= Solution
At a noncritical point a <holomorphic map> has a local holomorphic inverse. Fibres of a nonconstant map between compact connected <Riemann surfaces> are finite: zeros are isolated and compactness precludes infinitely many. Around a nonbranch value $w$, choose disjoint inverse neighbourhoods of every point in its fibre. Compactness of the remaining source ensures a sufficiently small disc about $w$ has no additional preimages. Its preimage is therefore a disjoint union of open sets each mapped biholomorphically onto that disc. This proves a finite-sheeted unbranched <covering map> after removing the branch fibres.
Here the word “regular” must mean locally unbranched or evenly covered, rather than the stronger modern normal-cover condition. A general holomorphic map does not give a normal covering, and the example below itself shows the distinction.
Lift a loop based at $w$ uniquely from each point of $f^{-1}(w)$. Its endpoint defines a permutation; reversing the loop gives the inverse permutation, and concatenation gives composition. Homotopic loops yield the same endpoints. The <monodromy group of a covering> consists of these permutations. Removing finitely many points from a connected surface leaves it path-connected. Any two points in the same fibre can therefore be joined upstairs; the projection of that path is a loop whose lift joins them. Thus \b[the monodromy action is transitive].
For the rational map of degree three, differentiation gives
$$
f'(z)=\frac{z^2(3-z^2)}{(1-z^2)^2}.
$$
At $z=0$, the local degree is three, so a small loop around its value $0$ acts as a 3-cycle. At $z=\pm\sqrt3$, the local degree is two; their distinct branch values are $\mp3\sqrt3/2$, and a loop around either acts as a transposition. The poles $\pm1$ and the point at infinity are simple and unramified. A 3-cycle and a transposition generate the full symmetric group, so
$$
\boxed{H=S_3}.
$$
The corresponding degree-three connected cover is not normal: a normal cover would have a regular monodromy action, whereas $S_3$ on three letters has nontrivial point stabilizers. Thus the stronger interpretation of the printed terminology cannot hold.
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