By T. E. Venkata Balaji

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Then we obtain a conjugate of G in the fundamental group of M . Hence an isomorphism class of coverings [p : M −→ M ] uniquely determines a conjugacy class [G] of subgroups of π1 (M ). If p : M −→ M is Galois, G turns out to be normal in π1 (M ) and hence all of its conjugates coincide. , the set of orbits of G in M . Endow M with the quotient topology. The covering map p : M −→ M induces a continuous map p : M −→ M and it is checked that this is a covering of M . Then this covering determines a subgroup of the fundamental group of M as above and this subgroup is found to be conjugate to G.

Since p is proper, discrete and open, the set B := p(A) is also closed and discrete. B is called the set of critical values of p. Now let M1 denote the complement in M of the inverse image of B under p, and M1 the complement of B in M . , it has no branch points. 5). 4)) is defined and is finite. Let this number be n. This means that every value in M1 is taken by p exactly n times. But what about values in B? It turns out that if we consider the (local) multiplicities with which p takes a value in B, then p also takes each value in B, counting multiplicities, exactly n times.

We next study the relationship between Kleinian and discrete subgroups of PSL(2, C). 4 Discrete Subgroups of PSL(2, C) Recall that an abstract subgroup G of PSL(2, C) is said to be discrete if its underlying set is a discrete subset of the underlying topological space of PSL(2, C). 1 Lemma. Any Kleinian subgroup G ⊂ PSL(2, C) is discrete. Proof. If G is finite, it is discrete. 4). If G had an accumulation point g ∈ PSL(2, C), then we may find a sequence {gn } of distinct points of G which tends to g.