# Total complex homology exact sequence

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.

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 $y \in E^1_{p,q+1}$.
Note: your understanding of differentials in $\mbox{Tot}(E)$ is not right.
• I think you mean $d_{p,q}^v(b)=0$ on the third line numerator. – Jānis Lazovskis Oct 7 '15 at 23:54
• @WWK why $x,y$ are not in $E^{0}_{p-1,p+2}, E_{p,q+1}^{0}$, but in $E^{1}_{p-1,p+2}, E_{p,q+1}^{1}$? – user124697 Jun 4 '16 at 12:14
• 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 $d^{v}(a) \neq 0$; – user124697 Jun 4 '16 at 12:20