Homeomorphisms on a Hausdorff space

Question

Let $M$ be a topological space. I want to show the equivalence of the following statements.

1. There is an open subset $U\subset M$ and a homeomorphism $\Psi:U\rightarrow\mathbb{R}^n$.
2. There are open subsets $U^\prime\subset M$, $V\subset\mathbb{R}^n$ and a homeomorphism $\varphi:U^\prime\rightarrow V$.
3. There is an open subset $U^{\prime\prime}\subset M$, an open ball $B\subset \mathbb{R}^n$ and a homeomorphism $\phi:U^{\prime\prime}\rightarrow B$.

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• 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.

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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?

Answer 1

You had a good idea because there is always a homeomorphism between $R^n$ and the open ball of $ R^n$. Think of the one-dimensional case in which you would like to map $R$ to $(-1,1)$: In this case the good map can be $f(x)=\frac{2}{\pi}arctan(x)$ . Oviously you can generalize this concept to a higher size.

Answer 2

Since any open subset $B\subset \mathbb{R} ^n$ is homeomorphic to $\mathbb{R} ^n$ by some $\phi^\prime$ we can define the map $\Psi:=\phi^\prime\circ\phi$. As a composition of homeomorphisms it is itself a homeomorphism.

The proposition is therefore valid, q.e.d..