Removing a point from a sphere

Let $\Sigma^n$ be a smooth manifold homeomorphic to an $n$-sphere (possibly exotic) and let $p\in\Sigma$ be a point. Then the complement $\Sigma^n\setminus \{p\}$ is homeomorphic to $\mathbb{R}^n$.

Is the complement $\Sigma^n\setminus\{p\}$ also a smooth manifold? If so, then I think it must be diffeomorphic to $\mathbb{R}^n$.

The case $n=4$ seems to be a special case because $\mathbb{R}^4$ does not have a unique smooth structure. Can $\Sigma^4\setminus\{p\}$ be diffeomorphic to an exotic $\mathbb{R}^4$, or is it always standard $\mathbb{R}^4$?

Also, if you add a point to an exotic $\mathbb{R}^4$, can you get an exotic $S^4$? I guess this isn't possible, as someone would have already thought of it.

An open subset $U$ of a smooth manifold inherits a smooth structure (if $U_\alpha$ was a chart for the larger manifold, take $U \cap U_\alpha$ to be a chart of the smaller space). So yes, $\Sigma^n \setminus \{p\}$ inherits a smooth structure, and yes, because $\Bbb R^n$ has a unique smooth structure in high dimensions, it is diffeomorphic to $\Bbb R^n$. This gives a nice corollary (known long before this piece of smoothing theory): Every exotic sphere can be defined by exactly two charts.

In fact, every exotic sphere can be defined by gluing together two standard smooth $n$-discs $D^n$ along their boundary, but the diffeomorphism with which we identify $\partial D^n_1 \cong \partial D^n_2$ (note that these are both spheres of the same dimension) may be nonstandard: you glue them by some choice of diffeomorphism $\varphi: S^{n-1} \to S^{n-1}$. The resulting manifold doesn't change if you modify $\varphi$ by a (smooth) homotopy through diffeomorphisms, but it is not clear (in fact, not true) that every orientation-preserving diffeomorphism is isotopic to the identity.

It is a theorem of Cerf (essentially, the pseudoisotopy theorem) that in fact, in dimension $n \geq 6$, the smooth mapping class group of oriented diffeomorphisms of $S^{n-1}$ modulo isotopy is isomorphic to the group of oreinted exotic $n$-spheres.

In dimension 4, exotic $\Bbb R^4$s may be partitioned in many ways, but relevant to your question is whether or not they are "standard at infinity". If $\mathcal R$ is a smooth manifold homeomorphic to $\Bbb R^4$, and there is a smoothly embedded $S^3 \hookrightarrow \mathcal R$ so that the unbounded component of its complement is diffeomorphic to $(0, \infty) \times S^3 \cong \Bbb R^4 \setminus \{0\}$, then we say that $\mathcal R$ is standard at infinity (or has a "cylindrical end smoothing"). If this is true, we can just add a new chart around the point at infinity to cook up a smooth structure on the 1-point compactification; if $\mathcal R$ was exotic, so would be this new sphere.

So the Poincare conjecture is equivalent to finding exotic $\Bbb R^4$s with a cylindrical end smoothing. This is indeed, as you expect, wide open.

• Just to avoid confusion: The conjecture you refer to in the end is the smooth Poincaré conjecture which says "(the topological manifold) $\mathbb{S}^4$ has no exotic smooth structures". This is what is "wide open". The standard Poincaré conjucture "every simply connected, closed 3-manifold is homeomorphic to the 3-sphere" has been solved, in the affirmative, by Perelman. Wikipedia: generalized Poincaré conjecture Jul 3 '18 at 10:09
• @Mike Miller: Thanks for your nice answer. To make sure I understood your second-to-last sentence, is the following statement correct? If $\Sigma^4$ is an exotic $S^4$, then for all $p\in \Sigma^4$, the complement $\Sigma^4\setminus \{p\}$ is diffeomorphic to an exotic $\mathbb{R}^4$. Jan 11 '19 at 17:36
• @smnas Yes, but this requires more argument than I offered here. One must argue that every oriented embedding of the closed 4-ball in $\Sigma$ is isotopic - without moving the basepoint if desired (a lemma of Cerf-Palais), or equivalently that any two cylindrical end smoothings are isotopic. This implies that if $\Sigma \setminus \{p\} \cong \Bbb R^4$, one may modify this by an isotopy so that this preserves cylindrical ends, and then this would extend to a diffeomorphism $\Sigma \cong S^4$.
– user98602
Jan 11 '19 at 18:03