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?

9$\begingroup$ How about this: the compact manifold admits a Morse function, the gradient flow as usual provides a CWstructure. The top cell is diffeomorphic to $\mathbb{R}^n$, and the complement is a closed subset. Choosing a diffeomorphism from the top cell to $\mathbb{R}^n$ should provide a chart whose closure is $M$. $\endgroup$– Matthias WendtAug 2, 2014 at 15:19

1$\begingroup$ The tag is "differential geometry", I would say that the author of the question is assuming the smooth structure... $\endgroup$– Ilias A.Aug 2, 2014 at 16:03

4$\begingroup$ The answer is yes. Start with a handle decomposition of the manifold and take a maximal forest in the 1skeleton. That's a collection of discs in the manifold, and you can inflate its interior to be dense in the manifold, since its complement is a regular neighbourhood of the dual (n1)skeleton. This is a pretty common observation in courses where you study handle decompositions, the hcobordism theorem and such. $\endgroup$– Ryan BudneyAug 2, 2014 at 18:24

1$\begingroup$ @MatthiasWendt Possibly dumb question: You seem to be assuming just one top cell. It seems obvious to me that a compact connected manifold has a Morse function with only one local maximum, but how do you prove it? $\endgroup$– David E SpeyerAug 3, 2014 at 2:00

5$\begingroup$ As valeri says in the comment below every $n$manifold contains an open dense subset diffeomorphic to an $n$disk: equip the manifold with a complete Riemannian metric, fix any point, and note that the complement to the cut locus at the point is diffeomorphic to a disk. The cut locus is nowhere dense. See e.g. Sakai's "Riemannian geometry''. $\endgroup$– Igor BelegradekAug 3, 2014 at 2:53
2 Answers
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 nball by some identification of points on its boundary, the nsphere. )
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 $n1$. 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.

$\begingroup$ Yes, for a sphere I'd get two hemispheres and I can map them e.g. to two disjoint open disks such as $B(0,1)$ and $B(3,1)$. $\endgroup$ Aug 2, 2014 at 17:07

3$\begingroup$ since the tag is differential geometry, how about this argument: take a point and issue from this point a geodesic g(t) in every direction v until (some t<t_0) it is minimal. Then the subset of all tv in the tangent space is homeomorphic to R^n and its image under exponential map is almost all M^n as required. $\endgroup$– valeriAug 2, 2014 at 17:16

$\begingroup$ I did not notice your comment Valeri. It really says the same as my answer abover.m. $\endgroup$ Aug 7, 2014 at 13:12

$\begingroup$ And Igor Belegradek provided a reference for proof of the fact that cut locus is nowhere dense in another comment. For example, I've never even heard that term during 3 semesters of differential & Riemannian geometry. $\endgroup$ Aug 7, 2014 at 20:16