# 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 $X$.

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There's no reason to suspect a priori that this is true. If you expect two things to be homeomorphic, you probably have some particular map between them in mind that you expect is a homeomorphism. What map could that be in this case? – Qiaochu Yuan Feb 25 '12 at 1:10

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.

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what if $Y$ is a quotient space $Y=X/G$ from the action of a group $G$ on $X$ and $f:X\to X/G$ is the quotient map, do we have this propriety in this case? – palio Dec 7 '11 at 13:13
What assumptions are you making about the action? For example, is it a continuous action? – Zev Chonoles Dec 7 '11 at 13:18
i don't have any assymptions on the action but what happens if it is continuous ? – palio Dec 7 '11 at 13:19
@palio: I have constructed a counterexample even in the case that it is continuous, and added it to my answer. I am pretty sure that some sufficiently nice combination of conditions could force the property that you want, but I don't know which conditions off the top of my head. – Zev Chonoles Dec 7 '11 at 13:36

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.

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