Characterization of $\mathbb{R}^n$?

Solution 1:

Yes, this is true. First, orient your $n$-manifold $M$ (your hypotheses imply that $M$ is contractible, so this is possible).

First, by your hypothesis, you obtain an increasing exhaustion $M_k \subset M$ of compact sets, diffeomorphic to the $n$-ball, so that each $M_k$ is contained in the interior of $M_{k+1}$.

This is enough; here is the idea. Write $B(k)$ for the unit ball of radius $k$ in $\Bbb R^n$. We may construct, for each $k$, some oriented diffeomorphism $\phi_k: M_k \to B(k)$. If we had $\phi_k \big|_{M_{k-1}} = \phi_{k-1}$, then by taking the increasing union of these $\phi_k$, we define a bijection $M \to \Bbb R^n$ which is a diffeomorphism on the interior of any compact set, and hence is a global diffeomorphism.

In practice, each successive $\phi_k$ has nothing to do with the previous one. Here is how we will resolve this.

Consider the map $g_k: \phi_{k+1}\phi^{-1}_k: B(k) \to B(k+1)$. All we know about this map is that it is a smooth oriented embedding into the interior.

Lemma: There is only one oriented embedding of the $n$-disc into any oriented smooth open $n$-manifold $M$, up to isotopy.

This is a lemma of Cerf and Palais, independently; see here. The idea is to take any smooth embedding to the linear embedding in a chart given by the derivative at zero. In particular, we may find an ambient isotopy $f_t: B(k+1) \to B(k+1)$ which is the identity near the boundary, so that $f_0 = \text{Id}$ and $f_t g_k$ is a smooth isotopy between $f_0 g_k = g_k$ above and $f_1 g_k$ the canonical inclusion map.

Therefore, the map $f_1 \phi_{k+1}$ restricts to $\phi_k$ on $M_k$. So we choose this to be our given diffeomorphism $M_{k+1} \to B(k+1)$. Proceeding inductively, we have our desired result.