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.)



What are the possible images of a square under an area-preserving map?
Yes, you can smooth all corners. 
Say, in the class of maps $(x,y)\mapsto (f(x),g(x,y))$.


Moreover, one can map open unit square to any domain bounded by convex smooth curve
by a map of the above type.


I'm sure that any simply connected domain of the same area can be also obtained as an image...


