In Appendix B of Matsumura's "Commutative Ring Theory," in Theorem B1, he asserts that if a double complex
$$K_{**} =: \{K_{pq} \}$$
satisfies $\{ K_{p*} \}$ is exact (except maybe at $q=0$) for all $p$, then the homology of $K_{**}$ is isomorphic to the homology of the complex
$$X_* =: \{ H_0(K_{p*}) \}$$
He then shows what the obvious morphism should be, and proves its surjectivity, all of which I understand. Then, he says something to the effect of "the proof for injectivity is similar." Despite my efforts, being brand new to homology, I am unable to finish this proof.
Using the text's notation, I know we need to take an $x \in K_n$, the direct sum over the $K_{pq}$ such that $p+q = n$, with $dx = 0$ ($d$ is the differential for $\{K_n\}$), and show that if $\Phi(x) = 0$, where $\Phi$ is the morphism in question, then $x = 0$. After that, I'm losing myself in the terminology and all of the diagrams and relationships I'm building.
See Theorem 1.6 here for more background and information:
It is not injective at the level of cycles, but only after taking homology. So what you have to prove fot that $x$ is that if $\Phi(x)$ is zero in homology (i.e. it is of the form $\Phi(x) = d_Xy$ for some $y$), then $x=d_Kz$ for some $z$.