Given a closed surface in $\mathbb R^3$, is it necessarily an "embedded surface"? I think it is true, but that is just because I can't think of a closed surface for which we cannot construct a smooth parametrization, though of course that is not a valid argument!
Closed / embedded surface
3
$\begingroup$
geometry
algebraic-geometry
differential-geometry
definition
-
0@jerrysciencemath, Thanks! – 2012-04-30