This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MAT 325: Topology Professor Zoltan Szabo Problem Set 8 Rik Sengupta rsengupt@princeton.edu April 13, 2010 1. Munkres, p. 334, problem 1 A subset A of R n is said to be star convex if for some point a of A , all the line segments joining a to other points of A lie in A . (a) Find a star convex set that is not convex. (b) Show that if A is star convex, A is simply connected. Solution. We will solve the two parts one by one. (a) Consider the subset A of R 2 bounded by the parametric polar curve { ( r ( t ) , ( t )) : r ( t ) = sin(5 t ) + 1 . 2 , ( t ) = t, t < 2 } . Figure 1: The subset A R 2 This subset is star convex if we let a be the origin notice that it even looks like a star. This is because the line segment from the origin to any other point in A is contained in A . This is not convex, because the line segment between, say, (1 , 0) and (0 , 1) is not contained in A . 1 (b) Assume that A is star convex. We need to show that A is path connected, and also that 1 ( A,a ) is trivial for the point a in the definition of star convex. Let a 1 ,a 2 A . We can create a path from a 1 and a 2 by first traveling along the line segment from a 1 to a , and then along the line segment from a to a 2 . By the definition of star convex, this path is contained entirely within A . Therefore, A is path connected. Next, consider a loop g at a . Define H : I I A as the straight line path homotopy H ( x,t ) = ta + (1 t ) g ( x ). We know H is continuous, H ( x, 0) = g ( x ), and H ( x, 1) = a . Therefore, H is a path homotopy from g to the constant loop at a . We may now conclude that 1 ( A,a ) = { } , which is the trivial group, and so, A is simply connected. 2. Munkres, p. 341, problem 3 Let p : E B be a covering map; let B be connected. Show that if p 1 ( b ) has k elements for some b B , then p 1 ( b ) has k elements for every b B . In such a case, E is called a kfold covering of B . Solution. We present two different solutions. Solution 1. Since p 1 ( b ) = k , we can find U such that p 1 ( U ) = k G i =1 V i and p V i = p i : V i U is a homeomorphism, i. We assume that there exists some b 1 B such that p 1 ( b 1 ) = j 6 = k , and we will contradict the fact that B is connected. Define C = { b B : p 1 ( b ) = k } D = { b B : p 1 ( b ) 6 = k } . Then, we have b C , so C 6 = , and b 1 D , so D 6 = . Also, it is clear that C D = , and C D = B , from the way they are defined. So it just remains to show that C and D are open, for the contradiction....
View
Full
Document
 Fall '09
 ZoltánSzabó
 Topology

Click to edit the document details