B_{r + 1}(E^{pq}_r) \subseteq B_{r + 2}(E^{pq}_r)$ and $Z_{r + 2}(E^{pq}_r) \subseteq Z_{r + 1}(E^{pq}_r)$. But I don't see a way to show this, there is just no morphisms between $E^{pq}_r$ and $E^{pq}_{r + 1}$. $\endgroup$
    – Jxt921
    May 7 '20 at 21:50
  • $\begingroup$ You're right, the way I wrote my base case is stupid, I meant $k-r = 0$, not $-1$, in which case $B_r(E_r^{pq}) = 0$ so it is obviously included in $B_{r+1}(E_r^{pq})$. Similarly, $Z_r(E_r^{pq}) = E_r^{pq}$ $\endgroup$ May 7 '20 at 21:54
  • $\begingroup$ I thought about it, but it's not that easy. To apply the construction starting from $k - r = 0$ we need to prove that $B_{r + 1}(E^{pq}_r)$ is obtained as a pullback from $B_{r + 1}(E^{pq}_{r + 1})$. But I don't know any maps from $B_{r + 1}(E^{pq}_r) \to B_{r + 1}(E^{pq}_{r + 1}) = E^{pq}_{r + 1}$. $B_{r+1}(E^{pq}_r)$ is a subobject of $E^{pq}_r$, and there are still no maps from $E^{pq}_r$ to $E^{pq}_{r + 1}$. $\endgroup$
    – Jxt921
    May 7 '20 at 22:06
  • $\begingroup$ Why do you write $B_{r+1}(E^{pq}_{r+1}) = E_{r+1}^{pq}$ ? It is defined to be $0$ $\endgroup$ May 7 '20 at 22:15

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