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I'm been trying to do this problem (Problem 5.1.1) from Weibel's Introduction to Homological Algebra but I can't really see how to finish it. The statement of the problem is summarized as follows:

Suppose that we have a double complex $E$ consists of only two columns $p$ and $p-1$. Let $T_n = Tot(E)$ be the total complex then show that there is an exact sequence \begin{equation} 0 \rightarrow E^2_{p - 1,q + 1} \rightarrow H_{p + q} (T) \rightarrow E^2_{p,q} \rightarrow 0 \end{equation}
So what I've tried so far is attempting to calculate each of the object in the sequence and show that \begin{equation} E^2_{p,q} \cong H_{p+q}(T)/E^2_{p-1,q+1} \end{equation} or something. So I did the calculation and I got \begin{equation} E^2_{p-1,q+1} \cong \mbox{ker} (d^v_{p-1, q+1})/\mbox{im}(d^h_{p, q+1}) \\ H_{p + q}(T) \cong \frac{(\mbox{ker}(d^h_{p, q}) \cap \mbox{ker}(d^v_{p,q})) \oplus \mbox{ker} (d^v_{p-1,q+1})}{\mbox{im} (d^v_{p,q+1}) \oplus (\mbox{im} (d^h_{p,q+1}) + \mbox{im} (d^v_{p-1,q+2}))} \\ E^2_{p,q} \cong \mbox{ker} ({d^h_{p,q}}_{\star}) \end{equation} Where ${d^h_{p,q}}_{\star}: E^1_{p,q} \rightarrow E^1_{p-1,q}$ is the induced horizontal differential map after taking the vertical homology. However, \begin{equation} H_{p+q}(T)/E^2_{p-1,q+1} \cong \frac{(\mbox{ker}(d^h_{p, q}) \cap \mbox{ker}(d^v_{p,q}))}{\mbox{im} (d^v_{p,q+1})} \stackrel{?}{\cong} \mbox{ker} ({d^h_{p,q}}_{\star}) \end{equation} and I can't convince myself that the second equality is true. Am I on the right track (or close to)? If someone could point out the mistake or guide me to the right direction that would be great. Thank you.

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The thing is in the first page $E^1_{*,*}$, the vertical map is actually zero. Now \begin{equation} E^2_{p-1,q+1} \cong \mbox{ker} (d^h_{p-1, q+1})/\mbox{im}(d^h_{p, q+1}) \cong E^1_{p-1,q+1}/\mbox{im}(d^h_{p, q+1}) \\ E^2_{p,q} \cong \mbox{ker} ({d^h_{p,q}})\\ H_{p + q}(T) \cong \frac{\{(a,b) | d^v_{p-1,q+1}(a)+d^h_{p,q}(b)=0;d^v_{p,q}(b)=0 \}}{\{(a,b) | a=d^v_{p-1,q+2}(x)+d^h_{p,q+1}(y); b=d^v_{p,q+1}(y);\}}=\frac{(E^1_{p-1,q+1},\mbox{ker} ({d^h_{p,q}}))}{(\mbox{im} (d^h_{p,q+1}),0)} \\ \end{equation} Here $x \in E^1_{p-1,q+2}$ and $y \in E^1_{p,q+1}$.

Hence you get the desired exact sequence.

Note: your understanding of differentials in $\mbox{Tot}(E)$ is not right.

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  • $\begingroup$ I think you mean $d_{p,q}^v(b)=0$ on the third line numerator. $\endgroup$ – Jānis Lazovskis Oct 7 '15 at 23:54
  • $\begingroup$ @jlv Yes, you are right. $\endgroup$ – WWK Oct 8 '15 at 1:18
  • $\begingroup$ @WWK why $x,y$ are not in $E^{0}_{p-1,p+2}, E_{p,q+1}^{0}$, but in $E^{1}_{p-1,p+2}, E_{p,q+1}^{1}$? $\endgroup$ – user124697 Jun 4 '16 at 12:14
  • $\begingroup$ I'm curious that why there is no case such that $d^{v}_{p-1,q+1}(a) + d^{h}_{p,q}(b)=0$ but not $d^{v}(a) \neq 0$; $\endgroup$ – user124697 Jun 4 '16 at 12:20

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