Algebraic Topology Double Complexes I am going through Bott and Tu and trying to do Exercise 9.13 which says

When a homomorphism $f: K \rightarrow K'$ of double complexes induces $H_d$-isomorphism, it also induces $H_D$-isomorphism.

This is Step 2 of a proof for Kunneth formula. The exercise seems to be for general double complexes, but it is only used where the double complex is $(C^p(\mathscr{U}, \Omega^q), d, \delta)$.
While trying to prove it in general, I keep wanting that the rows be exact, which is true in context of the proof, but not in general. So I'm wondering if the statement in general is not true and the exercise is to actually prove it in context.
I'm assuming Bott and Tu worded it correctly and it is true in general, but I'm afraid of wasting time since I seem to be stuck, so does anyone have any clue about this statement?
Notes:
Let $f_D: H_D^n(K) \rightarrow H_D^{n+1}(K')$ be the induced $H_D$ homomorphism. Define $f_d$ similarly.
We want to show if $[w] \in ker f_D$ then $[w] = 0$, i.e. $w \in im D$
Suppose $w \in K^{0,0}$. We have $f(w) \in im D: 0 \rightarrow K^0$, so $f(w) = 0 \in im d$. Since $f_d$ is an isomorphism ($f^{-1}(im d) \subset im d$), $w \in im d: 0 \rightarrow K^{0,0}$. Thus $w = 0 \in im D$, so the statement is true in this case.
Now suppose $w = w_0 + w_1 \in K^1 = K^{0,1} \oplus K^{1,0}$. Similarly we can show $w_0 \in im d$, say $w_0 = d\beta$. Say $f(w_0) = d\alpha$ and, thus, $f(w_1) = \delta\alpha$; then $f(w_1) = f(d\beta) = df(\beta) = d\alpha)$, so $f(\beta) - \alpha \in ker d$, so $f(\delta\beta) - \delta\alpha = f(\delta \beta - w_1) \in ker d$ so $\delta \beta - w_1  = k \in ker d$. We want to show $k$ is in the image of $\delta$. We can show $\delta(f(k)) = 0$ since $f(w) = 0$, then similar to the first case, we can say $\delta k \in im d: 0 \rightarrow K^{1,0}$ so $\delta k = 0$ so $k \in ker \delta$. If the rows are exact, then $k \in im \delta$ and we're done in this case.
So it looks like we have this is only true in this second case if $K^{0,0} \rightarrow K^{1,0} \rightarrow K^{2,0}$ is exact at $K^{1,0}$ which I'm guessing is not true in general.
 A: As far as I can tell, this statement is true in the following generality:
Suppose $K$ and $K'$ are bounded double (cochain) complexes and $f: K \rightarrow K'$ is a map of double complexes inducing an isomorphism either after taking vertical or horizontal cohomology of the complexes. Then $f$ induces an isomorphism on $H^*(Tot(K)) \rightarrow H^*(Tot(K'))$. 
(Here I'm pretty sure $H^*(Tot(K))$ is what you are calling "$H_D(K)$". "Bounded" means that along each diagonal there are finitely many nonzero terms... this is definitely true in our case since everything lives in the first quadrant.)
The easiest proof uses a spectral sequence argument, but maybe that's not kosher here since Bott and Tu don't introduce spectral sequences until a little later. 
The proof without spectral sequences is really a diagram chase where you sort of push things over to the edge of the double complex where life is good and things are zero and then do some inductive argument. I'll give a quick reduction and then a reference:
It suffices to show that: If $C$ is a bounded complex with exact rows (or columns) then $Tot(C)$ has zero homology.
Indeed, just take the mapping cone of the map $f: K \rightarrow K'$. Then, by the long exact sequence on cohomology, we get the result we want.
The lemma is sometimes called the "Acyclic assembly lemma." You can find an elementary proof in Algebra: Chapter 0, but it is probably best done in the privacy of your own home.
