# An open set in $\mathbb{R}^{2}$ is not homeomorphic to an open set in $\mathbb{R}^{n}$ for $n > 2$

I'm taking an introduction to algebraic topology course and this is one of the assigned problems. We have computed the fundamental group of the circle and we have proved that $S^{n}$ is simply connected for $n \geq 2$. This way we can prove the result if we replaced "set" with "ball". Indeed, an open ball in $R^{n}$ is homeomorphic to $R^{n}$, and $R^{2}$ can't be homeomorphic to $R^{n}$ for $n > 2$ because $R^{2} - \{x\}$ has non-trivial fundamental group while $R^{n} - \{y\}$ is simply connected.

So I have been trying to prove that a given homeomorphism $f : U \rightarrow V$, $U \subset R^{2}, V \subset R^{n}$ open sets, can be restricted to give a homeomorphism between balls. We start by taking a point $u \in U$. Take a ball $B_{1}$ around $u$. Since $f$ is continuous, there exists a ball $A_{1}$ around $v = f^{-1}(u)$ such that $f(A_{1}) \subset B_{1}$. Since $f^{-1}$ is continuous, there exists an open ball $B_{2}$ around $u$ such that $f^{-1}(B_{2}) \subset A_{1}$. Thus clearly $B_{2} \subset f(A_{1})$. However, I'm not sure how to proceed here. It just seems like I'll keep going forever, getting smaller and smaller balls.

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Let $A$ be a ball in $V$.
Then $A$ and $f^{-1}(A)$ will be homeomorphic and for $a\in A, \ A-\{a\}$ is simply connected while the fundamental group of $f^{-1}(A)-\{f^{-1}(a)\}$ is non-trivial (is $\pi_1\left(f^{-1}\left(A\right)\right)* \mathbb Z$).
To see this, suppose $\pi_1\left(f^{-1}(A)-\{f^{-1}(a)\}\right)=0$ to conclude that $\pi_1\left(\mathbb R^2-\{f^{-1}(a)\}\right)=0$ ↯.
But why is the fundamental group of $f^{-1}(A) - \{f^{-1}(a)\}$ non-trivial? –  Pedro Jan 27 '13 at 20:13
@Pedro Look at a small neighborhood of $f^{-1}(a)$: it is a disk minus its center. The fundamental group is the same as for the circle, that is, $\mathbb Z$. (Punctured disk can be retracted to a circle). –  user53153 Jan 27 '13 at 21:49
The fundamental group will be $\pi_1(f^{-1}(A)) \ast \mathbb Z$, not $\pi_1(f^{-1}(A)) \times \mathbb Z$. –  Alexander Thumm Jan 28 '13 at 7:53