Are Minkowski and Schwarzschild spacetimes diffeomorphic? Another mathematical question, arising from GR. Some days ago I wrote, in an answer to 1, that they are. Then @magma commented they are not. He promised a proof, but none appeared. After magma's comment I have some doubts about my intuition and I would like to see a proof, in one sense or the other, or at least a reliable reference.
 A: They aren't. That is because Kruskal-Schwartzschild is diffeomorphic to $S^2 \times R^2$, while Minkowski is $R^4$. This  is roughly  similar to the difference between a 2D  cylinder $S^1\times R$  and $R^2$.
The formal reason is that pointed out by G. Golfetti in a comment above: the second homotopy groups are different, as is expected by the presence of the factor $S^2$. In M. space you can shrink two-spheres continuously into points, in K-S space you cannot: when you encounter the bifurcation surface (the 2D event horizon) you cannot go further and you pass from the right wedge to the left wedge. The standard global Kruskal coordinates map K-S manifold to $S^2\times R^2$. There is no way to make vanishing the radial coordinate. It would vanish on the singularity, but it does not belong to the manifold. At the center of the manifold it takes the Schwarzshild radius value which  is strictly positive. There is no center of those spheres. Similar coordinates can be defined on Minkowski spacetime but they are not global since they do not cover the set $r=0$ which is part of the manifold.
