325 HW 08

# 325 HW 08 - MAT 325 Topology Professor Zoltan Szabo Problem...

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Unformatted text preview: MAT 325: Topology Professor Zoltan Szabo Problem Set 8 Rik Sengupta [email protected] 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 ) ,θ ( t )) : r ( t ) = sin(5 t ) + 1 . 2 ,θ ( t ) = t, ≤ t < 2 π } . 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 π 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 π 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 k-fold 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....
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325 HW 08 - MAT 325 Topology Professor Zoltan Szabo Problem...

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