Free objects in the category of dg modules Suppose that $A$ is a dg algebra, does the category of dg modules over $A$ where morphisms are degree zero maps that commute with differential have a free object ( in general)? I have been reading a lot about semi-free, but nobody ever seems to talk about free and right now I think that the answer is no. But this seems strange.
 A: The apparent conflict between Martin Brandenburg's comments and Najib Idrissi's answer can be resolved by observing that "free object" is ambiguous until you specify what exactly you want free objects on; more precisely you should think of free functors as left adjoints to forgetful functors, and so you need to specify a forgetful functor. 
Najib Idrissi's answer takes the forgetful functor to be the underlying graded set functor, and Martin Brandenburg's comments take the forgetful functor to be the underlying chain complex functor, but other choices are justifiable. It depends on what you want to do. (With Martin's choice of forgetful functor, for example, free objects aren't necessarily projective, so you don't necessarily get nice objects to construct resolutions out of. The point of looking at semi-free modules seems to be to construct resolutions.) 
A: The differentials actually do make the problem more complicated, as far as I'm aware. For example suppose that the free dg-module on a graded set $X$ were the formal linear combinations of elements of $X$. Then the only sensible way to define a differential would be to make it zero. But then if $f : X = \{x\} \to M$, $f(x) = m$ extends to $\varphi : F(X) \to M$, this would imply $d(f(x)) = \varphi(dx) = 0$, and this for all elements $m \in M$!
In fact if $X$ is a graded set, you need to adjoin formal elements $\mathrm{d}x$ to $F(X)$, where $\deg(\mathrm{d}x) = \deg(x)-1$. The differential of $x$ is set to be $\mathrm{d}x$, and of course the differential of $\mathrm{d}x$ is zero. Then the differential of $ax$ is set to be $(d_Aa) x \pm a \mathrm{d}x$, and the differential of $a \mathrm{d}x$ to be $(d_A a) \mathrm{d}x$ (and then you extend by additivity). You then get 
$$\bar{F}(X) = \left\{ \sum_{x \in X} \alpha_x x + \beta_x \mathrm{d}x : \alpha_x, \beta_x \in A, \text{the sum is finite} \right\}.$$
Then any graded map $f : X \to M$ where $M$ is a dg-module extends to a dg-module map $\varphi: \bar{F}(X) \to M$ by setting $\varphi(x) = f(x)$ and $\varphi(\mathrm{d}x) = d_M f(x)$ (this is an easy check).
