Mathematics 730, Fall 1998 Fundamental Concepts of Topology Final Exam Solutions Dr. Rosenberg

1.

(a) There are many equivalent ways for formulate the proof. For example, suppose {x_{α}} is a net in A converging to x_{0} ∈ X. By continuity of f and g, f(x_{α}) → f(x_{0}) and g(x_{α}) → g(x). But since x_{α} ∈ A, f(x_{α}) = g(x_{α}) for all α. Thus f(x) and g(x) are limits of the same net. Since Y is Hausdorff, this forces f(x) = g(x), so x ∈ A. That shows A is closed in X.

(b) For a simple example, take X to be the unit interval [0, 1], and let Y = {a, b} where {a} is closed but not open and {b} is open but not closed. (Thus Y is T_{0} but not T_{1}, certainly not Hausdorff.) Define f(x) = a for all x, g(x) = b for x < 1, g(1) = a. Then f and g are both continuous but f(x) = g(x) on A = [0, 1), which is not closed in X. One can get a similar example with Y the “interval with two endpoints,” which is T_{1} but not Hausdorff.

2. For convenience in the proof, let Z = (X × {0}) ∪ (A × [0, 1]), with the subspace topology from X × [0, 1].

(i) (a) ⇒ (b). If we assume (a), we have a retraction r: X ×[0, 1] → Z. Given the situation of (b), g and h together define a continuous map k: Z → Y. (That’s because g is continuous on X × {0}, h is continuous on A × [0, 1], and they agree on the overlap, A × {0}.) So let H = k ∘ r. This extends g on X × {0} and extends h on A × [0, 1]. It’s continuous because it’s the composite of two continuous maps.

(i) (b) ⇒ (a). Apply (b) with Y = Z. Then H is the desired retraction. (ii) It’s enough to check (a). If X = [0, 1] and A = {0}, then Z = ([0, 1] × {0}) ∪ ({0} × [0, 1]).

One can define the desired retraction [0, 1] × [0, 1] → Z by

r(x, y) = \{ \begin{matrix}(x − y, 0) & x ≥ y, \\ (0, y − x) & y ≥ x. \end{matrix}

3.

(a) The key points to notice are that T_{1} and T_{2} are smooth 4-manifolds with boundary, and that since A has determinant 1, f is a diffeomorphism ∂T_{1} → ∂T_{2}, with the inverse diffeomorphism given by the inverse matrix,

( \begin{matrix}1 & −1 & 1 \\ 0 & 1 & −1 \\ 0 & 0 & 1 \end{matrix} ) .

Since no identifications take place over the interiors of T_{1} and T_{2}, manifold-with-boundary charts for T_{1} and T_{2} serve as manifold charts for points in the interiors of T_{1} and T_{2}. For points on ∂T_{1} ≅ ∂T_{2}, one gets manifold charts for X by combining a chart for T_{1} with a chart for T_{2} via the diffeomorphism f on the boundaries.

(b) Note that X = T_{1} ∪ T_{2}, and that T_{1} ∩ T_{2} = ∂T_{2} ≅ T^{3}. Now T_{1}, T_{2}, and T_{1} ∩ T_{2} are closed but not open in X; nevertheless, Van Kampen’s Theorem applies since each is the deformation retract of a small open neighborhood. We have π_{1}(T_{1} ∩ T_{2}) = π_{1}((S^{1})^{3}) = π_{1}(S^{1})^{3} ≅ ℤ^{3} and π_{1}(T_{1}) ≅ π_{1}(T_{2}) ≅ ℤ^{2}. Since the maps π_{1}(T_{1} ∩ T_{2}) → π_{1}(T_{j}), j = 1, 2, are surjective, Van Kampen’s Theorem implies that π_{1}(X) is the quotient of π_{1}(T_{1} ∩ T_{2}) by the relations introduced by the maps π_{1}(T_{1} ∩ T_{2}) → π_{1}(T_{2}) and π_{1}(T_{1} ∩ T_{2}) → π_{1}(T_{2}). We compute these as follows. Identify π_{1}(T_{1} ∩T_{2}) with π_{1}(∂T_{2}) = ℤ^{3}. The map ∂T_{2} → T_{2} or S^{1} ×S^{1} ×S^{1} → S^{1} ×D^{2} ×S^{1} corresponds to killing off the second standard basis vector e_{2} of ℤ^{3}. Similarly, the map ∂T_{1} → T_{1} or S^{1} × S^{1} × S^{1} → D^{2} × S^{1} × S^{1} corresponds to killing off the first standard basis vector e_{1} of ℤ^{3}. And under f_{∗} = A, e_{1} maps to e_{1}. So π_{1}(X) is therefore ℤ^{3} divided by the relations e_{1} = 0, e_{2} = 0, and is thus ℤ (with e_{3} ∈ π_{1}(∂T_{2}) as the generator.

4.

(a) S^{2} ∨ S^{1}, the one-point union of S^{2} and S^{1}.

(b) many examples, for instance the “comb” space shown on page 135 of Bredon. Any small neighborhood of x_{0} has infinitely many components, yet x_{0} can be joined by a path to any point on one of the “teeth” of the comb by going down, across, and up.

(c) If f(x, y, z) = x^{2}y + 2xz, then df at (x, y, z) is the linear map ℝ^{3} → ℝ with matrix ∇f = (2(xy + z), x^{2}, 2x). The critical points are just points where ∇f = (0, 0, 0), or points of the form (0, y, 0). Since f(0, y, 0) = 0, there is exactly one critical value, 0. So if w ∈ ℝ, w ≠ 0, w is a regular value of f. By the Implicit Function Theorem, f^{−1}(w) must in this case be a smooth manifold of dimension 3 − 1 = 2. In fact, for any w ≠ 0, f^{−1}(w) is diffeomorphic to (ℝ \ {0}) × ℝ via the map g: (x, y, z) ↦ (x, z). This space can also be identified with the disjoint union of two copies of ℝ^{2}. The map g is bijective since if x^{2}y + 2xz = x(xy + 2z) = w ≠ 0, then x can take any non-zero value, z can take any real value, and x and z uniquely determine y via the equation xy = \frac{w}{x} − 2z. The map g is obviously smooth with the smooth inverse

(x, z) ↦ (x, \frac{1}{x} (\frac{w}{x} − 2z) , z) .

(d) For a path-connected space X which has a universal space, such as a manifold, the connected two-to-one coverings are classified by subgroups of index 2 in the fundamental group Γ. (That’s because if X̃ ⟶^{p} X is the universal cover of X, then this is a Galois covering with group of covering transformations Γ, so X = X̃/Γ, and if Y ⟶^{q} X is a connected two-to-one covering space, then by the universal property of X̃, Y = X̃/H for some subgroup H of Γ, and H must have index 2 since the fibers of q are the cosets Γ/H.) But any subgroup of index 2 is necessarily normal with quotient group cyclic with two elements. Thus the subgroups H of index two in any group Γ are exactly the kernels of surjective homomorphisms Γ ↠ {±1}. For the Klein bottle X, the fundamental group Γ = ⟨a, b | aba^{−1} = b^{−1}⟩. Any homomorphism γ: Γ ↠ {±1} is determined by γ(a) = ±1 and by γ(b) = ±1, and γ is surjective if γ(a) and γ(b) are not both +1. So there are at most three possibilities: a ↦ 1 and b ↦ −1, a ↦ −1 and b ↦ −1, and a ↦ −1 and b ↦ 1. All of these satisfy the relation in Γ, so there are in fact exactly three connected two-to-one coverings of X. The covering space Y = Γ/ ker γ is the torus when H preserves orientation on X̃ = ℝ^{2} and is the Klein bottle otherwise. Now a preserves orientation and b does not, so H reverses orientation when it contains words involving b to an odd power and preserves orientation when it contains only words involving b to an even power. So Y is a torus in the two cases where γ sends b to −1 and is the Klein bottle in the remaining case where b lies in H = ker γ.