2
$\begingroup$

I'm trying to prove the following assertion in Weibel's Homological Algebra page 125, 5.2.8

Given a homology spectral sequence, we see that each $E^{r+1}_{pq}$ is a subquotient of the previous term $E^r_{pq}$. By induction on $r$, we see that there is a nested family of subobjects of $E^a_{pq}$: $$0 =B^a_{pq}\subseteq \cdots \subseteq B^r_{pq} \subseteq B^{r+1}_{pq} \subseteq \cdots \subseteq Z^{r+1}_{pq}\subseteq Z^r_{pq} \subseteq \cdots \subseteq Z^a_{pq} =E^a_{pq}$$ such that $E^r_{pq}\cong Z^r_{pq}/B^r_{pq}$

If we are working in the category of $R$-modules, then this proposition is trivially seen to be a consequence of this diagram

diagram

Because if $B^{r+1}_{pq}$ is contained in $E^r_{pq}$, then it has to contain $0$ which represents $B^{r}_{pq}$. Clearly $B^r_{pq}\subseteq Z^r_{pq}$ and $Z^r_{pq}\subseteq Z^{r-1}_{pq}$ o/w we wouldn't have $Z^{r+1}_{pq}\hookrightarrow E^r_{pq}$.

But I can't generalise this reasoning to work in any abelian category. The diagram still holds, but I really don't know how to prove (using the sub-objects) the claim Weibel does. (Clearly I don't want to use the embedding in $R$-Mod)

Can someone provides me some hints on what is the general argument for abelian cats?

ADDENDUM actually I fear that even in the $R$-Mod case, this doesn't make much sense. In fact, if $B^{r+1}_{pq}\hookrightarrow Z^r_{pq}/B^r_{pq}$, then how come $B^r_{pq}\hookrightarrow B^{r+1}_{pq}$? in fact it'd be clearly a submodule of the kernel of the inclusion, but the inclusion is injective and hence it'd be trivial but in general $B^r_{pq}\neq 0$. So what does Weibel mean here?

$\endgroup$
1
  • $\begingroup$ Perhaps it would be helpful to look at §9.4 here, particularly 9.4.12(v). $\endgroup$
    – Zhen Lin
    Dec 9, 2015 at 0:11

1 Answer 1

4
$\begingroup$

The object $B^{r+1}_{pq}$ is not defined to be the image of $d_{p_1q_1}$. Rather, it is the preimage of $\operatorname{im}(d_{p_1q_1})\subseteq E^r_{pq}$ under the quotient map $\pi:Z^r_{pq}\to E^r_{pq}$. So it is not a subobject of $E^r_{pq}$ but of $Z^r_{pq}$ (and hence, by induction, of $E^a_{pq}$). Similarly, $Z^{r+1}_{pq}$ is the preimage of $\ker(d_{pq})$ under $\pi$. The thing you have to check is then that $Z^{r+1}_{pq}/B^{r+1}_{pq}\cong \ker(d_{pq})/\operatorname{im}(d_{p_1q_1})$. This follows from a general fact about pulling back subobjects in abelian categories that should not be too hard to prove: if you have an object $X$ with subobjects $U\hookrightarrow V\hookrightarrow X$, and an epimorphism $\pi:Y\to X$, then $\pi^{-1}(V)/\pi^{-1}(U)\cong V/U$.

$\endgroup$
4
  • $\begingroup$ ok I'll work out what you wrote! it seems what I need, but now is kind of late here in Europe, so I'll check tomorrow. Thanks a lot in advance though! $\endgroup$
    – Luigi M
    Dec 8, 2015 at 23:12
  • $\begingroup$ are there some references to prove this last result? because I'm having an hard time figure it out.. $\endgroup$
    – Luigi M
    Dec 9, 2015 at 16:22
  • $\begingroup$ My main problem is to show that the map (induced by pullback) $\pi^{-1}(U)\to U$ is epic known that $\pi$ is epic. Morally it's obvious, but I need a way to formalise it $\endgroup$
    – Luigi M
    Dec 9, 2015 at 18:06
  • $\begingroup$ Let $W=\operatorname{im}(\pi^{-1}(U)\to U)$, and consider the maps $Y\to X\to X/U$ and $Y\to X\to X/W$. These are both epic with kernel $\pi^{-1}(U)$, and so there is an isomorphism $X/U\cong X/W$ that commutes with the maps from $Y$. Epicness of $Y\to X$ then implies that this isomorphism also commutes with the maps from $X$, which implies $U=W$. $\endgroup$ Dec 9, 2015 at 18:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .