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Let $X$ and $Y$ be Hausdorff topological spaces, and let $f : X \to Y$ be a Borel-measurable function. Suppose that $D \subseteq X$ is dense, that the image $f(D) \subseteq Y$ is dense, and that $f$ is a homeomorphism from $D$ to $f(D)$.

Does it follow that $f$ is a homeomorphism from $X$ to $Y$?

If not, is $f$ a measurable isomorphism from $X$ to $Y$?

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    $\begingroup$ The inclusion $f: (0, 1] \to [0, 1]$ with $D = X$?? $\endgroup$
    – Todd Trimble
    Commented Oct 13, 2013 at 8:36
  • $\begingroup$ @ToddTrimble: thanks, that's a clean counterexample. $\endgroup$
    – Tom LaGatta
    Commented Oct 13, 2013 at 8:38
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    $\begingroup$ Even continuity is not enough to deduce homoemorphism: look at the inclusion of a dense interval to the circle. For the second question, look at the inclusion of the rational numbers in the set of real numbers. $\endgroup$
    – Benoît Kloeckner
    Commented Oct 13, 2013 at 8:39
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    $\begingroup$ Something is strange: take $X=Y=\mathbb R$, $D=\mathbb Q$. Put $f(x)=x$ on $D$ and $f(x)=\sqrt 2$ on $X\setminus D$. I'm pretty sure you didn't have this idiotic example in mind, so what conditions are missing? $\endgroup$
    – fedja
    Commented Oct 13, 2013 at 8:39
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    $\begingroup$ Take the Fourier transform from $L^p$ on the circle to $\ell^q$ ($p<2<q=\frac{p}{p-1}$) then. Looks like you'll need to disclose everything before we run out of stupid counterexamples :-). $\endgroup$
    – fedja
    Commented Oct 13, 2013 at 8:58

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