Cohomology of Double Complex (extreme case) Let $K^{p,q}$ be a double complex with ascending indices, i.e. the differentials are $d':K^{p,q} \rightarrow K^{p+1,q}$ and $d'':K^{p,q} \rightarrow K^{p,q+1}$. Suppose that $K^{p,q}=0$ if $p<0$ or $q<0$ and that $H^n(K^{p \cdot})=0$ when $n>0$ for any $p$. Define $X^p=ker(K^{p,0} \rightarrow K^{p,1})$. Then $X$ is a complex with differential induced by $d'$. How can we show that $H^n(K) = H^n(X)$?
I am interested in an argument that does not involve any spectral sequence theory.
Reference: Matsumura, Commutative Ring Theory, p. 277.
 A: Consider the inclusion $X\to K$ and let $L$ be the quotient, so that we have a short exact sequence of first-quadrant double complexes $$0\to X\to K\to L\to 0$$
Now show by hand that $L$ is exact. Suppose $n>0$ and we have an $n$-cocycle in $L$, which is a sequence $(x_0,\dots,x_n)$ with $x_i\in L^{i,n-1}$ for each $i$ such that $d'x_i=d''x_{i+1}$ if $0\leq i<n$, $d''x_0=0$ and $d'x_n=0$. In particular, $x_0$ is a cocycle for the vertical differential. The hypothesis on exactness implies there is a $y_0$ such that $d''y_0=x_0$. Then $d''(d'y_0-x_1)=d'd''y_0-d''x_1=d'x_0-d''x_1=0$, and $d'y_0-x_1$ is a cocycle for the vertical differental. Exactness implies there is a $y_1$ such that $d''y_1=d'y_0-x_1$, so that $x_1=d''y_1-d'y_0$.
In this way we go down the $(n-1)$th diagonal. When we reach the last step, though, we have to use the fact that $L$ is constructed in a specific way.
It is the sort of boring argument that is best done in detail by no one other that by yourself! 
(And which spectral sequences perfectly package...)
