Let $M$ be a smooth manifold and $N$ a closed embedded submanifold. Assume that they have the same dimension. In this case are they equal?
EDIT: M is connected.
[Math] Submanifolds – same dimension
differential-geometrysmooth-manifolds
differential-geometrysmooth-manifolds
Let $M$ be a smooth manifold and $N$ a closed embedded submanifold. Assume that they have the same dimension. In this case are they equal?
EDIT: M is connected.
Best Answer
Yes. If $N\subseteq M$ is an embedded submanifold (without boundary) of the same dimension as $M$, then the inclusion map $N\hookrightarrow M$ is a smooth embedding, which means that its differential at each point is bijective. It follows from the inverse function theorem that its image is open in $M$. Since $N$ is both open and closed, it is all of $M$.