# How does one get the short exact sequence in a two-column spectral sequence?

In a two-column double complex, one gets from the associated spectral sequence short exact sequences $0\to E_2^{1,n-1}\to H^n\to E_2^{0,n}\to 0$, where $H^n$ is the cohomology of the total complex, but I have never seen the construction of this sequence. Any text I've seen merely states it as a fact, or leaves it as an exercise which I have had no luck trying to solve. Can anyone give a construction or good reference?

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See remark 1.4 in these notes. – Bruno Stonek Jun 24 '15 at 13:22

This follows precisely from the very definition of convergence of the spectral sequence, once one has identified the $\infty$-term. It is done with some details in McLeary's User Guide---which is, in my opinion, a very good reference for both the technicalities and the pragmatics of dealing with spectral sequences.

Now, if you are starting with a two column double complex (as opposed to starting with an arbitrary double complex whose spectral sequence has two contiguous columns), you can get the short exact sequences very much 'by hand'.

Indeed, suppose your double complex is $T^{\bullet,\bullet}=(T^{p,q})_{p,q\geq0}$ and that $T^{p,q}\neq0$ only if $p\in\{0,1\}$. If we define complexes $X^\bullet$ and $Y^\bullet$ with $X^q=T^{0,q}$ and $Y^q=T^{1,q}$, with differentials coming from the vertical differential $d$ of $T^{\bullet,\bullet}$, then the horizontal differential of $T^{\bullet,\bullet}$ can be seen as a map of complexes $\delta:X^\bullet\to Y^{\bullet}$.

Now, in the spectral sequence induced by the filtration by columns we clearly have $E_1^{0,q}=H^q(X^\bullet)$, $E_1^{1,q}=H^q(Y^\bullet)$ and the differential on the $E_1$ page is induced by the horizontal differential in $T^{\bullet,\bullet}$. In other words, the $E_1$ page is more or less the same thing as the map $H(\delta):H(X^\bullet)\to H(Y^\bullet)$. It follows that we have short exact sequences $$0\to E_2^{0,q}\to H^q(X^\bullet)\xrightarrow{H^q(\delta)} H^q(Y^\bullet)\to E_2^{1,q}\to 0,$$ and the spectral sequence dies at the second act for degree reasons.

On the other hand, there is a short exact sequence of complexes $$0\to Y[-1]^\bullet\to\mathrm{Tot}\;T^{\bullet,\bullet}\to X^\bullet\to 0,$$ from which we get a long exact sequence $$\cdots\to H^{q-1}(Y^\bullet)\to H^q(\mathrm{Tot}\; T^{\bullet,\bullet})\to H^q(X)\to H^q(Y)\to\cdots,$$ in which you can compute directly that the map $H^q(X)\to H^q(Y)$ is precisely $H^q(\delta)$. Since the first four-term exact sequence identifies for us the kernel and the cokernel of $H^q(\delta)$, exactness of the second exact sequence provides the short exact sequence $$0\to E_2^{1,q-1}\to H^q(\mathrm{Tot}\; T^{\bullet,\bullet})\to E_2^{0,q}\to 0$$ that you wanted.

(It is an extremely instructive exercise to try to see what can one do in this spirit for a three-column double complex, and fighting with this is a great prelude to an actual exposition to spectral sequences...)

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Thanks, that really helps! – Ketil Tveiten Jan 21 '10 at 10:46