Prove that there are exactly 2n symmetries of a regular sided polygon, $P_n$ .

Question

I'm going through my lecture notes and didn't really understand a section of the proof for showing there are exactly $2n$ symmetries of a regular sided polygon. The proof states: It suffices to show that any symmetry of $P_n$ maps adjacent vertices to adjacent vertices. Isometries are injective. Take $g \in Sym(P_n)$ and claim the boundary of $P_n$ denoted by $\delta P_n$ is mapped to itself, i.e $g(\delta P_n) \subseteq \delta P_n$. If this is not true, then there is $x \in \delta P_n, \rho > 0$ such that $D_\rho(x)$ (disc centred at $x$ with radius $\rho$) is mapped inside $P_n$. But then $g(P_n) = P_n$ implies that for any $y \in D_\rho(x) \setminus P_n, g(y)$ has at least two preimages under $g$ contradicting injectivity of the isometry $g$. The proof then continues, but I understand the rest of it. My main difficulty is understanding the proof by contradiction to show $g(\delta P_n) \subseteq \delta P_n$, and would really appreciate someone walking me through it (I can't really seem to visualise what's going on. Thanks!

Answer 1

Let me reformulate the proof, hopefully it will be clearer.

Since $g$ is bijective, $g$ not only maps $P_n$ to $P_n$, it also maps $\mathbb{R}^2\setminus P_n$ to $\mathbb{R}^2\setminus P_n$. If $x\in\delta P_n$, a neighbourhood of $x$ (say $D_\rho(x)$) contains both elements of $P_n$ and elements of $\mathbb{R}^2\setminus P_n$, so that $g(D_\rho(x))$ also contains both elements of $P_n$ and elements of $\mathbb{R}^2\setminus P_n$. But if $g(x)\notin \delta P_n$, one can choose $\rho$ small enough so that $g(D_\rho(x))$ is strictly contained in $P_n$: contradiction.