Involution on Cantor space with exactly one fixed point Let $X=\{0,1\}^{\mathbb{N}}$ be the Cantor space. What is an example of a continuous map $\sigma : X \to X$ with $\sigma^2=\mathrm{id}$ and $\# \{x \in X : \sigma(x)=x\} = 1$?
This has to exist, since $X$ is homeomorphic to $\mathbb{Z}_p$ by Brouwer's Theorem, and there we can take $x \mapsto -x$.
 A: Define $\sigma:(x_1,x_2,\dots)\mapsto(y_1,y_2,\dots)$ so that 
$y_n=x_n\Leftrightarrow x_1+x_2+\cdots+x_{n-1}=0$.
In words, flip everything after the first $1$.
A: Consider the homeomorphism from $\{0,2,3\}^{\mathbb{N}}$ to $\{0,1\}^{\mathbb{N}}$
 mapping a sequence $(a_n)$ to the concatenation of binary writings of the $a_n$'s.  
A: $\newcommand\N{\mathbb{N}}$Let $p=3$ for simplicity (what follows could work for any $p$, not that we really need it). First we define a homeomorphism $f : \{0,1,2\}^\N \to \{0,1\}^\N$. $f(x)$ is defined by concatenating the $\varphi(x_n)$ where:
$$\varphi(u) = \begin{cases}
(0) & u = 0 \\
(1,0) & u = 1 \\
(1,1) & u = 2
\end{cases}$$
This is continuous: we need to check that the preimage of $\{x | x_n = k \}$ (where $k,i$ are fixed) is open, but there's only a finite number of beginning of sequences satisfying that. The inverse map is defined "inductively": if we see a zero we add a zero to the image, otherwise if we see $1*$ we add either $1$ or $2$. The map is therefore bijective, and since both spaces are compact (Tychonoff) and Hausdorff, this is a homeomorphism.
Now the map $\{0,1,2\}^\N \to \mathbb{Z}_3$ defined by $x \mapsto \sum_n x_n 3^n$ is a homeomorphism (it's easy, I won't prove it but I won't actually need it), which lets us intuit the involution. Define $\sigma : \{0,1,2\}^\N \to \{0,1,2\}^\N$ by:
$$\sigma(x)_n = \begin{cases}
0 & x_n = 0 \\
2 & x_n = 1 \\
1 & x_n = 2
\end{cases}$$
This corresponds to $y \mapsto -y$ on $\mathbb{Z}_3$. Then $\sigma$ is a self-homeomorphism (bijective is obvious, each component map is the composite of a projection and $(12) \in \mathfrak{S}_2$, finally both spaces are compact Hausdorff so it's a homeomorphism), and the zero sequence is the only fixpoint. Finally $f \sigma f^{-1}$ is the involution we wanted.
