Does semi-free behave well under totalization Suppose I have a dg algebra $(A,d)$ and a chain complex $M^\bullet$ of semi-free $(A,d)$ modules. I am hoping it is true that $ Tot^\coprod (M^\bullet)$ is again a semi-free $(A,d)$ module. Is this so? And if yes, what filtration should one choose?
The definition of semi-free I am using is: $X$ is semi-free if $X$ is the union of  an increasing sequence of dg $A$-modules $X_i$ , such that $ i \in \mathbb{N}$, $X_0 = 0$ and $ X_{i+1}/X_i$ is a free dg $A$-module. 
 A: If $M_\bullet$ is not bounded above, then the totalization may not be semi-free.
For example, let $k$ be a field, and $A=k[x]/(x^2)$, considered as a DG algebra concentrated in degree zero, with trivial differential. Then the chain complex
$$\dots\to A\to A\to A\to A\to\dots,$$
where all the differentials are multiplication by $x$, is not semi-free considered as a DG $A$-module, as it has no free DG submodules.
A: So here is a partial answer, If $M_\bullet$ chain complex of dg $A$ modules then $\text{Tot}^\coprod M$ can be realized as a dg $A$ module whose underlying graded module is given by
$ \coprod\limits_{i \in \mathbb{Z}} [-i] M_i $
and whose differential is 
$d = d_{[-i]M_i} + d^i_{M_\bullet}$ 
Now suppose that $M_\bullet$ is bounded above, i.e. of the form:
$\cdots \rightarrow M_{-i} \rightarrow M_{-i+1} \rightarrow \cdots \rightarrow M_0 \rightarrow 0$
Then you can build the desired filtration where $X_0 = 0$ and for $ j \geq 1$ 
$ X_j : = \text{Tot}^{\coprod} ( \cdots \rightarrow 0 \rightarrow M_{-j+1}\rightarrow M_{-j+2} \rightarrow \cdots \rightarrow M_0) $
Note that the quotients in this filtration will be semi-free and not free but this enough to show that the totalization is semi-free.
