Prof. Dr. M. Wolf WS 2018/19 M. Heinze Sheet 1 Differential Topology: Exercise Sheet 1 Exercises (for Oct. 24th and 25th) 1.1 Examples for quotient topologies Draw the following topological spaces: (a) Let V := [−1; 1] × [0; 1] ⊂ R2 with the subspace topology and ∼ the equivalence relation such that (x; 0) ∼ (x; 1) 8x 2 [−1; 1] and all other (x; t), 0 < t < 1 are only equivalent to themselves. Consider V= ∼. (b) Let W := [−1; 1] × [0; 1] ⊂ R2 with the subspace topology and ∼ the equivalence relation such that (x; 0) ∼ (−x; 1) 8x 2 [−1; 1] and all other (x; t), 0 < t < 1 are only equivalent to themselves. Consider W= ∼. (c) Define the group action (Z × Z) × R2 ! R2 as (m; n) · (x; y) 7! (x + m; y + n) for m, n 2 Z, x,y 2 R. Consider R2=(Z × Z). By this, we mean the quotient space of R2 with respect to the equivalence relation 2 (u; v) ∼ (x; y) if 9(m; n) 2 Z×Z such that (m; n)·(u; v) = (x; y) for (x; y); (u; v) 2 R : (d) Identify S1 ' fz 2 Cj jzj = 1g. Let Y := S1 × [0; 1] and ∼ the equivalence relation such that (z; 0) ∼ (z; 1) 8z 2 C and all other (z; t), 0 < t < 1 are only equivalent to themselves. Consider Y= ∼. Solution: (a) Cylinder (b) Moebius strip (c) Torus (d) Klein bottle 1.2 Hausdorff property of quotient spaces In this exercise you will show that the Hausdorff separation property of a given topological space does generally not extend to a quotient space thereof. Consider X := [−1; 1] × f0; 1g ⊂ R2 with the subspace topology (inherited from the natural topology on R2) that is Hausdorff and the equivalence relation (x; 0) ∼ (x; 1) 8x > 0 : (1) Show that the quotient space X= ∼ with the quotient topology is not Hausdorff. Solution: Set x = (0; 0) and y = (0; 1). Consider the disjoint points [x] and [y] in the quotient space X= ∼. Let Ux and Uy be open neighborhoods of [x] and [y], respectively. Their pre-images (under the quotient map q) are open subsets of X and hence we can 2 find open balls (the open sets in R ) centered around the points x and y with radii x and −1 −1 y, respectively, such that Bx (x) \ X ⊂ q (Ux) and By (y) \ X ⊂ q (Uy). Now choose > 0 such that < x and < y, then −1 −1 (, 0) 2 Bx (x) \ X ⊂ q (Ux) ; (, 1) 2 By (y) \ X ⊂ q (Uy) : (2) In X= ∼, the elements [(, 0)] 2 Ux and [(, 1)] 2 Uy. Due to the equivalence relation, we have (, 0) ∼ (, 1) and therefore [(, 0)] = [(, 1)] 2 Ux \ Uy. 1.3 Another basic notion from topology Show that a topological space X is connected if and only if every continuous g : X 7! f0; 1g is constant, where f0; 1g has the discrete topology. Note that continuity of a map between two topological spaces will be treated in the lecture on Oct. 22nd. Solution: Let X be connected, assume that there is a continuous non-constant g : X 7! f0; 1g. Then U = g−1(f0g), V = g−1(f1g) are open, disjoint and their union is X, which is a contradiction. Assume now that every continuous g : X 7! f0; 1g is constant and suppose that X is not connected. Then there are disjoint open U, V ⊂ X such that their union is X and ( 0 x 2 U g(x) = 1 x 2 V is a continuous function, a contradiction. 1.4 Infinitude of prime numbers In this exercise you will show that there are infinitely many prime numbers. Consider the set Z and its affine subsets S(a;b) := fan + bjn 2 Zg = aZ + b for a; b 2 Z and a 6= 0. We define the open sets as those that are either empty or arbitrary unions or finite intersections of sets S(a;b) = fan + bjn 2 Zg. Show that: (a) This defines a topology on Z; (b) No non-empty finite subset of Z is open; Hint: show that every intersection of two { or finitely many { affine subsets contains an (infinite) affine subset; (c) Each affine subset S(a;b) = fan + bjn 2 Zg is both open and closed; (d) There are infinitely many prime numbers by analyzing the set [ S := S(p;0) where p prime ; (3) p and using the above properties (b) and (c) of the topology. Solution: This is a proof originally introduced by Furstenberg.¨ (a) The empty set is in the topology by definition. The whole set can be written as Z = S(1;0) and hence also is. Unions and finite intersections of sets S(a;b) are open by definition. (b) Assume that the intersection of affine sets S(ai;bi) for i = 1; : : : ; k is non-empty. Then there is an x 2 S(ai;bi) for all i = 1; : : : ; k. In other words, we can find n1; : : : ; nk 2 Z such that x = aini + bi for all i = 1; : : : ; k. For some n 2 Z let us define yn = na1 ··· ak, i.e. it is a multiple of the ai, then x + yn 2 S(ai;bi) for all i = 1; : : : ; k. Since this holds for all n 2 Z we found an affine (therefore infinite) subset S(a1:::ak;x) of all sets S(ai;bi) for i = 1; : : : ; k. As a consequence any (non-empty) open set contains an affine subset and hence is infinite. In other words, no non-empty finite subset of Z is open. (c) Let us consider an affine set S(a;b) that is open by definition. Its complement is open since it can be written as the union of affine and hence open sets, i.e. as a−1 [ Z n S(a;b) = S(a;b+j) : (4) j=1 Hence S(a;b) is also closed. (d) Assume that the number of primes is finite, i.e. there is an m 2 N such that the prime numbers are p1; : : : ; pm. Then the set S would be a finite union of the sets S(p1;0);:::;S(pm;0). Consider the complement of S that can be written as Z n S = Z n S(p1;0) \···\ Z n S(pm;0) (5) which is a finite intersection of the sets Z n S(p1;0); ··· ; Z n S(pm;0). Since the affine sets S(p1;0);:::;S(pm;0) are closed due to property (c), their complements are open. Hence ZnS is open as the finite intersection of open sets. Its complement S is therefore closed. On the other hand, the complement of S is the set {−1; 1g (any other integer can be written as n · p for n 2 Z and p prime due to prime factoring). This is a finite non-empty subset of Z. Property (b) hence implies that it cannot be open. Therefore its complement, the set S = Z n {−1; 1g, cannot be closed which is a contradiction..