Why does spectral sequence $E^\infty$ need AB4*

Question

So in Weibel, he states

Warning: In an unbounded spectral sequence, we will tacitly assume that $B^{\infty}$, $Z^{\infty}$, and $E^{\infty}$ exist! The reader who is willing to only work in the category of modules may ignore this difficulty. The queasy reader should assume that the abelian category A satisfies axioms {AB4) and (AB4*).

But, I don't understand the requirement of AB4* at all. I know AB3 and AB3* are needed for sure for the limits and colimits and I believe that with AB5, we can get that they are all subobjects with the correct containment relations. Then, why does Weibel insist on AB4*?

Answer 1

$\require{AMScd}\newcommand{\ab}{{\mathsf{AB4}^\ast}}\newcommand{\abb}{{\mathsf{AB5}}}$There is some subtlety here, in large part due to your comment from four years ago about unions of subobjects. We can always talk about unions of subobjects, assuming cocompleteness, but they are in general quotients of the colimits and not necessarily equal to them. They lack the same universal property of a colimit however they crucially still have the universal property of a union i.e. largest subobject containing the other subobjects. For convenience's sake $\ab$ really is necessary for many essential first properties, to answer your original question. $\ab$ is the axiom that products of epimorphisms are epimorphisms, and $\abb$ is the axiom that filtered colimits are exact (equivalently, the colimit of a monomorphism of filtered diagrams is a monomorphism of the filtered colimits). Actually, it is perhaps the case we never really need the full strength of $\abb$ and rather only require all unions to equal the relevant colimits; I don't know if that's equivalent to $\abb$ or not. Being very generous, it may be the case Weibel was right all along and only $\mathsf{AB4}$ is needed in the stead of $\abb$ but I am not sure if it proves unions = colimits. Does anyone know about that?

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I claim the following:

• The spectral sequence of a bounded below, exhaustive filtration exists and converges to the homology of the complex, always; this requires nothing on the Abelian category. You just have to somewhat painfully re-argue everything in a general manner, doing the necessary diagram chases and universal property yoga to construct what we need to construct. From now on, I will assume the Abelian category is bicomplete $(\mathsf{AB3},\mathsf{AB3}^\ast)$ in order to talk about the unbounded cases at all. If you have a bounded below filtration or more generally a bounded below spectral sequence then the mapping theorems discussed below all work automatically, with no need for any $\mathsf{AB}n$ axioms, because the limits/colimits involved are all finite.
• The mapping theorem for spectral sequences sort of does and sort of doesn't require additional axioms on the Abelian category:

Let $E,E'$ be two spectral sequences of homological type starting with pages $a,a'$. Let $f:E\to E'$ be a morphism of spectral sequences. If $f^r$ is an isomorphism of pages for some $r\ge\max(a,a')$ then as per Weibel's (presumably canonical) construction of the $\infty$-pages we find the induced map $f^\infty:E^\infty\to(E')^\infty$ is an isomorphism.

Suppose $E,E'$ both converge (completely and Hausdorffly) to the filtered graded objects $H_\ast,H'_\ast$ and suppose $h:H_\ast\to H'_\ast$ is compatible with $f$. If $f^\infty$ is an isomorphism (usually requiring $\ab$ to hold) then $h$ is an isomorphism if the Abelian category satisfies $\abb$. Alternatively, if there also exists $h_0:H'_\ast\to H_\ast$ and $f_0:E'\to E$ compatible with one another such that $f^\infty,f_0^\infty$ are mutually inverse then $h,h_0$ are mutually inverse isomorphisms (with no need to cry for $\abb$).

In the instance that $E,E'$ start on different pages and/or $f$ is not an isomorphism on the initial page, some caution is required. We need to pass to a higher order page, but Weibel defines $E^\infty$ in terms of the starting page only. Can it be equivalently defined by a higher page? Yes, but you need to be careful. In the original version of this post I thought it required axiom $\ab$ but with thanks to Thorgott I now realise that that is not required. It boils down to the canonical map $\left(\bigcap_{s\ge r}Z^s\right)/B^r\to\bigcap_{s\ge r}(Z^s/B^r)$ being an isomorphism. I originally proved this by finding lifts one at a time, one for each $Z^s/B^r$, and forming a limiting element (an argument which did need $\ab$) but we can make do without that. It's clearly a monomorphism, so we need to show it is an epimorphism.

Simply take $x\in_m\bigcap_{s\ge r}Z^s/B^r$, include it into $E^a/B^r$; there is some member class $y\in_m E^a$ which projects to $x$. In particular, the projection of $y$ fits into all $Z^s/B^r$ which implies $y$ lives in the union $Z^s+B^r=Z^s$ for all $s$; thus $y$ factors through the subobject intersection $\bigcap_{s\ge r}Z^s$ so $x$ is in the image of $\left(\bigcap_{s\ge r}Z^s\right)/B^r\to\bigcap_{s\ge r}Z^s/B^r$. If one actually checked the details of this argument you'd realise it's essential that we're working with subobjects here. To expect limits to commute with quotients like this in general I expect one needs $\ab$ or something like it.

Why do I care about this map? Well, it allows you to reduce to the case where both sequences start on the same page and $f$ is an isomorphism of the starting pages. If $E^{r\ge a}$ is a spectral sequencec then for fixed $r_0\ge a$ there is of course a spectral subsequence $E_0:=E^{r\ge r_0}$ with all the same pages and differentials in large degree... but to say their infinity pages are the same as per Weibel's construction, realising $Z^r(E_0)\cong Z^r(E)/B^{r_0}$ and $B^r(E_0)\cong B^r(E)/B^{r_0}$ as subobjects of $E_0^{r_0}$ (the starting page!) you will need $\left(\bigcap_{r\ge r_0}Z^r\right)/B^{r_0}\overset{\cong}{\to}\bigcap_{r\ge r_0}(Z^r/B^{r_0})$.

The reason for the fuss about $\abb$ is explained well here. We can get around this issue if we have some kind of ambient maps $H\to H'\to H$ compatible with $E^\infty\cong(E')^\infty\cong E^\infty$. In fact, the way we prove $f^\infty$ is an isomorphism in the mapping theorem uses the fact there is an ambient map $E^a\cong(E')^a$ to avoid the otherwise thorny issue of whether or not $f$ induces an isomorphism $B^\infty\to(B')^\infty$.

Here, $A,A'$ would be the colimits $\varinjlim_t F_tH/F_sH$ and $C,C'$ would be the quotients, the images in $H,H'$, $\bigcup_t F_tH/F_sH\cong H/F_sH$ by exhaustivity. If $\abb$ holds then $A=C,A'=C'$ with no need to worry. Maybe $\mathsf{AB4}$ also gives $A=C$, it's plausible, but I don't see it at the moment.

• The complete convergence theorem: we need the Abelian category to have enough injectives, and to satisfy $\ab$, in order to justify certain Mittag-Leffler arguments, certain derived inverse limit arguments; in order to justify Weibel's proof of Boardman's criterion, really. As far as I can tell, $\abb$ is not required. That said, a major reason why you care about complete convergence is the implications it has when connected with the mapping theorem which does seem to require $\abb$. Although we can have a good notion of $\varprojlim^1$ for towers in the absence of enough injectives, to have a notion of $\varprojlim^1$ for double towers seems to require the general injective resolution machinery and to make a certain calculation, Weibel requires an exact sequence which seemingly only comes from a Grothendieck spectral sequence which itself involves injective double-complex Cartan-Eilenberg resolutions; I doubt we can avoid needing that.