By A. F. Beardon

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**Additional info for A Primer on Riemann Surfaces**

**Sample text**

D zq is topologically a The two copies D_. of distinct from zq by non-intersecting simple 2(m+l)-gon, D each a. ,a : they are joined to each other 1 m is odd and they are joined to themselves across if is even. Using Euler's formula we can compute the genus resulting surface. For example, if m = 2 g of the we identify the edges of two (topological) squares.

We take these maps as y < 0 in each Pfc) Select distinct points € U c P , V t U and V being open in P^ u and and the sets in S S cannot is Hausdorff. and choose sets U f(U), f(V) S Pg V respectively. Then = f(v) € f(V) are open in can be chosen so that f(U) S. s = t, Pg (=pt ) : we take then f (U) U and V and f (V) We need to show, then, that and f(V) For ease of reference, write If v S € V c P , and u — f(u) 6 f(U), V (C) . with, say, U and Pfc with it is clear that have a countable base. It remains only to show that V is a contains an uncountable collection of pairwise disjoint open sets (the half-spaces and thus ffc : Pfc -*■ the inverse charts (after, of course, identifying each As Pg : E are disjoint.

We shall discuss two ways of cutting First, we construct a simple arc [z^,z^3 (C^ as described earlier. z^ to z^, from then a simple Cz 0,z_] from z_ to z_ and so on, ending with Cz ,°°]. Together, z 3 2 3 m these are constructed to give a simple arc from z 1 to 00 and we take two arc copies z1 and of the complementary domain. The nature of implies that we join (topologically) D1 to f near (twice) across 54 so that two circuits of are required to return original value. °] D1 iz^,z^\, Lz^ z^] D2 and if and only if m Let and z^ M be (and in effect, this means that need not have been made.