# Easy spectral sequence

This is a question in Weibel's Homological Algebra.

Suppose that a spectral sequence converging to $H_\ast$ has $E^2_{pq} = 0$ unless $p = 0,1$. Show that there are exact sequences $$0 \rightarrow E^2_{1,n-1} \rightarrow H_n \rightarrow E^2_{0,n} \rightarrow 0.$$

And what I have done: It is easy to see that $E^\infty_{pq} = E^2_{pq}$. It is also not hard to prove that each $H_n$ must have a filtration of the form $$0 = F_{-1}H_n \subseteq F_0H_n \subseteq F_1H_n = H_n.$$

It then follow that $F_0H_n/F_{-1}H_n = F_0H_n = E^2_{0n}$ and $F_1H_n/F_0H_n = H_n/E^2_{0n} = E^2_{1,n-1}$. But these give maps $H_n \rightarrow E^2_{1,n-1}$ and $E^2_{0n} \subseteq H_n$. I don't know how to get the maps in the other direction.

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Maybe I'm going completely crazy, but the fact that it converges to $H_*$ should mean that $H_n$ is isomorphic to the direct sum $\oplus_{p+q=n}E_{pq}^\infty$ which as you've pointed out is just $H_n\simeq E^2_{0,n}\oplus E^2_{1, n-1}$. I'm not sure if you get some "natural maps" coming from the spectral sequence, but there is definitely some exact sequence after choice of such an isomorphism just given by inclusion then projection. – Matt Nov 16 '12 at 20:17
Hi Matt! The way Weibel has defined "converging" is to have the isomorphisms $F_kH_n/F_{k-1}H_n = E_{k, n-k}$. So $H_n/E^2_{0n} = E^2_{1,n-1}$ as I wrote above, but I don't see why this splits into $H_n = E^2_{0n}\oplus E^2_{1,n-1}$. – sumo Nov 17 '12 at 2:52
You're right. I guess it doesn't have to split. Could this be a typo in the book? I guess for a cohomological spectral sequence the natural edge maps go the other way which might be a reason for the typo. All the places immediately before and after this problem say what you are saying. So unless for some reason it really does split in this case I'm confused with you. – Matt Nov 17 '12 at 23:50