Is there a Borel-measurable projection to a closed subgroup

Question

Suppose that $G$ is a compact metrizable group and let $H$ be a closed subgroup of $G$.

Is it true that there must exists a Borel-measurable projection map $p:G\rightarrow H$ with the property that for every $h\in H$ and $g\in G$ one has that $p(gh)=p(g)\cdot h$? (In particular $p(h)=h$).

Answer 1

The answer is a yes! (Thanks to GEdgar who pointed out the relation with "selection theorems")

In the book "A Course on Borel sets" page 186 they proved a claim that implies the following:

Theorem: Let $H,G$ be as in the question. Then there exists a "cross-section" $s:G/H\rightarrow G$ such that $s\circ p = id$ where $p:G\rightarrow G/H$ the quotient map $p(g)=gH$.

Proof: Let $\Pi = \{a_i H : a_iH \text{ is a coset of H} \}$ then by the theorem in the book there exists a set Borel set $S$ such that $S\cap a_i H = \{s_i\}$.

This defines a (Borel) map $s:G/H\rightarrow G$ such that $s(u) = s_i$ if $u\in a_i H$. (The map is Borel because $s^{-1}(U)=s^{-1}(U\cap S) = p(U\cap S)$ and quotient map is open so it sends Borel to Borel), which proves the theorem.

This theorem leads to our result, as one can can define a map $\varphi:G\rightarrow H$ by $\varphi(g) = g\cdot (s(p(g))^{-1}$.

Since for every $h\in H$ one has that $p(gh)=p(g)$ we have that $\varphi(gh)=\varphi(g)\cdot h$.

This also answers positively my (now obviously related) other question:

$G\cong G/H\times H$ measureably