Images of functions with the same sets of fibers Let $f,g:\mathbb{R}\to\mathbb{R}$ be two continuous functions with the same sets of fibers, i.e., $\{f^{-1}[\{y\}]:y\in\mathbb{R}\}=\{g^{-1}[\{y\}]:y\in\mathbb{R}\}$. Does it follow that their images $\{f(x):x\in\mathbb{R}\}$, $\{g(x):x\in\mathbb{R}\}$ are homeomorphic?
 A: Let $Y_f=f[\Bbb R]$ and $Y_g=g[\Bbb R]$. Let $h:Y_f\to Y_g$ be defined so that $\{h(y)\}=\left\{g\left[f^{-1}[\{y\}]\right]\right\}$, i.e., so that the $f$-fibre of $y$ is the $g$-fibre of $h(y)$. Clearly $h$ is a bijection, and it suffices to prove that $h$ is continuous, since $h^{-1}$ has the same definition with $f$ and $g$ interchanged. Say that a subset of $\Bbb R$ is saturated if it is a union of fibres.
Suppose that $y\in Y_f$ and $\langle y_n:n\in\Bbb N\rangle$ is a sequence in $Y_f$ converging to $y$. Let $F=f^{-1}[\{y\}]$, and for $n\in\Bbb N$ let $F_n=f^{-1}[\{y_n\}]$. Then every saturated open nbhd of $F$ contains all but finitely many of the fibres $F_n$. But this is precisely what it means for $\langle h(y_n):n\in\Bbb N\rangle$ to converge to $h(y)$ in $Y_g$, so $h$ is continuous, and $Y_f$ and $Y_g$ are homeomorphic.
Added: That last step may require a little more justification. If $\langle h(y_n):n\in\Bbb N\rangle$ does not converge to $h(y)$, there is an open nbhd $U$ of $h(y)$ such that $M=\{n\in\Bbb N:h(y_n)\notin U\}$ is infinite. But then $F_n\cap g^{-1}[U]=\varnothing$ for each $n\in M$, and $g^{-1}[U]$ is therefore a saturated open nbhd of $F$ that misses infinitely many of the fibres $F_n$. This is impossible, so $\langle h(y_n):n\in\Bbb N\rangle\to h(y)$. (This actually shows that the maps $f$ and $g$ are necessarily quotient maps, and since they plainly correspond to the same decomposition of $\Bbb R$, their ranges $Y_f$ and $Y_g$ are necessarily homeomorphic.)
A: Let $f:\mathbb{R}\to\mathbb{R}$ be continuous, $y\in f[\mathbb{R}]$, and  $y_n\in f[\mathbb{R}]$ for every $n\in\mathbb{N}$. Denote $F=f^{-1}[\{y\}]$, $F_n=f^{-1}[\{y_n\}]$ for every $n$. Let us call a set $X\subseteq\mathbb{R}$ saturated if $X=f^{-1}[Y]$ for some $Y\subseteq\mathbb{R}$.
We prove that $y_n\to y$ if and only if every open saturated set containing $F$ contains all but finitely many $F_n$'s.
Let $U$ be an open saturated set such that $F\subseteq U$. Denote $M=\{n\in\mathbb{N}:F_n\nsubseteq U\}$. Clearly, $F_n\subseteq\mathbb{R}\setminus U$ for $n\in M$. Assume that $y_n\to y$. Then there exist $a,b$ such that $\{y_n:n\in\mathbb{N}\}\subseteq [f(a),f(b)]$. Without loss of generality assume $a<b$. For each $n$, find some $x_n\in [a,b]\cap F_n$ using Intermediate Value Theorem. If $M$ is infinite then, by compactness, the sequence $\langle x_n:n\in M\rangle$ has a limit point $x\in [a,b]\setminus U$, but then also $f(x)=\lim y_n=y$, a contradiction. So if $y_n\to y$ then $M$ must be finite.
If $y_n\nrightarrow y$ then there exist an open nbhd $V$ of $y$ such that $N=\{n\in\mathbb{N}:y_n\notin V\}$ is infinite. Then $U=f^{-1}[V]$ is an open saturated set such that $\{n\in\mathbb{N}:F_n\nsubseteq U\}=N$ is infinite, q.e.d.
Now, if $f,g$ are two continuous functions with the same sets of fibers then
$h:f[\mathbb{R}]\to g[\mathbb{R}]$, defined so that $g(x)=h(f(x))$ for every $x\in\mathbb{R}$, is a homeomorphism, since $y_n$ converges to $y$ in $f[\mathbb{R}]$ if and only if $h(y_n)$ converges to $h(y)$ in $g[\mathbb{R}]$. Moreover, both $f[\mathbb{R}]$ and $g[\mathbb{R}]$ are homeomorphic to the quotient space $X/E$, where $E$ is the equivalence relation defined by $xEy$ iff $f(x)=f(y)$, $X/E$ is the set of all equivalence classes, and the topology on $X/E$ consists of all $\mathcal{U}\subseteq X/E$ such that $\bigcup\mathcal{U}$ is open in $\mathbb{R}$.
