Is every Compact $n$-Manifold a Compactification of $\mathbb{R}^n$?

I read the result that every compact $n$-manifold is a compactification of $\mathbb{R}^n$.

Now, for surfaces, this seems clear: we take an n-gon, whose interior (i.e., everything in the n-gon except for the edges) is homeo. to $\mathbb{R}^n$, and then we identify the edges to end up with a surface that is closed and bounded.

We can do something similar with the $S^n$'s ; by using a "1-gon" (an n-disk), and identifying the boundary to a point. Or we can just use the stereo projection to show that $S^n-\{{\rm pt}\}\sim\mathbb{R}^n$; $S^n$ being compact (as the Alexandroff 1-pt. -compactification of $\mathbb{R}^n$, i.e., usual open sets + complements of compact subsets of $\mathbb{R}^n$). And then some messy work helps us show that $\mathbb{R}^n$ is densely embedded in $S^n$.

But I don't see how we can generalize this statement for any compact n-manifold. Can someone suggest any ideas?

Thanks in Advance.

Assuming $M$ is a differentiable manifold, this can be shown by choosing normal coordinates about any point $P\in M$. The following is expanding on the method suggested by Jason DeVito in the comments.

Putting a metric on the manifold, we can define the exponential map $\exp_P\colon T_PM\to M$ and then, choosing an orthonormal basis $e_1,\ldots,e_n$ for $T_PM$, we get a map $$ \begin{align} f\colon\mathbb{R}^n&\to M,\\ (x^1,\ldots,x^n)&\mapsto\exp_P(x^ie_i) \end{align} $$ (using the summation convention). As the manifold is compact, this is an onto mapping (only completeness is required). The idea, then, is that there is a circled open subset $U\subseteq\mathbb{R}^n$ on which this becomes a coordinate map with dense image. That $U$ is circled (aka balanced) means that the line segment joining any point $x\in U$ to the origin is in $U$. It can be seen that any circled open set is diffeomorphic to $\mathbb{R}^n$ (also see star domain).

Let $D$ be the set of $x\in\mathbb{R}^n$ such that $[0,1]\to M$, $t\mapsto f(tx)$ is a minimizing geodesic. Equivalently, $D$ is the set of $x\in\mathbb{R}^n$ with $d(P,f(x))=\Vert x\Vert$, from which we see that it is closed. Also, let $U$ be the interior of $D$. The set $\partial D\equiv D\setminus U$ is called the cut locus (more precisely, the cut locus is the subset of $T_PM$ corresponding to $\partial D$).

Fixing some $x\in\mathbb{R}^n\setminus\{0\}$, let us consider the line $t\mapsto f(tx)$ for $t\ge0$. Define $r > 0$ to be the maximum real number with $rx\in D$. For any $0 < r_0 < r$, we can show the following.

1. On $[0,r_0]$, $t\mapsto f(tx)$ is the unique minimizing geodesic joining $P$ to $f(r_0x)$.
2. The derivative of $f$ at $r_0x$ is invertible, so $f$ is nonsingular at $r_0x$.
3. $r_0x$ is in $U$.
4. $f(D)=M$ and $f(\partial D)\subseteq M$ has zero measure.

Property (3) implies that $U$ is circled, so it is diffeomorphic to $\mathbb{R}^n$. Property (1) says that $f$ is one-to-one on $U$ and, by (2), it is a diffeomorphism. Then, by (4), $f(U)\supseteq M\setminus f(\partial D)$ is dense, so $f$ gives a diffeomorphism from $U$ to a dense subset of $M$.

I'll give proofs of these statements now, although they do seem quite standard. See also these notes and, in particular, Lemma 5.3 for the proofs of (1) and (2) and Lemma 5.4 for the proof of (4).

Proof of (1): Consider a minimizing geodesic $\gamma\colon[0,r_0]\to M$ joining P to $f(r_0x)$, which will have length no more than $r_0\Vert x\Vert$. Then, we can extend $\gamma(t)$ to $r_0\le t\le r$ by setting $\gamma(t)=f(tx)$. As this curve joins $P$ to $f(rx)$ and is of length no more than $r\Vert x\Vert$, it must be a geodesic. So, $\gamma(t)=f(tx)$ for all $t\le r$, and $t\mapsto f(tx)$ is a unique minimizing geodesic on $[0,r_0]$.

Proof of (2): Choose a nonzero $y\in\mathbb{R}^n$ and consider the vector field $t\mapsto Y_t$ given by $Y_t=t\nabla_y f(tx)=\frac{\partial}{\partial s}f(t(x+sy))\vert_{s=0}$. By geodesic deviation, this is a Jacobi field. Also, $Y_0=0$ so, if $\nabla_yf(tx)$ was zero, $P$ and $f(r_0x)$ would be conjugate points along $t\mapsto f(tx)$. Then, it is a standard result that $t\mapsto f(tx)$ is not a minimizing geodesic on $[0,r]$ for any $r > r_0$ (see characterization of the cut locus), contradicting the choice of $r$. So, $\nabla_yf(r_0x)\not=0$, and $f$ is nonsingular at $r_0x$.

Proof of (3): If not, then there would be a sequence $x_i\not\in D$ tending to $r_0x$. By the definition of $D$, there exist $y_i\in\mathbb{R}^n$ with $f(y_i)=f(x_i)$ and $\Vert y_i\Vert < \Vert x_i\Vert$. Passing to a subsequence, we can suppose that $y_i$ tends to a limit $y$. So, by continuity, $\Vert y\Vert\le r_0\Vert x\Vert$ and $f(y)=f(r_0x)$. By (1), this means that $y=r_0x$. But, then, setting $a_i=\Vert y_i-x_i\Vert$, (2) contradicts the limit $\nabla_{(y_i-x_i)/a_i}f(r_0x)\sim (f(y_i)-f(x_i))/a_i=0$.

Proof of (4): By completeness, for any $Q\in M$, there is at least one minimizing geodesic joining $P$ to $Q$. So, $f(x)=Q$ for some $x\in D$, and $f(D)=M$. Next, (3) implies that, for any $x\in\mathbb{R}^n\setminus\{0\}$, the radial line $t\mapsto tx$ ($t\ge0$) intersects $\partial D$ at a single point. This means that $\partial D$ has zero measure. As $f$ is locally Lipschitz, it maps zero measure sets to zero measure sets, so $f(\partial D)$ has zero measure.

Finally, I'll admit that the details are a bit trickier than I initially thought when I commented that the method can be made to work "without too much trouble".

Since the question is tagged differential-topology, I assume we are talking about a differentiable manifolds.

One way to see that $M$ is a compactification of $\mathbb{R}^n$ is the following:

Take $f$ to be a Morse function on $M$ with one minimum.The flow of $\nabla f$ around the minimum will parametrize almost all of $M$ (the missing part will be the unstable manifold of critical points of smaller index).

Here is the answer in the context of topological manifolds:

1. For smooth manifolds you already have completely satisfactory answers.

2. If a compact topological manifold $M$ has dimension $n\ne 4$ then it admits a handlebody decomposition (see the discussion and references here). Once you have a handlebody decomposition $M=H_1\cup_S H_2$, you use the fact that the complements to some $n-1$-dimensional disjoint disks $\Delta_{1,j}, \Delta_{2j}$ in the interiors of $H_1, H_2$ are homeomorphic to $R^n$. Now, find an open $n-1$-disk $D\subset S$ disjoint from the boundaries of the disks disks $\Delta_{1,j}, \Delta_{2j}$. Then the union $$ int(H_1)\cup D \cup int(H_2) $$ is homeomorphic to $R^n$. This union is clearly dense in $M$.

3. There are 4-dimensional manifolds which do not admit a handle decomposition, but Frank Quinn in

Ends of Maps III: Dimensions 3 and 4, Journal of Differential Geometry vol. 17 (1982)

proved that all noncompact 4-dimensional manifolds are smoothable. Hence, removing a point from such a manifold results in a smoothable manifold which admits an open and dense subset homeomorphic to $R^4$.

The conclusion is then that every compact topological $n$-manifold is homeomorphic to a compactification of $R^n$.

My guess it that you could generalize your example of the n-gon to any triangulation (with higher simplices, instead of only triangles) of your manifold. Any manifold admits a triangulation.