Lecture 7: the quotient topology

Transcription

1 Lecture 7: the quotient topology Saul Glasman September 21, 2016 Why couldn t we say that a homeomorphism was just a continuous bijection? Why did we have to explicitly require the inverse to be continuous as well? Because we have to contend with examples like the following: Example 1. Consider the half-open interval [0, 2π) R and the circle S 1 R 2, each with its subspace topology. Define by φ : [0, 2π) S 1 φ(θ) = (cos θ, sin θ). Then φ is bijective, and it s continuous since cos and sin are continuous (which we know from calculus). But we don t want to consider a circle topologically the same as a half-open interval - if we did, something would have gone wrong. In fact, the inverse of φ isn t continuous. To see this, note that since φ is a bijection, φ 1 is continuous if and only if the image of any open set under φ is open, since (φ 1 ) 1 (U) = φ(u) (If φ has this property, we say φ is an open map). However, take the open subset Then [0, π) [0, 2π). φ([0, π)) = {(x, y) S 1 y > 0} {(1, 0)}. This fails to be open, since it contains no open neighborhood of (1, 0) (draw a picture). Right now we don t have many tools for showing that different topological spaces are not homeomorphic, but that ll change in the next few weeks. Let s continue to another class of examples of topologies: the quotient topology. 1

2 Definition 2. Let X be a topological space and let R be an equivalence relation on X. Let X/R denote the set of equivalence classes for R, and let q : X X/R be the function which takes each x to its equivalence class: q(x) = EC R (x). The quotient topology on X/R is the finest topology for which q is continuous. That is to say, a subset U X/R is open if and only q 1 (U) is open. (We leave it to the reader to prove that this is a topology.) Note that there s something slightly different here. When we defined the subspace and product topologies, we proved that they were the coarsest topologies making certain maps continuous, but the quotient topology is the finest topology making a certain map continuous. This asymmetry arises because the subspace and product topologies are defined with respect to maps out (the inclusion and projection maps, respectively), which force these topologies to be fine; the quotient topology is defined with respect to a map in, the quotient map, which forces it to be coarse. We proved theorems characterizing maps into the subspace and product topologies. Let s prove the corresponding theorem for the quotient topology. Theorem 3. Let T be a topological space and let Hom R (X, T ) be the set of continuous functions X T that collapse equivalence classes of R; that is, Hom R (X, T ) = {f Hom(X, T ) if xry then f(x) = f(y)}. Then composition with q gives a bijection q : Hom(X/R, T ) Hom R (X, T ). Proof. Again, we ll show that q is injective and surjective. For injectivity, suppose f 0, f 1 : X/R T are two continuous functions with f 0 q = f 1 q. Then for each x X, f 0 (q(x)) = f 1 (q(x)); that is, f 0 (EC R (x)) = f 1 (EC R (x)). So f 0 = f 1. For surjectivity, suppose f : X T collapses equivalence classes. For each equivalence class E X/R, choose x E and let g(e) = f(x); since f collapses equivalence classes, it doesn t matter which x E we choose. Then clearly g q = f and we just have to prove that g is continuous. Suppose U T is open; then q 1 (g 1 (U)) = f 1 (U) is open. Thus g 1 (U) is open by the definition of the quotient topology. Having the quotient topology further expands our repertoire of topological spaces. We ll retrieve a couple of examples of equivalence relations from earlier in the course: Example 4. Let Q be the following equivalence relation on R: xqy if x y Z. 2

3 Example 5. Let R be the following equivalence relation on [0, 1]: xry if x = y or x, y {0, 1}. That is, 0 is equivalent to 1 and all the other equivalence classes consist of single points. In fact, these two equivalence classes give rise to homeomorphic quotient spaces! Lemma 6. R/Q and [0, 1]/R are homeomorphic. Proof. This is going to be quite hard work, but hopefully it will help you get used to quotient topologies, so bear with me. By the theorem on functions out of a quotient space, to give a continuous function R/Q [0, 1]/R, we need to give a continuous function f : R [0, 1]/R that collapses equivalence classes for Q. Define f as follows: let r R. If r / Z, there is a unique r (0, 1) such that r r Z. Let f(r) = {r } [0, 1]/R. If r Z, let f(r) = {0, 1} [0, 1]/R. We need to prove that f is continuous, and this requires us to describe a basis for the topology on [0, 1]/R. In general it s not easy to explicitly describe a basis for a quotient topology, but in this case we can do it with a little bit of thought. Apart from at the endpoints, the topology on [0, 1]/R is basically the same as the topology on [0, 1] - in particular, the open intervals inside (0, 1) are open in [0, 1]/R. What are the open sets containing the nontrivial equivalence class {0, 1}? If V [0, 1]/R is open, then q 1 (V ) [0, 1] is an open set containing 0 and 1. Every such open set contains a subset of the form [0, a) (b, 1]. So as a basis for the quotient topology on [0, 1]/R, we can take B = {(a, b) 0 a < b 1} {{{0, 1}} (0, a) (b, 1) 0 < a < b < 1}. (Draw a picture.) Now we have to show that for U B, f 1 (U) is open in R. Now f 1 (a, b) is the set of real numbers whose fractional part is between a and b, and f 1 ({{0, 1}} (0, a) (b, 1)) is the set of real numbers which are integers, or whose fraction part is between 0 and a or between b and 1. (Draw pictures of both). These are both clearly open. 3

4 So we have a function f : R [0, 1]/R that collapses equivalences classes for Q and which is continuous; therefore, it gives a continuous function g : R/Q [0, 1]/R. We claim that g is a homeomorphism. Since g is continuous, we must show that it s bijective and open. It should already be fairly clear that g is bijective; if we define by h : [0, 1]/R R/Q h(a) = EC Q (a) = a + Z if a (0, 1), h({0, 1}) = Z, then h is an inverse to g. Finally, we need to prove that g is open. Since every open set in R/Q is the image of an open set in R, it suffices to prove that f : R [0, 1]/R is open. But the image of any open interval in R is an element of our basis B; if the interval doesn t contain an integer, it s of the first type, and if it does contain an integer, it s of the second type. This concludes our proof. You might have guessed that both of these spaces are just circles. (Draw a picture.) But wait - we already have a way of constructing the circle as a topological space, using the subspace topology in R 2. If our theory is sane, these spaces had better be the same. Theorem 7. Let X = [0, 1]/R with its quotient topology, and let Then X and Y are homeomorphic. Y = {(x, y) R 2 x 2 + y 2 = 1}. Proof. We ll see that this is really a fix for our non-homeomorphism continuous bijection example from before. Combining the theorem about functions out of a quotient space with the theorem about functions into a subspace, we know how to define a continuous function from X to Y : we just have to write down a continuous function f : [0, 1] R 2 which collapses equivalence classes for R and whose image is in Y. We ll use f(x) = (cos 2πx, sin 2πx). This is continuous because cos and sin are continuous. To say f collapses equivalence classes for R is the same as saying f(0) = f(1), which is true: f(0) = f(1) = (1, 0). And of course, since cos 2 + sin 2 = 1, the image of f belongs to the circle Y. This gives us a continuous function g : X Y. Once more, to show that g is a homeomorphism, we must show that it is bijective and open. This is basically visually clear, and the rigorous verification is tedious analysis, so I ll omit it here. 4

5 I d like to regard this theorem as the climax of the course up until now. We came up with various ways of constructing topological spaces; we constructed some spaces that felt like they should be the same but came from different places; we were able to rigorously show that they were the same, showing that we have a robust and flexible theory. 5