Let $R$ be a reasonable ring (maybe I mean a PID, or $\mathbb{Z}$, and when sufficiently desperate, a field). Now consider fixed sequences $C_n$ and $H_n$ of $R$-modules, which are tame in every possible sense -- finite rank, zero for large enough $n$, etc. Here's the problem:

What is the set (space? category?) of all $R$-linear maps $d_n:C_n \to C_{n-1}$ so that $(C_\bullet,d_\bullet)$ forms a chain complex with homology groups $H_n$?

I've tried to figure things out in baby cases without much luck in terms of seeing the bigger picture. You can easily create situations where no $d_\bullet$s will suffice, but most fascinating are the cases where several different sequences of boundary operators will work!

A very general thing to do is fix a basis of all chain groups in sight and examine conditions that entries of various $d_\bullet$-representing matrices have to satisfy. Just requiring the $d_\bullet$s to produce a chain complex yields an algebraic variety generated by a system of multivariate quadratic equations (which sounds nightmarish). So unless forcing the homology to be prescribed by $H_n$s greatly simplifies things, I should not expect a very explicit answer. However, I'd like to know what the obstructions to getting such an answer might be.

In the cases of interest, all my $d_\bullet$s can be assumed to have an upper-triangular form as $R$-matrices, although I'd also be interested in a solution that ignores such structural constraints.

**Update**: Thanks to David Speyer's answer, which provides a wealth of information on the problem when $R$ is a field. I'm still seeking answers for more general choices of $R$.

quasi-isomorphicto a trivial complex. Two complexes are quasi-isomorphic if there exists a morphism of complexes between them that induces an isomorphism in homology. $\endgroup$ – Liviu Nicolaescu Sep 16 '15 at 17:14