Homeomorphism between $\mathbb{R}^n$ and $\mathbb{R}^n$ I am trying to solve a problem relating manifolds. I was trying to define charts and the homeomorphisms, so I want to know if the map defined by $P_{\sigma}:\mathbb{R}^n \to \mathbb{R}^n$ defined by $$P_{\sigma}(x_1,x_2 \dots x_n)=(x_{\sigma (1)},x_{\sigma (2)} \dots x_{\sigma (n)} )$$
is a homeomorphism or not for any $\sigma \in S_n$.
The map $P_{\sigma}$ is clearly bijective. If it is continuous, then we are done because we can see $$(P_{\sigma})^{-1}=P_{\sigma^{-1}}$$
So I have issues with continuity only. I think it is going to be continuous. if $U \subset \mathbb{R}^n $ then $P_{\sigma}^{-1}(U)$ is an open set. Probably because for any $x \in P_{\sigma}^{-1}(U) \Rightarrow P_{\sigma}(x) \in U$ and we can use an open set around $P_{\sigma}(x)$ to get the open set around $x$.
 A: it's linear and all linear maps are continuous with respect to the metric topology on $\mathbb{R}^n$.
To see linear maps are continuous, let $T:\mathbb{R}^n\to\mathbb{R}^n$ be linear. Given any $x,y\in\mathbb{R}^n$
$$\|Tx-Ty\|\leq\|T(x-y)\|\leq\|T\|\|x-y\|$$
here, $\|T\|$ is the norm of the map.  Continuity follows since we can take $\delta<\frac{\epsilon}{\|T\|}$.
Now, try and convince yourself that the map you defined is linear and you are done.
As a note, the matrix of the map you wrote down is an example of what we call a permutation matrix.  These are interesting for one reason since they provide a representation of the symmetric group.
A: Your map is an isometry, so it is continuous:
$$\begin{array}{rcl}d(P_\sigma(x), P_\sigma(y))&=&\sqrt{(y_{\sigma(1)}-x_{\sigma(1)})^2+(y_{\sigma(2)}-x_{\sigma(2)})^2+\ldots+(y_{\sigma(n)}-x_{\sigma(n)})^2}\\ &=&\sqrt{(y_1-x_1)^2+(y_2-x_2)^2+\ldots+(y_n-x_n)^2}\text{  (the terms reordered)}\\&=&d(x,y)\end{array}$$
for any $x=(x_1,x_2,\ldots,x_n)\in\mathbb R^n$, $y=(y_1,y_2,\ldots,y_n)\in\mathbb R^n$ and $\sigma\in S_n$.
