Let S be the open unit square in R^2: the set of points (x,y) with 0 < x < 1 and 0 < y < 1. Consider an area-preserving smooth map S --> R^2, that is, a map whose Jacobian has determinant 1 at every point. What can the image of S look like?
Can the image of the square have a smooth boundary? I think you can smooth out the two corners on top with a transformation of the form (x,y) --> (x,y+f(x)), but this makes the other two corners sharper.
(I had added, and now removed, some nonsense about Gromov's nonsqueezing theorem, which does not hold in dimension 2, and doesn't say what I thought it said besides.)