By A. F. Beardon
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This version additionally gains extra difficulties and instruments emphasizing fractal functions, in addition to a brand new solution key to the textual content routines.
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Additional info for A Primer on Riemann Surfaces
5. For any surface S, the following conditions 37 are equivalent: (1) the topology on S has a countable base? (2) S has a countable open cover of parametric discs? 'K 2' •• • of s with K. c K 0 c ... L , UK n Z = S. Proof. }. For each j, choose, wherever possible, a single parametric disc Q . with B . c q .. } is a countable collection of open 3 3 3 3 parametric discs. ) and there is some B . , B . c q . Thus x 3 3 3 3 and so uQ^ = S which proves (2). As each parametric disc is Qj homeomorphic to a disc in I, it has a countable base.
Is any set of the form and Y A x b where respectively. Clearly Z itself A and B are is an open rectangle: also, a finite intersection of open rectangles is an open rectangle. 1 that the class of open rectangles is a base for some topology on Z: we call this the product topology on Z. There are natural coordinate maps P 1 : (x,y) of Z onto X and Y h- x It follows that if h- y respectively and these are continuous because (p )_ 1 (A) = A x Y Plf , P 2^« P 2 : (x,y) , , f : W -*■ Z (p2)_ 1 (B) = X * B .
Unfortunately, as a general procedure this does not always yield the correct quotient space (it does here but in general, we need extra hypotheses) so we must seek a more explicit procedure Each point of the plane can be expressed uniquely in the form aA + by let S xs where S xs a and b are real. Let S be the unit circle in be the corresponding product space. There is map of (D and onto given by . 2). 2. 1), we see that G/G is homeomorphic to S*S. As a torus is obtained by rotating Sxs a circle around an axis, it can be parametrised by so G/G is topologically a torus.