Does every compact manifold exhibit an almost global chart Let $M$ be a compact connected manifold.
Is there a chart $\Psi:U \to \mathbb{R}^n$ such that the closure of $U$ is $M$?
This is true for $S^n, T^n, K$, all compact surfaces, etc.
If it is not true in general, what is the obstruction?
 A: The exponential map for any  Riemannian metric on your compact manifold $M$, based at any point $p$ of $M$,  maps the tangent space $T_p M$, an ${\mathbb R}^n$, onto $M$ and  is a diffeo inside the cut locus.   Back on the tangent space, this `inside' of  the  cut locus is a  star shaped domain relative to the origin, so defines a domain $V$ in ${\mathbb R}^n$ which is mapped diffeomorphically onto an open set $U$ whose closure is $M$. 
(The closure of the domain $V$ is homeomorphic to the closed ball in the tangent space, so this same argument shows that every compact manifold is the quotient of the n-ball by some identification of points on its boundary, the n-sphere. ) 
A: Take a covering $\mathcal{U}_0 = \{U_0^\alpha \,|\, \alpha < \kappa \}$ of $M_0 = M$ by some charts. Define $V_0 = U_0^0$ and consider $M_1 = M_0 \setminus \overline{U_0}$. Then $U_1^\alpha = U^\alpha_0 \cap M_1$ is a covering of $M_1$. Proceed by (transfinite) induction to obtain $V_\alpha$. (If $M$ is compact then you can assume that the covering $\mathcal{U}_0$ was finite and hence you have a finite set $\{V_\alpha\}$ of open subsets of $M$.) Now $M\setminus \bigcup_\alpha V_\alpha$ is a collection of boundaries of $U_\alpha$ which are manifolds of dimension $n-1$. Hence the closure of $\bigcup_\alpha V_\alpha$ is the whole $M$. If $U_\alpha$ were domains of charts $\varphi_\alpha$, then one obtains, translating the image of $\varphi_\alpha$ if necessary, a well defined chart $\varphi$ on $\bigcup_\alpha V_\alpha$ just by restriction $\varphi|_{V_\alpha} = \varphi_\alpha$, since the sets $V_\alpha$ are disjoint. 
I haven't thought about the noncompact case so I'm not sure the transfinite induction will go through the limit ordinals. If it is even true, can one use a partition of unity to obtain a uniform proof?
Finally, a better notion suited for studying this kind of problems is the Lusternik Schnirelmann category.
