Submanifolds - same dimension

Question

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.

Answer 1

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 $M$.

Answer 2

As the original poster added later, you require connectedness:

As Jack argued, the inclusion is an embedding. Since the manifolds have same dimension, it is an open map. Thus the image is open. By assumption it is also closed. Therefore, the image is a connected component of $M$---but it need not be $M$ itself, unless $M$ has only one component.