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Unformatted text preview: MAT 325: Topology Professor Zoltan Szabo Problem Set 9 Rik Sengupta [email protected] April 24, 2010 1. Munkres, p. 366, problem 2 For each of the following spaces, the fundamental group is either trivial, infinite cyclic, or isomorphic to the fundamental group of the figure eight. Determine for each space which of the three alternatives holds. (a) The “solid torus,” B 2 × S 1 . Solution. For ( x,y ) ∈ B 2 × S 1 , the homotopy H (( x,y ) ,t ) = ((1 t ) x,y ) shows that S 1 is a deformation retract of the solid torus, so π 1 ( B 2 × S 1 ) = π 1 ( S 1 ) = Z . (b) The torus T with a point removed. Solution. We have, I 1 { (1 / 2 , 1 / 2) } has its boundary as a deformation retract, by the straight line homotopy from (1 / 2 , 1 / 2). Under the quotient map which makes I 2 into the torus, the boundary ∂ ( I 1 { (1 / 2 , 1 / 2) } ) becomes a figure eight. (c) The cylinder S 1 × I . Solution. We have, π 1 ( S 1 × I ) = Z , because the cylinder has S 1 as a deformation retract by the homotopy H (( x,y ) ,t ) = ( x, (1 t ) y ) . (d) The infinite cylinder S 1 × R . Solution. This is easy. We have, π 1 ( S 1 × R ) = Z by the same reason (and with the same homotopy) as part (c). (e) R 3 with the nonnegative x , y and z axes deleted. Solution. Consider that any loop in this space is homotopic to some combination of α,β,γ and their inverses, as depicted in the figure on the next page. Note that α * β = γ , so that the fundamental group can be generated without γ . Also, note that β * α ’ ¯ γ 6 = γ , and so the fundamental group is not abelian. So this has a fundamental group isomorphic to that of the figure eight. 1 The following subsets of R 2 : (f) { x : k x k > 1 } Solution. This space has 2 S 1 = { x ∈ R 2 : k x k = 2 } as a deformation retract by the straight line homotopy H ( x,t ) = (1 t ) x +2 tx/ k x k . Now, 2 S 1 is obviously homeomorphic to S 1 by x 7→ x/ 2, and so the fundamental group is Z . (g) { x : k x k ≥ 1 } Solution. This space has S 1 as a deformation retract by the straight line homotopy H ( x,t ) = (1 t ) x + tx/ k x k . So, the fundamental group is Z . (h) { x : k x k < 1 } Solution. By H ( x,t ) = (1 t ) x , we have the fundamental group of this to be trivial. (i) S 1 ∪ ( R + × 0) Solution. This fundamental group is trivial, by H ( x,t ) = (1 t ) x + tx/ k x k . (j) S 1 ∪ ( R + × R ) Solution. Again, this fundamental group is trivial, by H ( x,t ) = (1 t ) x + tx/ k x k . (k) S 1 ∪ ( R × 0) Solution. By the homotopy H ( x,t ) = x if k x k ≤ 1 , (1 t ) x + tx/ k x k if k x k > 1 we obtain the theta space of Example 3 in the text. This has the fundamental group of the figure eight, as both are retracts of the double punctured plane, as we find by applying Theorem 58.3 to Example 2 in the text....
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This note was uploaded on 05/07/2010 for the course MAT 325 taught by Professor Zoltánszabó during the Fall '09 term at Princeton.
 Fall '09
 ZoltánSzabó
 Topology

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