Does $A$ homeomorphic to $B$ imply $f^{-1}(A)$ is homeomorphic to $f^{-1}(B)$? Let $X$ and $Y$ be topological spaces, and $f:X\to Y$ a continuous map. Is the following true: 

If  $A$ and $B$ are two homeomorphic subspaces of $Y,$ then $f^{-1}(A)$ and $f^{-1}(B)$ are homeomorphic subspaces of $X$.

 A: This is false. 
Let $C$ and $D$ be any two non-homeomorphic topological spaces, let $X=C\coprod D$ with the disjoint union topology, let $Y=\{0,1\}$ with the discrete topology, let $A=\{0\}$ and $B=\{1\}$, and define the function $f:X\to Y$ by 
$$f(x)=\begin{cases}0\text{ if }x\in C,\\ 1\text{ if }x\in D.\end{cases}$$
Then $f$ is continuous, and $A$ is homeomorphic to $B$, but $f^{-1}(A)=C$ is not homeomorphic to $f^{-1}(B)=D$.

This is false even if we assume that $Y=X/G$ where $G$ is a topological group acting continuously on $X$, and $f:X\to Y$ is the quotient map. For example:
Let $D=\{0,1\}$ with the discrete topology, let $C$ be any non-empty space, let $X=C\coprod D$ with the disjoint union topology, let the group $G=\mathbb{Z}/2\mathbb{Z}=\{\overline{0},\overline{1}\}$ with the discrete topology act on $X$ by 
$$\begin{align*}\overline{0}\cdot x&=x\text{ for all }x\in X,\\\\
\overline{1}\cdot x&=\begin{cases}x\text{ if }x\in C,\\
0\text{ if }x=1,\\1\text{ if }x=0,\end{cases}\end{align*}$$
which is continuous. Then $X/G\cong C\coprod\{\star\}$, where $\star$ represents the orbit $D$ of the action of $G$. For any $c\in C$, we have that $A=\{c\}$ and $B=\{\star\}$ are homeomorphic, but $f^{-1}(A)=\{c\}$ and $f^{-1}(B)=D$ are not homeomorphic.
A: As an even simpler counterexample, take $f:\mathbb R\to\mathbb R$ with the usual topology:
$$f(x)= x^2$$
and take $A=[1,2]$ and $B=[0,1]$.
$A$ and $B$ are obviously homeomorphic, but $f^{-1}(A)$ consists of two separate intervals $[-\sqrt 2,-1]\cup[1,\sqrt 2]$ but $f^{-1}(B)$ is the single interval $[-1,1]$. Since one of the preimages is connected but the other isn't, they cannot be homeomorphic.
