For a practice-exam exercise I am trying to understand why $X/ \mathbb{Z}$ is homeomorphic to $S^1$. Here, $X = (-1,\infty)$, and $\mathbb{Z}$ is acting as an additive group on $X$ with the action:
$\phi_n: X \to X$, $\phi_n(t)= 2^n(t+1) -1$ .
I have tried to find a surjective continuous function $f:X \to S^1$ such that $f(x) = f(y) \iff y = \phi_n(x)$ for some $n$. This would then imply the existence of an bijective continuous function from $X/\mathbb{Z}$. Then I would use that $X/\mathbb{Z}$ is Hausdorff (since $X$ is) and $S^1$ is compact, so $f$ is a homeomorphism.
However, I've been staring at this for an hour and still haven't found an explicit $f$. I hope someone can help!
Suggestion: Put $Y = (0, \infty)$ and define $t:X \to Y$ by $t(x) = x+1$. Now look at the action of the integers on $Y$ by $(n, y) \mapsto 2^{n}y$.
Or (motivated by the preceding paragraph), define $\psi:X \to \mathbf{R}$ by $\psi(x) = \log_{2}(1 + x)$, and convince yourself that $\phi_{n}$ corresponds (under conjugation by $\psi$) to translation by $n$. :)