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Let $X$ be a topological space. Does $f\colon X\to X$ with $f^{-1}(U)=U$ for all $U$ open imply $f$ is homeomorphism?

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  • $\begingroup$ To clarify, the condition is, $f^{-1}(U)=U$ for every open set $U \subseteq X$? $\endgroup$ – Mike F Nov 21 '15 at 3:59
  • $\begingroup$ Do you mean for all $U$? $\endgroup$ – copper.hat Nov 21 '15 at 3:59
  • $\begingroup$ Yes, for all $U$ open $\endgroup$ – Qixiao Nov 21 '15 at 4:01
  • $\begingroup$ For a Hausdorff topology (or even for some weaker separation axioms), this condition implies $f = \mathrm{id}_X$, which is certainly a continuous function. Do you see why? $\endgroup$ – Mike F Nov 21 '15 at 4:03
  • $\begingroup$ @MikeF I believe the relevant requirement is "sober". (T2 $\implies$ sober $\implies$ T0) $\endgroup$ – JustAskin Nov 21 '15 at 4:05
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This is true if and only if $X$ is $T_0$, in which case any such $f$ is the identity. First, suppose $X$ is $T_0$. For each $x\in X$, let $U_x$ be the intersection of all open sets containing $x$. If $f:X\to X$ is such that $f^{-1}(U)=U$ for all open sets $U$, it follows that $f^{-1}(U_x)=U_x$ for each $x$. For any $x\in X$, $f(x)\in U_{f(x)}$, so $x\in f^{-1}(U_{f(x)})=U_{f(x)}$. But also $x\in U_x=f^{-1}(U_x)$, so $f(x)\in U_x$. That is, $x$ is in every open set containing $f(x)$, and $f(x)$ is in every open set containing $x$. Since $X$ is $T_0$, this implies $x=f(x)$. Thus $f$ is the identity.

Conversely, suppose $X$ is not $T_0$; say there are distinct points $x,y\in X$ such that an open set contains $x$ iff it contains $y$. Define $f:X\to X$ by $f(z)=z$ if $z\neq y$, and $f(y)=x$. Then it is easy to see that $f^{-1}(U)=U$ for all open $U$, but $f$ is not a homeomorphism because it is not a bijection. More generally, you can see that $f:X\to X$ satisfies $f^{-1}(U)=U$ for all open $U$ iff $f(x)\sim x$ for all $x\in X$, where $\sim$ is the "topologically indistinguishable" equivalence relation.

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No. Let $X$ be any topological space with the indiscrete topology and more than one element. Let $a$ be one of those elements. Then define $f(x)=a$ for all $x\in X$.

Since $X$ has the indiscrete topology the only open sets are $X$ and $\emptyset$.

Therefore, $f^{-1}(X)=X$ and $f^{-1}(\emptyset)=\emptyset$ but since $f$ maps more than one element to one element it cannot be a homeomorphism.

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  • $\begingroup$ Constant maps are always continuous, regardless of the topology chosen on the domain and target spaces... $\endgroup$ – Mike F Nov 21 '15 at 4:07
  • $\begingroup$ @MikeF Yes, but a homeomorphism requires more than continuity. $\endgroup$ – John Douma Nov 21 '15 at 4:07
  • $\begingroup$ Yes, sorry, I just looked back and saw $f$ was supposed to be a homeomorphism and not merely a continuous map.... I'm a bit curious whether such $f$ can be discontinuous. I suspect it is not possible. Thoughts? $\endgroup$ – Mike F Nov 21 '15 at 4:08
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    $\begingroup$ Hi again. To answer my own question, of course such an $f$ is continuous. Indeed, we are demanding $f^{-1}(U) =U$ for all open sets $U$ and so, in particular, the inverse image of every open set is open. Duh.... $\endgroup$ – Mike F Nov 21 '15 at 4:12
  • $\begingroup$ @MikeF It has to be continuous because if $f^{-1}(U)=U$ then $f^{-1}(U)$ is open. Also, I think your point about Hausdorff spaces above implies that the theorem would be true if we required the topology to be Hausdorff. $\endgroup$ – John Douma Nov 21 '15 at 4:12
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No. Consider this topology $\{\{\Phi\},\{1\},\{1,2,3\}\}$ on $X=\{1,2,3\}$ and $f:X\rightarrow X$ by $f(1)=1,f(2)=3,f(3)=3$.

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