Let $M$ be a topological space. I want to show the equivalence of the following statements.
- There is an open subset $U\subset M$ and a homeomorphism $\Psi:U\rightarrow\mathbb{R}^n$.
- There are open subsets $U^\prime\subset M$, $V\subset\mathbb{R}^n$ and a homeomorphism $\varphi:U^\prime\rightarrow V$.
- There is an open subset $U^{\prime\prime}\subset M$, an open ball $B\subset \mathbb{R}^n$ and a homeomorphism $\phi:U^{\prime\prime}\rightarrow B$.
Let (1) be given. Then for any open subset $V\subset\mathbb{R}^n$ let $U^\prime:=\Psi^{-1}(V)$. Since $\Psi$ is a homeomorphism, the map $\varphi:=\Psi|_{U^\prime}$ is also a homeomorphism.
Let (2) be given. Since $V$ is open, it contains an open ball $B$. Now let $U^{\prime\prime}:=\varphi^{-1}(B)$. Then again, $\phi:=\varphi|_{U^{\prime\prime}}$ is a homeomorphism.
Now, how do I show the last implication from (3) to (1)? I thought of showing that any open ball is homeomorphic to $\mathbb{R}^n$ itself. But that's false. Then I thought of composing a homeomorphism onto $\mathbb{R}^n$ out of homeomorphisms $\phi_i$ onto disjoint open balls $B_i$, such that $\cup B_i=\mathbb{R}^n$. But my intuition tells me that that's impossible.
Any thoughts?