Deformations of a complex trivial up to quasi-isomorphism Let $(C, d_0)$ be some cochain complex over commutative ring $R$ and $R$ is an algebra over a field $k$, then I could consider complex $(C[t], d_0+td_1)$ over ring $R \otimes k[t]$ where $d_1$ is some perturbation such that $(d_0+td_1)^2=0$. 
What conditions guarantees that this deformation has the same cohomology groups? In other words I want $H^i(C, d_0+td_1)$ to be "constant in t" i.e. $k[t]$-module $H^i(C, d_0+td_1)$ is free. I also could hope for quasi-isomorphism $\phi_t$between $(C, d_0)$ and $(C[t], d_0+td_1)$, but it might be to much to ask.
General deformation theory suggest that if I take $d_1$ that is trivial in $Ext^1(C,C)$ then I get a complex isomorphic to $C$ but this is too restrictive.
This question reminds me situation of homological perturbation lemma, but in homological perturbation lemma we start with some rich data and produce some explicit formulas and here I want only to know that cohomology groups are the same.
 A: In general we can consider a differential $d_t$ which depends polynomially on $t$ and whose value at $t = 0$ is our original differential. Simple examples show that we can pick up extra cohomology at special values of $t$. For example we can take a complex of the form $C^0 \xrightarrow{d_t} C^1$ with $d_t = \left[ \begin{array}{cc} 1 & 0 \\ 0 & t \end{array} \right]$. The dimension of $H^0(C) = \text{ker}(d_t)$ jumps from $0$ to $1$ at $t = 0$ and so does the dimension of $H^1(C) = \text{coker}(d_t)$. 
However, I claim that at least for complexes of finite-dimensional vector spaces we can never lose cohomology at special values of $t$ (I'll be more specific about what I mean by this). In particular, if the complex is acyclic at special values of $t$ then it is acyclic everywhere. The example I have in mind, which I learned about very recently, is a particular complex associated to the family of algebras
$$U_t(\mathfrak{g}) = T(\mathfrak{g})/(X \otimes Y - Y \otimes X - t [X, Y])$$
where $\mathfrak{g}$ is a Lie algebra. $U_t(\mathfrak{g})$ is isomorphic to the universal enveloping algebra for $t \neq 0$ but isomorphic to the symmetric algebra for $t = 0$, and the corresponding complex is isomorphic to the Chevalley-Eilenberg complex of $\mathfrak{g}$ for $t \neq 0$ but isomorphic to the Koszul complex of $S(\mathfrak{g})$ for $t = 0$. The claim above implies that the acyclicity of the latter implies the acyclicity of the former. (In order to think of my complex as the same complex with a varying differential as $t$ varies I need PBW.)
So, here's what I mean by special: the action of the differential $d_t$ in a particular degree $C^n \to C^{n+1}$ induces a map $\Lambda^k(C^n) \to \Lambda^k(C^{n+1})$ for all $k \ge 0$. The dimension $\dim \text{im}(d_t)$ is the largest $k$ such that this map is nonzero. The condition that this map is nonzero, for fixed $k$, is a Zariski open condition in $t$ since it is defined by the nonvanishing of at least one $k \times k$ minor of $d_t$. Hence $\dim \text{im}(d_t)$ generically takes on a single value, and at a possibly empty exceptional set of points defined by the vanishing of the $k \times k$ minors of $d_t$ for $k = \dim \text{im}(d_t)$ it takes on a smaller value. And of course $\dim \text{ker}(d_t) + \dim \text{im}(d_t) = \dim C^n$ is fixed, so $\dim \text{ker}(d_t)$ generically takes on a single value and takes on a larger value on a possibly empty exceptional set. 
In the above example the behavior for $t \neq 0$ is uniform so the only possible special point is $t = 0$. In general it might be harder to say what the special points are. 
