A covering map from a differentiable manifold Let $p: C \to X$ is a covering map. Suppose that $C$ is a differentiable manifold.
Is X - differentiable manifold?
More precisely, I am interested in the case where $C$ is Submanifold of Lie algebra, $p$ is the exponential map, and $X= Im \, p$.
 A: Apparently it is not the case that $X$ must be a differentiable manifold.  In 

D. Ruberman.  Invariant Knots of free involutions of $S^4$, Top. Appl. 18 (1984), 217-224

Ruberman shows, among other things, the existence of a topological manifold $X$ which is homotopy equivalent but not homeomorphic to $\mathbb{R}P^4$.  Further, $X$ is not smoothable.
On the other hand, the universal cover $C$ of $X$ will be a compact simply connected topological manifold with $\pi_2(C) = \pi_2(X) = 0$ (since $X$ is homotopy equivalent to $\mathbb{R}P^4$ and $\pi_2(\mathbb{R}P^4) = 0$).  By Freedman's classification, this implies $C$ is homeomorphic to $S^4$.  In particular, $C$ can be given the structure of a smooth manifold.
As far as you "more precisely", I'm not sure what you mean.  Generally, $\exp$ is not a covering map.
A: 
$\color{red}{\text{Beware:}}$ I am quite unsure what I had in mind when I wrote this. Right now I cannot see how to deal with Neal's objection in the comment! I'll leave this here to record the gap.

Yes, there is a canonical smooth structure on $X$ characterized by the fact that the map $p$ becomes, with respect to it, a local diffeomorphism.
You can construct its charts by composing charts out of $C$ with local sections of $p$.
