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Download Math 525 Notes for sec 22 Final Topologies Let Y be a set, {(X i,τi
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Math 525 Notes for sec 22 Final Topologies ¯ ¯ Let Y be a set, {(Xi , τi ) ¯ i ∈ I} a collection of topological spaces, and {gi ¯ i ∈ I} a corresponding family of functions, where gi : Xi → Y for all i ∈ I. Let µ be the topology on Y defined by U ∈ µ iff gi−1 (U ) ∈ τi for all i ∈ I. Then µ is a topology on Y , and it is the finest topology on Y which makes all of the maps gi : (Xi , τi ) → (Y, µ) continuous. The topology µ is called the final topology on Y induced ¯ ¯ by {gi ¯ i ∈ I} and {(Xi , τi ) ¯ i ∈ I}. Quotient Spaces Suppose we let g : (X, τ ) → Y be a surjective map, and let µ be the final topology on Y ¯ induced by (X, τ ) and g. Then µ = {U ⊆ Y ¯ g −1 (U ) ∈ τ } is called the quotient topology on Y , and (Y, µ) is called the quotient space of (X, τ ) relative to g. As discussed earlier, the quotient topology µ is the finest topology on Y making g : (X, τ ) → (Y, µ) continuous. To understand how “quotient spaces” work, consider a quotient (surjective) map g : (X, τ ) → (Y, µ). Let ∼ be the equivalence relation defined on X defined by x ∼ y iff g(x) = g(y). ¯ The set {[x] ¯ x ∈ X} of equivalence classes relative to ∼ will be denoted by X̂. Note that [x] = g −1 (g(x)). Prop A38 In the notation of the preceding paragraph, σ : X̂ → Y defined by σ([x]) = g(x) is a bijection. Let p : X → X̂ be the natural projection map p(x) = [x]. Then, in the g ✲Y diagram below, σ ◦ p = g. X p❘ ✒ σ X̂ Proof : For any x ∈ X, σ(p(x)) = σ([x]) = g(x), so σ ◦ p = g. To show σ is a bijection, note that σ is surjective because g is a quotient map (and therefore surjective), so that for every y ∈ Y , we have y = g(x) = σ([x]) for some x ∈ X. σ is injective because g(x1 ) = g(x2 ) ⇒ [x1 ] = [x2 ]. ¥ In the preceding discussion, the map p is clearly surjective. Let µ̂ be the quotient topology on X̂ induced by (X, τ ) and p. Then: Prop A39 In Prop A38, σ : (X̂, µ̂) → (Y, µ) is a homeomorphism. Proof : We already know from A38 that σ is a bijection. Now: Claim 1: σ is continuous. To show this, let U ∈ µ. Then g −1 (U ) ∈ τ (since g : (X, τ ) → (Y, µ) is continuous), and by A38, we have p−1 (σ −1 (U )) = g −1 (U ) ∈ τ . By definition of µ̂, this means that σ −1 (U ) ∈ µ̂. Thus, σ is continuous. Claim 2: σ −1 is continuous. To show this, let V ∈ µ̂. Then by definition of µ̂, p−1 (V ) ∈ τ . but by A38, p−1 (V ) = g −1 (σ(V )). So σ(V ) ∈ µ, by definition of µ. Thus, σ −1 is continuous. So σ is a continuous bijection whose inverse is continuous, so it is a homeomorphism. ¥ From a topological standpoint, two homeomorphic spaces are regarded as copies of the same space. So we can, up to homeomorphism, consider every quotient space to be defined on a quotient set. Considering (X̂, µ̂) as a quotient space of (X, τ ), it is interesting to note that [ if A ⊆ X̂, then p−1 (A) = {x}. Thus a set A is open in the quotient topology on X̂ iff [x]∈A [ {x} ∈ τ . Compare this with Munkres’ definition of a quotient map on page 137. [x]∈A A common way to construct quotient spaces is to start with a given space, and identify certain points to form one or more equivalence classes, imposing the quotient topology on the resulting quotient set. Example 1 Let (X, τ ) be [0, 1] with its usual topology (that inherited as a subspace of (R, τu )). Define a simple equivalence relation on [0, 1] by making 0 ∼ 1, and making all other points equivalent only to themselves. Let a = [0] = [1] = {0, 1}. Then we can write X̂ = (0, 1) ∪ {a}, where we identify any x ∈ (0, 1) with {x}. Then p : X → X̂ is defined by a if x = 0 or x = 1 p(x) = . x if x ∈ (0, 1) If x ∈ X̂ and x 6= a, then a basic nbhd of x in X̂ still has the form (x − ǫ, x + ǫ), where ǫ is small enough so that (x − ǫ, x + ǫ) ⊆ (0, 1). A basic nbhd of a has the form W = U ∪ V , where U is a basic τ -nbhd of 0 and V is a basic τ -nbhd of 1. Note that p−1 (W ) looks like: ) 0 ( 1 Since 0 and 1 are identified in X̂, we can visualize X̂ as a circle obtained by bending the line segment [0, 1] into a circle and gluing the endpoints together. Topologically, this quotient space is homeomorphic to a circle, considered as a subspace of R2 . Example 2 Let (X, τ ) be the closed unit square [0, 1] × [0, 1], considered as a subspace of R2 . We define a quotient set X̂ by identifying pairs of points on the two vertical boundaries which are the same height above the x-axis (that is, identify (0, y) with (1, y) for all y ∈ [0, 1]). The other points of the square are equivalent only to themselves. We can visualize X̂ as a cylinder obtained by gluing together the vertical boundaries of X. When the quotient topology is imposed on X̂, the resulting quotient space is homeomorphic to a cylinder embedded in R3 . Example 3 Let (X, τ ) be as in Example 2, and define an equivalence relation by identifying points diametrically opposed on the left and right vertical boundaries, That is, identify the point (1, y) with the point (0, 1 − y) for all y ∈ [0, 1]. The resulting quotient space is homeomorphic to a Möbius strip. Example 4 Define (X, τ ) as before, identify point (0, y) with point (1, y) for all y ∈ [0, 1], and identify point (x, 0) with point (x, 1) for all x ∈ [0, 1]. The resulting quotient space is a torus. Example 5 For the same space (X, τ ), suppose boundary points are identified as follows: (0, y) with (1, y) and (x, 1) with (1 − x, 0). The resulting quotient space is called a Klein bottle. None of the topological properties we’ve studied thus far are preserved by quotient maps, in general.