Bott and Tu Proposition 12.1 Question about the proof of Bott and Tu's Proposition 12.1: Given any double complex $K$, if $H_\delta H_d(K)$ has entries only in one row, then $H_\delta H_d$ is isomorphic to $H_D$.
Remark: $\delta$ is the horizontal, and $d$ the vertical, delta map.

I do not understand the highlighted statement. From what I understand, if $H^{p+2,q-1}_\delta H_d = 0$ this means the following $\ker(\delta: H^{p+2,q-1} \to H^{p+3, q-1}) = \text{Im}(\delta:H^{p+1,q-1} \to H^{p+2,q-1})$. However, it is clear that $\phi_1$ is not $d$ closed, so it does not represent a $H_d$ cohomology class, so it surely cannot represent a $H_\delta H_d$ cohomology class, right? Any help would be appreciated.
 A: If I understand your question (and I am not screwing up!): 
(Up to sign errors... ) $D''\delta \phi_1 = \delta D''\phi_1 = \delta (-\delta \phi ) = 0$. Hence $\delta \phi_1$ is $D''$-closed.
Addendum:
Therefore $\delta\phi_1$ is both $d$ and (visibly) $\delta$ closed, and thus represents a class in $H_\delta H_d$, which is, by assumption, $0$ in the appropriate bi-degree. Hence there exists $\hat \phi_1$, such that $d \hat \phi_1 =0$, and $\phi_2$, such that 
$$ \delta \phi_1 = \delta \hat \phi_1 \pm d \phi_2,$$
or $$\delta (\phi_1 -\hat\phi_1) =  \pm d\phi_2.$$
Since $\hat\phi_1$ is $d$-closed, we can replace $\phi_1$ with $\phi_1 -\hat \phi_1 $ to keep $\delta \phi + D'' \phi_1 = 0$, and also to obtain the desired $ \delta \phi_1= \pm d\phi_2$. 
A: Ok, thanks to peter a g and some of my colleagues, I now understand this problem, and I am going to post a full solution, starting from scratch. I believe that the proof in Bott and Tu is slightly misleading. I am not really going to worry about the big $D$ notation and just stick to $\delta$ and $d$ for sanity's sake. The point of this proof is to start with a $H_\delta H_d$ cocycle, do some diagram chasing and arrive at a $H_D$ cocycle.
We start with an element $\phi$ in the bicomplex $K$ such that $\phi \in K^{p,q}$. If $\phi$ were to represent a cohomology class in $H_\delta H_d$, it would have to be the case that first it represented a class in $H_d$, which means that $d \phi = 0$. Now for $[\phi]_d \in H_d$ to also represent a cohomology class in $H_\delta H_d$ it must be the case that $\delta [\phi]_d = [\delta \phi]_d = [0]_d$, which means that $\delta \phi = d \phi_1$ for some $\phi_1$. Now it is clear that $\phi_1$ cannot represent a $H_\delta H_d$ cohomology class because it is not $d$ closed. 
Thus we look at $\delta \phi_1$. This is $d$ closed because $d \delta \phi_1 = \delta d \phi_1 = \delta \delta \phi = 0$. Thus $[\delta \phi_1]_d$ represents a cohomology class in $H_d$. Also $\delta [\delta \phi_1]_d = [\delta \delta \phi_1]_d = [0]_d$ so $\delta \phi_1$ also represents a cohomology class in $H_\delta H_d$. But $\delta \phi_1 \in (H_\delta H_d)^{p+2,q-1} = 0$, which means that since it is a $\delta$ closed cohomology class it is also a $\delta$ exact cohomology class, i.e. $[\delta \phi_1]_d = \delta [\phi'_1]_d = [\delta \phi'_1]_d$ for some $\phi'_1$ that represents a $d$ cohomology class, i.e. $d \phi'_1 = 0$. Now $[\delta(\phi_1 - \phi'_1)]_d = [0]_d$ means that $\delta(\phi_1 - \phi'_1) = d \phi_2$ for some $\phi_2$. Thus $\delta(\phi_1 - \phi'_1)$ represents a $H_d$ class and it is clear that it also represents a $H_\delta H_d$ class. We also see that $d (\phi_1 - \phi'_1) = \delta \phi$.
Now we can do the same process to $\phi_2$, namely look at $\delta \phi_2$, notice it represents a $H_\delta H_d$ cohomology class, but in the position cohomology is trivial and thus $[\delta \phi_2]_d = \delta [\phi'_2]_d$, and replace $\phi_2$ with $\phi_2 - \phi'_2$. Keep doing this process until you get to the end of the bicomplex where we have $\phi_n$ and $\delta \phi_n = 0$. Now we want to replace $[\phi]_{\delta d}$ with $[\phi + (\phi_1 - \phi'_1) + (\phi_2 - \phi'_2) + \dots + \phi_n]_{\delta d}$, which will now be a $D$ cocycle, which is easy to check. Thus we have crafted a map from $H_\delta H_d$ to $H_D$.
