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?

• See remark 1.4 in these notes. Jun 24, 2015 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.
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.