Profinite groups are Galois groups

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I tried to understand the proof on page 195 (in PDF page 17) here.

At the beginning $T$ is denoted as disjoint union of sets $G/U$. Is it important for the proof that they are disjoint ?

After the definition of $G_{k}$ are some $t_{i}\in G/U_{i}$ taken, that appear in the rational expression of $k$. What does that mean? Take they some $U_{i}$ so that $k$ mod $U_{i}$ equal $k$ or what is meant by that? And why are there finally many $U_{i}$.

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The simplest relevant example is $$\Bbb{Z_p} =\varprojlim \Bbb{Z/p^kZ}$$

For each $k$ and $a\bmod p^k$ create a variable $T_{a\bmod p^k}$ and let $$K = \Bbb{Q}(\{ T_{a\bmod p^k},k\ge 1, a\in 0,\ldots p^k-1\})$$ which is a field of rational functions in infinitely many variables. Each $b\in \Bbb{Z}_p$ acts naturally as an automorphism $g_b\in Aut(K)$, switching the variables as $$g_b(T_{a\bmod p^k})=T_{a+b\bmod p^k}$$ It remains to look at $L$ the subfield fixed by $G=\{ g_b,b\in \Bbb{Z}_p\}\subset Aut(K)$ and find that $Gal(K/L)=G$ which is done in your linked text and cannot be made simpler in my opinion, the key argument being that for any $y\in K$ the Galois orbit $G.y = \{g_b(y),b\in\Bbb{Z}_p\}$ is finite so that $L$ is generated over $\Bbb{Q}$ by the coefficients of the polynomials $$f_y(X)=\prod_{z\in G.y} (X-z) \in L[X]$$