Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/ii/paper-1/20f/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 , choose disjoint inverse neighbourhoods of every point in its fibre. Compactness of the remaining source ensures a sufficiently small disc about 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 uniquely from each point of . 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 the monodromy action is transitive.
For the rational map of degree three, differentiation gives
At , the local degree is three, so a small loop around its value acts as a 3-cycle. At , the local degree is two; their distinct branch values are , and a loop around either acts as a transposition. The poles and the point at infinity are simple and unramified. A 3-cycle and a transposition generate the full symmetric group, so
The corresponding degree-three connected cover is not normal: a normal cover would have a regular monodromy action, whereas on three letters has nontrivial point stabilizers. Thus the stronger interpretation of the printed terminology cannot hold.

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