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Homework 1, Algebraic Topology, 215A, Fall 2006 (1) Products. Let X and Y be topological spaces. We define their product X × Y to be the set of pairs (x, y), x ∈ X, y ∈ Y , with the following product topology: A subset W ⊂ X × Y is defined to be open if for every point (x, y) ∈ W there exist open neighborhoods U of x and V of y such that U × V ⊂ W . It is not hard to check that this indeed defines a topology on X × Y . Equivalently, we could define the product topology on X × Y as the topology whose base is given by all products U × V of open sets U ⊂ X, V ⊂ Y . (a) Show that for all m, n ∈ N we have Rm × Rn ∼ = Rm+n . (b) Show that a map (f1 , f2 ) : Z −→ X × Y is continuous if and only if the maps f1 : Z −→ X and f2 : Z −→ Y are continuous. (c) Conclude that this implies the following universal property of the product X ×Y : Given maps fi , i = 1, 2, as in (b) there exists a unique map g such that the following diagram commutes: X ×Y projX . f1 ↑g projY & f2 X ←− Z −→ Y (d) Use this to show that there is a bijection of sets Hom(Z, X × Y ) ∼ = Hom(Z, X) × Hom(Z, Y ), where Hom(Z, W ) denotes the set of continuous maps from Z to W . Note: A particularly nice aspect of the bijection of part (d) is that it is natural in the following sense: For all continuous maps φ : Z 0 −→ Z, ψ1 : X −→ X 0 , and ψ2 : Y −→ Y 0 we get a commutative diagram ∼ = Hom(Z, X × Y ) −→ Hom(Z, X) × Hom(Z, Y ) ↓ ↓ ∼ = Hom(Z 0 , X 0 × Y 0 ) −→ Hom(Z 0 , X 0 ) × Hom(Z 0 , Y 0 ). Here the vertical arrows are defined by pre- and post-composition with the given functions φ and ψi (how exactly?). One uses the word natural here to indicate that the isomorphism in (d) does not reflect specific properties of the spaces X, Y , and Z, but rather describes a general feature of the product construction. (2) Manifolds. In this problem we want to define an important class of topological spaces, so-called manifolds, and look at some examples. Let M be topological space and recall that M is Hausdorff if for any two distinct points x, y ∈ M there exist disjoint open neighborhoods U of x and V of y, i.e. two points in M can be “separated” by open sets. M is second countable if there exists a countable base for the topology on M . We define a n-dimensional (topological) manifold to be a second countable topological Hausdorff space that is locally homeomorphic to Rn . By this we mean the following: For each point x ∈ M there exists an open neighborhood U of x and a ∼ = homeomorphism φ : U −→ Rn . We call such a φ a chart for M . (a) Show that every open subset of Rn is a manifold. This implies that we could replace ’Rn ’ in the above definition with ’an open subset of Rn .’ P 2 (b) Let S n := {x ∈ Rn+1 / |x|2 = n+1 i=1 xi = 1} be the n-dimensional sphere. Show n that for each x ∈ S we have a homeomorphism S n \{x} ∼ = Rn . Conclude that S n is an n-dimensional manifold. (c) Show that the product of two manifolds is a manifold. We see for example the torus/donut S 1 × S 1 is a manifold. (d) Give an example of a topological space M which is locally homeomorphic to Rn , but which is not Hausdorff. Note: The second countability axioms is useful when one wants to construct interesting functions and geometric structures on a manifold. The empty set is a manifold of any dimension (a surprisingly important fact), otherwise the dimension of a manifold is uniquely determined, see the comment on the invariance of dimension in the note on problem 3. (3) A Peano curve. The heart of this problem is the construction of a continuous map from the unit interval I = [0, 1] onto the square I 2 ⊂ R2 . Consider a continuous function f : R −→ [0, 1] such that ( 0 if 0 ≤ t ≤ 13 f (t) = 1 if 32 ≤ t ≤ 1 and f (t + 2) = f (t) for all t ∈ R. Set X X x(t) := f (32n−1 t)2−n and y(t) := f (32n t)2−n n≥1 n≥1 Define γ : I −→ I 2 , γ(t) := (x(t), y(t)) (a) Show that γ is continuous. Here you can use the fact that a uniform limit of continuous functions is again continuous. (b) Prove that γ is surjective. Hint: A point (x0 , y0 ) ∈ I 2 can be represented as X X x0 (t) := a2n−1 2−n , y0 (t) := a2n 2−n , n≥1 n≥1 P −k−1 where ak = 0 or 1. Then γ maps t0 := k≥1 2ak 3 to (x0 , y0 ). (c) Show that γ is not injective. (d) Conclude that there is a surjective map I −→ I n for all n ∈ N. Note: These examples raise the question whether there exists a continuous, bijective map from the interval I to the n-dimensional cube I n . However, it is known that such a map doesn’t exsist. The background is the so-called theorem on invariance of dimension: The existence of a homeomorphism φ : U −→ V between open subsets U ⊂ Rm and V ⊂ Rn implies m = n. (We’ll prove this theorem this semester.) Now assume there existed a continuous bijective map γ : I −→ I n . Then γ is automatically a homeomorphism (this follows using the compactness of I and I n ), and restriction of γ to the subset γ −1 (γ((0, 1))∩(0, 1)n ) would give a homeomorphism between open subsets of Euclidian spaces of different dimensions which is impossible according to the invariance of dimension. Please return in class on Tuesday, Sept. 5.