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.)