Consider two chain complexes $C_\bullet$ and $C'_\bullet$ of finite dimensional vector spaces over a common field so that there is a based inclusion $C_\bullet \hookrightarrow C'_\bullet$. What this means, precisely, is that one can extend a basis for each $C_n$ to a basis for the corresponding $C'_n$ and our chain map can be represented by the matrix with ones on the diagonal and zeroes elsewhere (although of course in general we could have $\dim C'_n > \dim C_n$, so the matrix connecting them need not be square: it might be rather thin and tall).
An immediate consequence of having a based inclusion is the following: letting $d_\bullet$ be the boundary map of $C_\bullet$ and similarly $d'_\bullet$ for $C'$, the initial $p_n \times p_n$ sub-block of $d_n$ coincides with that of $d'_n$, where $p_n$ is the minimum of $\dim C_n$ and $\dim C_{n-1}$. I'm fairly sure the following is also true, but I've been having an embarrassingly tough time finding a short argument:
If the homology vector spaces of $C$ and $C'$ are isomorphic, then any based inclusion $C \hookrightarrow C'$ is a quasi-isomorphism
I'm not claiming that based inclusions always lead to isomorphisms on homology, but rather that if the homologies are isomorphic already and a based inclusion exists, then it must induce an isomorphism. What's a slick, non-tedious proof of this fact?
Let $C$ be the complex in which every $C_i$ is $k$ and all maps are zero.
Let $C'$ be the complex in which every $C_i'$ is $k\oplus k\oplus k$ and all maps are $\pmatrix{0&1&0\cr 0&0&0\cr 0&0&0}$.
Map $C_i$ into $C_i'$ by $\pmatrix{1\cr 0\cr 0}$. This is a based chain map.
Then the homology of $C_i$ and the homology of $C_i'$ are both $k$ at every stage. But the induced map on homology is zero.