Let $X$ be a Polish space. Let $H(X)$ be the set of homeomorphisms $h \colon X \to X$, equipped with the "evaluation $\sigma$-algebra", namely $\sigma(h \mapsto h(x) : x \in X)$.

(Note that for any measurable space $(\Omega,\mathcal{F})$, a map $p \colon \Omega \to H(X)$ is measurable if and only if $(\omega,x) \mapsto p(\omega)(x)$ is measurable. Note also that, given a countable dense set $\{x_i\}_{i \in \mathbb{N}} \subset X$, the evaluation $\sigma$-algebra on $H(X)$ is precisely the Borel $\sigma$-algebra of $H(X)$ regarded as a subset of $X^\mathbb{N}$ by the identification $h \mapsto (h(x_i))_{i \in \mathbb{N}}$.)

Is the map $h \mapsto h^{-1}$ measurable?

Remark. To know that the answer to the above is yes, it is sufficient just to know that $H(X)$ is a standard Borel space: for any $y \in X$, the graph of the map $h \mapsto h^{-1}(y)$ is measurable (since it is the preimage of $\{y\}$ under $(h,x) \mapsto h(x)$); so if $H(X)$ is a standard Borel space then $\mathrm{graph}(h \mapsto h^{-1}(y))$ is itself standard, and so [since injective measurable maps between standard Borel spaces map measurable sets onto measurable sets] the map $h \mapsto h^{-1}(y)$ is itself measurable.

(In any case, due to the measurable projection theorem, the above reasoning shows that $h \mapsto h^{-1}$ is universally measurable - but still, I am interested in actual measurability.)

The first obvious-seeming way to proceed would be to try and show directly that $H(X)$ is a measurable subset of $X^\mathbb{N}$ under the above identification. (This is indeed equivalent to the statement that $H(X)$ is a standard Borel space.)

  • $\begingroup$ Well, it seems to be true when $X$ is compact, but that case is quite easy because "homeomorphism" is equivalent to "continuous, injective, and dense image". $\endgroup$ – Nate Eldredge Apr 29 '17 at 2:07
  • $\begingroup$ Yes; if $X$ is compact then $H(X)$ is a Polish space under the compact-open topology, whose Borel $\sigma$-algebra will coincide with the evaluation $\sigma$-algebra. (All this is also true whenever $X$ has the property that every point admits a compact connected neighbourhood.) $\endgroup$ – Julian Newman Apr 29 '17 at 13:37
  • $\begingroup$ So, the obvious approach might be to proceed in steps: show that the set of continuous maps is Borel in $X^{\mathbb{N}}$, then the injective continuous maps, then the surjective continuous maps, then those with a continuous inverse. Do you know if any of these steps is particularly easy or hard? $\endgroup$ – Nate Eldredge Apr 29 '17 at 13:57

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