Fundamental group of $\mathbb{R^\times}/\sim$ This is not a homework problem. I was curious about the fundamental group of  $X:=\mathbb{R}^\times /\sim$, where $\sim$ is generated by the relation $x \sim x^2+1$.
My attempts where to try to consider the "easier" case $Y:=\mathbb{R}^\times /\sim$, where $\sim$ is generated by $x\sim x^2$.
I think I managed to show that $Y$ is homotopy equivalent to a point, so $\pi_1(Y,y_0)=0$ for all $y_0 \in Y$.
Next I managed to construct a fairly complicated injective continous map $f:X\to Y$, however this didn't get me anywhere. Any help would be appreciated.
I'm also interested in the spaces you get, when you replace $\mathbb{R}$ with $\mathbb{C}$.
 A: Let's do a bit of analysis on $\sim$.
First, define $f : (0, \infty) \to (1, \infty)$ by $f(x) = x^2 + 1$. Note that $f$ is a bijection (in fact, a diffeomorphism).
Given $x, y \in (0, \infty)$, we say $x \sim' y$ iff either $\exists n \in \mathbb{N} (x = f^n(y))$ or $\exists n \in \mathbb{N} (y = f^n(x))$ (we allow $n = 0$).
Note that $\sim'$ is an equivalence relation.
I claim that for all $x, y \in \mathbb{R}^\times$, we have $x \sim y$ iff $|x| \sim' |y|$.
To prove this, first note that it is clear that $x \sim |x|$ for all $x$, since $x \sim x^2 + 1 = |x|^2 + 1 \sim |x|$.
Furthermore, it's clear that if $|x| \sim' |y|$ then $|x| \sim |y|$, and we would thus have $x \sim |x| \sim |y| \sim y$.
To go the other direction, we note that the relation $x \sim'' y :\equiv |x| \sim' |y|$ is an equivalence relation such that for all $x$, $x \sim'' x^2 + 1$. Therefore, $\sim \subseteq \sim''$.
Thus, in particular, it is clear that $\mathbb{R}^\times / \sim$ is homeomorphic to $\mathbb{R}^+ / \sim'$ (under the map induced by $x \mapsto |x| : \mathbb{R}^\times \to \mathbb{R}^+$.
With a bit of cleverness, it is possible to show that $\mathbb{R}^+ / \sim'$ is homeomorphic to the circle and thus has fundamental group $\mathbb{Z}$. I'll leave that as an exercise.
Edit: apparently, it is not sufficiently clear that $\mathbb{R}^+ / \sim' \cong \mathbb{S}^1$, so I will provide an explicit proof.
The definition we will use for $\mathbb{S}^1$ is the topological quotient group $\mathbb{R} / \mathbb{Z}$. Consider the quotient maps $\pi_1 : \mathbb{R}^+ \to \mathbb{R}^+ / \sim'$ and $\pi_2 : \mathbb{R} \to \mathbb{R} / \mathbb{Z}$. Recall that $\pi_2$ is an open map.
Consider the function $g : \mathbb{R}^+ \to \mathbb{R}$ defined by
$$g(x) = \begin{cases}
  g(x^2 + 1) - 1 & x < 1 \\
  x & 1 \leq x \leq 2 \\
  g(\sqrt{x - 1}) + 1 & 2 < x
\end{cases}$$
Note that $g$ is a continuous open map and that $g(f(x)) = g(x) + 1$. Thus, the range of $g$ is $\mathbb{R}^+$. We see that $g$ lifts uniquely to a continuous map $\bar{g} : \mathbb{R}^+ / \sim' \to \mathbb{R} / \mathbb{Z}$ such that $\bar{g} \circ \pi_1 = \pi_2 \circ g$.
Clearly, $\bar{g}$ is a bijection. For given $\pi_2(w) \in \mathbb{R} / \mathbb{Z}$, we can find some $y \in \mathbb{R}^+$ such that $y - w \in \mathbb{Z}$, and we can find some $x \in \mathbb{R}^+$ such that $g(x) = y$; then $\bar{g}(\pi_1(x)) = \pi_2(g(x)) = \pi_2(y) = \pi_2(w)$; this proves that $g$ is surjective. On the other hand, we can show that $g(f^n(x)) = g(x) + n$. Since $g$ is injective, this means that $g(a) = g(b) + n$ iff $a = f^n(b)$ for all $n \in \mathbb{N}$. This is enough to show that $\bar{g}$ is also injective; if $\bar{g}(\pi_1(a)) = \bar{g}(\pi_1(b))$, then $\pi_2(g(a)) = \pi_2(g(b))$; WLOG, take $g(a) - g(b) \in \mathbb{N}$, so we see that $a = f^n(b)$ and thus $a \sim' b$, and thus $\pi_1(a) = \pi_1(b)$.
Finally, we can show that $\bar{g}$ is an open map. First, note that since both $g$ and $\pi_2$ are open maps, $\pi_2 \circ g = \bar{g} \circ \pi_1$ is an open map. Now suppose that $U \subseteq \mathbb{R}^+ / \sim'$ is open. Then since $\pi_1$ is surjective, we see that $U = \pi_1(\pi_1^{-1}(U))$. Then $\bar{g}(U) = \bar{g}(\pi_1(\pi_1^{-1}(U))$. Since $\pi_1^{-1}(U)$ is open and $\bar{g} \circ \pi_1$ is an open map, we see that $\bar{g}(U)$ is also open.
Thus, $\bar{g}$ is a homeomorphism.
A: First notice that $(0,1]$ represents all equivalence classes. Now define $f:\mathbb{R}^\times\to \mathbb{R}$ in the natural way. For example $f$ is the identity on $(0,1]$, $f=\sqrt{x-1}+1$ on $(1,2]$, $f=...$ on $(2,5]$ etc... Now consider $\mathbb{R}^\times\to \mathbb{R}\to\mathbb{R}/\mathbb{Z}$ which factors over $\mathbb{R}^\times/\sim$ and the resulting map is bijective. Since all above maps are open and continuous we have that the quotient space is homeomorphic to the circle.
The function $f$ basically stretches the two copies of $\mathbb{R}$ such that two equivalent points are an integer distance apart.
A: First, notice that
$$-x \sim (-x)^2 + 1 = x^2 + 1 \sim x.$$
So, $\mathbb R^\times / \sim$ is the same space as $\mathbb R^+ / \sim$.  Now, divide up $\mathbb R^+$ into intervals in the following way:
$$\mathbb R^+ = (0, 1] \cup [1, 2] \cup [2, 5] \cup [5, 26] \cup \cdots.$$
Each interval is found by applying $x \mapsto x^2 + 1$ on the previous interval.  Define a function $f_n$ recursively on the $n$th closed interval in the following way:
$$f_0 : [1, 2] \to [1, 2] \text{ is the identity},$$
$$f_n : [a, b] \to [1, 2] \text{ is defined by } f_n(x) = f_{n-1}(\sqrt{x - 1}).$$
Also, define $f_{-1} : (0, 1] \to [1, 2]$ by $f_{-1}(x) = x^2 + 1$.
Each $f_n$ is continuous and is the maps elements of an equivalence class to other elements of that same equivalence class.  Thus, by the pasting lemma, there is a continuous map $f : \mathbb R^+ \to [1, 2]$ that preserves equivalence classes.  So, $\mathbb R^+ / \sim$ is homeomorphic to $[1, 2] / \sim$, which is obviously the circle.
To answer your original question,
$$\pi_1(\mathbb R^\times / \sim) = \pi_1(\mathbb S^1) = \mathbb Z.$$
