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I'm trying to solve the following problem:

Let U be a connected open set, $U \subset \mathbb{R^2}$, and consider $p_1,p_2,...,p_n \in U$ and $q_1,q_2,...,q_n \in U$. Show that exists an homeomorphism $\varphi: U \rightarrow U$ such that $\varphi(p_i)=q_i$.

This comes from a basic algebraic topology book.I tried to find a contradiction using the fact that the fundamental group of $U\setminus \{p_i\}$ is the free group on $n$ generators but I'm not concluding. I know that an open connected subset in $\mathbb{R}^2$ is also path connecting but the fundamental group approach is not working.

Any hints would be greatly appreciated.

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  • $\begingroup$ Is is not true that the fundamental group of $U \backslash \{p_i\}$ is the free group on $n$ generators. There are two problems: 1. you only remove one point, 2. $U$ is not assumed to be simply connected. What you can say is that if you remove $n$ points of $U$ there is a copy of the free group on $n$ generators in the fundamental group of this space. $\endgroup$ – user60589 Nov 26 '15 at 15:08
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Let $\varepsilon >0$ and let $x\in \mathbb{R}^N$, then for any $y\in B_{\varepsilon}(x)$ you have a homeomorphism $$ \tau_y^{x,\varepsilon} \colon B_{\varepsilon} (x) \to B_{\varepsilon} (x) $$ which maps $x$ to $y$ and extends to the identity on the boundary, hence is can be extended to $\mathbb{R}^N$ and we will call this map $T_y^{x,\varepsilon}$

Let $H$ be the set of homeomorphisms $f\colon U \to U$ that map $p_i$ to $q_i$ for $i= 1, \dots ,n-1$ and let $U'=\{ q \in U \mid f(p_n)=q, \ f \in H \}$. Then show that $U'$ is a open and closed subset of $U\backslash \{ q_1, \dots , q_{n-1}\}$. ($U\backslash \{ q_1, \dots , q_{n-1}\}$ is connected since $N >0$).

Let $q \in U'$ then choose $\varepsilon >0$ such that $q_i\notin B_{\varepsilon}(q)\subset U'$ for $i=1, \dots, n-1$ and for any $q' \in B_{\varepsilon}(q)$ we then know that $q' \in U'$ since $q' = T_{q}^{q', \varepsilon} f(p_n) $ for some $f \in H$ and $T_{q}^{q', \varepsilon}\circ f \in H$. This shows that $U'$ is open.

Similar for $q\notin U'$ we find a neighborhood $N$ of $q$ such that $N\cap U'= \emptyset$. Thus $U'$ is closed.

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