Is the $\Bbb{R}^2\setminus\{P_1,...,P_n\}$ plane homeomorphic to $\Bbb{R}^2\setminus\{Q_1,...,Q_n\}$? Let $K=\{P_1,...,P_n\}\in\Bbb{R}^2$ and let $\Bbb{R}^2\setminus K$ be the plane with $n$ holes. Show that for any $K':=\{Q_1,...,Q_n\}$, $\Bbb{R}^2\setminus K \cong \Bbb{R}^2\setminus K'$.
As I understood it, I may need to find an $f:X \to Y$ that is a homeomorphism. At first I thought of an $f$ that sends each $(x,y)\in\Bbb{R}^2\setminus(K\cup K')$ to $(x,y)$, and, assuming WLOG that $K\cap K'=\emptyset$, $f(Q_i)=P_i$. But I see this doesn't make a continuous map.
Then I thought I should take $f(x,y)=(x,y)+(x_{P_1}-x_{Q_1},y_{P_1}-y_{Q_1})$ and try to show that it is surjective by setting $(x_i,y_i)=(x_{P_i}-x_{P_1}+x_{Q_1},x_{P_i}-y_{P_1}+y_{Q_1})$ for $i\neq1$ and that way achieve $f(x_i,y_i)=P_i$. But what if $(x_i,y_i)=P_j$ where $f$ is not defined? I am confused. Maybe I didn't get it at all. I could use some help here.
 A: As John Samples comments, it suffices to find a homeomorphisms $h : \mathbb R^2 \to \mathbb R^2$ such that $h(P_i) = Q_i$. This is even a stronger result than $h(K) = K'$ which would also show that  $\Bbb{R}^2\setminus K \cong \Bbb{R}^2\setminus K'$.
Note that it suffices to consider the special points $Q_i = (i,0) \in X = \mathbb R \times \{0\} $.
We can do it by induction. As the base case we can take $n = 0$ (nothing to prove). But also for $n =1$ it is obvious: Take a translation shifting $P_1$ to $Q_1$.
Now assume that we have $h : \mathbb R^2 \to \mathbb R^2$ such that $h(P_i) = Q_i$ for $i \le n$. In other words, we may assume w.l.o.g. that $P_i = Q_i$ for $i \le n$. For $(a, b) \in \mathbb R^2$ and $r > 0$ let us define $$\psi_{a,b,r} : \mathbb R^2 \to \mathbb R^2, \psi_{a, b,r}(x,y) =  \begin{cases} (x,y) & \lvert x - a  \rvert \ge r \\ (x,y + \frac{(\lvert x - a \rvert - r)b }{r}) & \lvert x - a  \rvert \le r \end{cases}$$
This is well-defined continuous map which shifts $(a,b)$ to $(a,0)$ and $(a,0)$ to $(a,-b)$ and moreover keeps all points $(x,0) \in X$ with $\lvert x - a \rvert \ge r$ fixed. It is a homeomorphism whose inverse is $\psi_{a,-b,r}$ (since $\psi_{a,b,r} \circ \psi_{a,-b,r} = id$ and $\psi_{a,-b,r} \circ \psi_{a,b,r} = id$).

*

*$P_{n+1} \notin X$.
Write $P_{n+1} = (a, b)$ with $b \ne 0$. There is a unique linear map $\phi : \mathbb R^2 \to \mathbb R^2$ such that $\phi(1,0) = (1,0)$ and $\phi(a,b) = (n+1,b)$ since $(1,0)$ and $(a,b)$ form a basis of $\mathbb R^2$. Clearly $\phi$ is a linear isomorphism, hence a homeomorphism. It keeps $X$ fixed. Then $h_{n+1} = \psi_{n+1,b,1} \circ \phi$ keeps  the $P_i$ with $i \le n$ fixed and maps $P_{n+1}$ to $Q_{n+1}$.


*$P_{n+1} \in X$.
Write $P_{n+1} = (a, 0)$. We have $P_{n+1} \notin \{P_1,\dots,P_n\}$, hence there exists $r > 0$ such that for $\lvert x - a  \rvert \le r$ we have $(x,0) \notin \{P_1,\dots,P_n\}$. Hence $\psi_{a,1,r}$ keeps the $P_i$ with $i \le n$ fixed and shifts $P_{n+1}$ to $(a,-1) \notin X$. Now proceed as in 1.
By the way, this proof shows that $h$ can always be chosen to be orientation preserving.
