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I am having trouble figuring out the following question (3.10 in Kechris, Classical Descriptive Set Theory): If $X$ is completely metrizable, and $A\subseteq X$ with $f:A\to A$ a homeomorphism, then there is a $G_\delta$ set $G\subseteq X$ containing $A$ and an extension $h:G\to G$ of $f$ which is a homeomorphism of $G$.

Lavrentiev's Theorem gets us $G_\delta$ sets $G'$ and $H'$ containing $A$, and a homeomophism $g:G'\to H'$ extending $f$, but it doesn't seem to me that the proof of that result can be easily adapted to make $G'=H'$. The key fact in all of this is the set $\bar{A}\cap\{x:\mathrm{osc}_f(x)=0\}$, for any continuous $f:A\to X$, is a $G_\delta$ set in $X$ on which we can continuously extend $f$.

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Can't you simply work with $(f,f^{-1})\colon A \times A \to A \times A$ to make the situation symmetric modulo switching coordinates? –  t.b. Jun 1 '12 at 23:23
    
@t.b. I don't see what you mean by "modulo switching coordinates". Could you explain? –  Iian Smythe Jun 2 '12 at 1:17
    
I was a bit quick, sorry. The idea was that in the proof of Lavrentiev's theorem you intersect two graphs to get the mutually inverse homeomorphisms. I mistakenly assumed that you automatically get a symmetric situation when applying this to the mutually inverse homomorphisms $(f,f^{-1})$ and $(f^{-1},f)$ from $A\times A \to A \times A$. Unfortunately, this isn't quite the case and it seems that you need to perform a Schroeder-Bernstein type argument as given by Brian below to make this idea work. –  t.b. Jun 2 '12 at 10:15
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By Lavrentiev’s theorem $f$ can be extended to a homeomorphism $f_0:G_0\to H_0$, where $G_0$ and $H_0$ are $G_\delta$-sets in $X$ containing $A$. Let $G_1=G_0\cap H_0$, and let $H_1=f_0[G_1]$; clearly $G_1$ is a $G_\delta$ in $X$. Moreover, $G_1$ is a $G_\delta$ in $G_0$, and $f_0$ is a homeomorphism, so $H_1$ is a $G_\delta$ in $H_0$ and therefore in $X$. Finally, $f_1\triangleq f_0\upharpoonright G_1:G_1\to H_1$ is a homeomorphism extending $f$. Now let $H_2=G_1\cap H_1$, $G_2=f_1^{-1}[H_2]$, and $f_2=f_1\upharpoonright G_2$; $G_2$ and $H_2$ are $G_\delta$-sets containing $A$, and $f_2$ is a homeomorphism between them extending $f$.

Continue in this fashion. If $n\in\omega$ is even, $G_{n+1}=G_n\cap H_n$ and $H_{n+1}=f_n[G_{n+1}]$, while if $n$ is odd, $H_{n+1}=G_n\cap H_n$ and $G_{n+1}=f_n^{-1}[H_{n+1}]$, and $f_{n+1}=f_n\upharpoonright G_{n+1}$ in both cases. The sets $G_n$ and $H_n$ are $G_\delta$-sets in $X$, and each $f_n$ is a homeomorphism from $G_n$ onto $H_n$ extending $f$. Note that $$H_0\supseteq G_1\supseteq H_2\supseteq G_3\supseteq H_4\supseteq\dots\;.$$

Now let $$G=\bigcap_{n\in\omega}G_n=\bigcap_{n\in\omega}H_n\qquad\text{ and }\qquad\bar f=\bigcap_{n\in\omega}f_n=f_0\upharpoonright G\;;$$ clearly $G$ is a $G_\delta$ containing $A$, and $\bar f$ is a homeomorphism of $G$ onto $$\bar f[G]=\bigcap_{n\in\omega}f_n[G_n]=\bigcap_{n\in\omega}H_n=G\;.$$

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I thought that some kind of dovetailing like this might be involved. Thanks! –  Iian Smythe Jun 3 '12 at 3:04
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