When we define spectral sequnces (as Weibel's book) for example in the abelian category $R$-mod, they are a collection of objects $E_{pq}^r$ for $p,q$ and $r\geq a$ integers with a collection of morphisms $d_{pq}^r:E_{pq}\rightarrow E_{p-r,q+r-1}$ which are differentails in the sense that $d^r\circ d^r=0$ and such that there exist isomorphisms $\alpha_{pq}^r:E_{pq}^{r+1}\cong Z_{pq}^r/B_{pq}^r$ where $Z_{pq}$ is the kernel of $d_{pq}^r$ and $B_{pq}^r$ is the image of $d_{p+r,q-r+1}^r$, that is, $E_{pq}^{r+1}$ is the homology of $E^r$ at the node $E_{pq}^r$.

Then a morphism $f:E\rightarrow E'$ between spectral sequences is collection of morphisms $f_{pq}^r:E_{pq}^r\rightarrow {E'}_{pq}^r$ defined for all $p,q$ and large $r$ satisfying:

(1) $fd^r=d^rf$ (this means $f_{p-r,q+r-1}^r\circ d_{pq}^r=d_{pq}^r\circ f_{pq}^r$);

(2)$E_{pq}^{r+1}\xrightarrow{\alpha}Z_{pq}^r/B_{pq}^r\xrightarrow{\bar{f}_{pq}^r} {Z'}_{pq}^r/{B'}_{pq}^r= E_{pq}^{r+1}\xrightarrow{f_{pq}^{r+1}} {E'}_{pq}^{r+1}\xrightarrow{\alpha} {Z'}_{pq}^{r}/{B'}_{pq}^{r}$.

from this definition it is clear that if we have two morphism $f,g:E\rightarrow E'$ such that there exists $r_0$ such that $f_{pq}^{r_0}=g_{pq}^{r_0}$ for all $p,q$, then we have $f_{pq}^r=f_{pq}^r$ for all $r\geq r_0$ for all $p,q$.

Now, it seems to me that in the definition of spectral sequences there is not any kind of relation between the maps $d_{pq}^{r+1}$ and $d_{pq}^r$. This suggests that the following assertion may not be true: Suppose that for some $r_0$ we have maps $f_{pq}^{r_0}:E_{pq}^{r_0}\rightarrow {E'}_{pq}^{r_0}$ satisfying $d^{r_0}f=fd^{r_0}$ at each point $p,q$. Then there exists a morphism $F:E\rightarrow E'$ such that $F_{pq}^{r_0}=f_{pq}^{r_0}$ for all $p,q$.

Does anyone know a counterexample for this?, or if I am lucky, does anyone know a proof of this?.

Many thanks,



1 Answer 1


There is absolutely no reason why the map induced by your $f^{r_0}$ on $E^{r_0+1}$ will commute with the differential $d^{r_0+1}$.

Contruct an exampe as follows: Let $E_{p,q}^r=0$ except when $r=0$, $q=0$ and $p\in\{0,1\}$, and let $E_{0,0}^0=E_{1,0}^0=\mathbb Z$. To turn $E$ into a spectral sequence, we have to define the differentials: in view of the many zeroes we have, the only differential we have to give is $d_{0,0}^1:E_{0,0}^1\to E_{1,0}^1$, because all the others must be zero. I pick $d_{0,0}^1$ to be the identity of $\mathbb Z$.

Now every endomorphism of the bigraded $E^0$ commutes with the differential $d^0$. But only a few of them induce an endomorphism on $E^1$ which commutes with the differential $d^1$.

  • $\begingroup$ Many thanks Mariano! $\endgroup$
    – Diego
    Feb 7, 2013 at 20:39

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