Concerning the tangent space of an exotic $\mathbb R^4$ My geometric intuition is very poor, so my naive approach to this question is "if $M$ is an exotic $\mathbb R^4$, then $TM$ must be something like $\mathbb R^8$, which is not exotic". Of course, my statement "like $\mathbb R^8$ " is probably rubbish. So, my concrete question is: is the tangent space of an exotic $\mathbb R^4$ a "nice" manifold? I personally don't consider exotic manifolds "nice" (again, because of my poor geometric intuition, though certainly they are very respectable objects of study). I am open-minded about what manifolds a topologist would regard as "nice".

An exotic $\Bbb R^4$ (let us denote it by $\Sigma$) is homeomorphic to the usual $\Bbb R^4$, so $\Sigma$ is contractible. Any vector bundle over a contractible manifold is trivial, and hence the tangent bundle $T\Sigma$ is trivial: $T\Sigma \cong \Sigma \times \Bbb R^4$. Hence we have that $T\Sigma$ is homeomorphic to $\Bbb R^8$. Since it is known that there are no exotic $\Bbb R^n$'s for $n \
eq 4$, it follows that $T\Sigma$ is diffeomorphic to $\Bbb R^8$ for any exotic $\Bbb R^4$ $\Sigma$.