Submanifold of ball is entire ball

Question

I've seen the following used in 2 and 3 dimensions; I don't know how to prove it, and am wondering if it's true in all dimensions:

If $M$ is an $n$-dimensional smooth compact manifold embedded in the closed ball $\mathbb{B}^n$, and if $\partial M=\partial\mathbb{B}^n$, then $M=\mathbb{B}^n$.

Intuitively this seems true, at least in the low dimensions I can visualize, but I don't know how to go about proving it. All I can think of is to prove by contradiction: look at $\mathbb{B}^n\setminus M$, an open subset, which by compactness has finitely many components. I would like to think each component would contribute an extra boundary component to $M$, but that's where I get stuck. Any help is appreciated!

Answer 1

First note, by compactness, there is a $ \varepsilon > 0 $ such that the $ \varepsilon $-neighborhood of $ \partial \mathbb{B}^n $ (the set of points within $ \varepsilon $ of $ \partial \mathbb{B}^n $) lies entirely within $ M $. Call this $ \varepsilon $-neighbordhood $ N$. We will need it later.

Suppose (for "contradiction") $ M $ is not all of $ \mathbb{B}^n $. Since $ M $ is closed, $ \mathbb{B}^n \backslash M $ is open. Let $ x \in \mathbb{B}^n \backslash M $. Define $$ R := \sup S := \sup \{ r > 0 \ | \ B_r(x) \cap M = \emptyset\} $$ $ R $ is a well-defined positive real number since $ \mathbb{B}^n \backslash M $ is open (clearly $ S $ is bounded above by $ 1 $). It is not hard to check that $ B_R(x)\cap M = \emptyset $ while $ \overline{B_R(x)} \cap M \neq \emptyset $.

(To see the second fact, use compactness: if it weren't true, there would be a positive distance $ \varepsilon $ between $ \overline{B_R(x)} $ and $ M $, contradicting the definition of $ R $ since we would have $ R + \varepsilon/2 \in S $.)

Pick $ x \in \overline{B_R(x)} \cap M $. I claim it is in $ \partial M $ (obvious) and also is not in $ \partial \mathbb{B}^n $. Indeed if it were in $ \partial \mathbb{B}^n $ then $ B_R(x) $ would intersect $ N \subseteq M $, a contradiction of our earlier observation that $ B_R(x) $ does not touch $M $.

Answer 2

I just want to add an answer I eventually figured out for this. Because $\partial M=\partial\mathbb{B}^n$, the dimension of $M$ is $n$. Then $Int(M)=M\cap Int(\mathbb{B}^n)$ is open (via slice charts) and closed (via compactness) in $Int(\mathbb{B}^n)$. By connectedness then $Int(M)=Int(\mathbb{B}^n)$, so $M=\mathbb{B}^n$.