If $M$ is a $k$-dimensional manifold, show that every point of $M$ has a neighborhood homeomorphic to all of $\mathbb{R}^k$ .

Question

If $M$ is a $k$-dimensional manifold, show that every point of $M$ has a neighborhood homeomorphic to all of $\mathbb{R}^k$ . Therefore, charts can always be chosen with all of Euclidean space as their co-domains.

I'm confused because I thought that the first sentence was part of the definition of a $k$-dimensional manifold. I don't understand what I'm being asked to prove.

Answer 1

It is obvious that if the definition ( homeomorphic to $\mathbb{R}^k$ ) implies your definition.

Now suppose for every point $p \in M$, there exists homeomorphism $h \colon U \to U'$ where $U \subseteq M$ and $U' \subseteq \mathbb{R}^k$ are open. Take an open ball $V'$ centered at $h(p)$ and $V' \subset U'$. The restriction $h|_{h^{-1}(V')} \colon h^{-1}(V') \to V'$ is an homeomorphism. Then we know an open ball $V'$ is homeomorphic to $\mathbb{R}^k$.

Answer 2

Let $p\in M$,

then by definition of manifold there exist co-ordinate chart $(\phi,U)$ around $p$. Hence $\phi$ is homeomorphism from $U$ to some open subset $V$ of $\mathbb R^n$. As $V$ is open there exist $r>0$ such that $B(\phi(p),r) \subseteq V$.

Now let $W=\phi^{-1}(B(\phi(p),r))$.

As $W$ is homeomorphic to $B(\phi(p),r)$ and $B(\phi(p),r)$ is homeomorphic to $\mathbb R^n$, $W $ is homeomorphic to $\mathbb R^n$.