Lecture 18 - 10/5/2012

Home , Unit interval

Math 5801 General Topology and Knot Theory

Nathan Broaddus

Ohio State University

October 5, 2012

Nathan Broaddus General Topology and Knot Theory

Lecture 18 - 10/5/2012 Quotient Topology Course Info

Reading for Monday, October 8 Chapter 3.23, pgs. 147-152

HW 6 for Monday, October 8

I Chapter 2.19: 3, 6, 7, 8

I Chapter 2.20: 2, 3a-b, 11

Nathan Broaddus General Topology and Knot Theory Lecture 18 - 10/5/2012 Quotient Topology Quotient Topology

Deﬁnition 179 (The unit interval) The unit interval I is the topological (or metric) space I = [0, 1] with the subspace topology (resp. subset metric) inherited from R.

Example 180 (The 2-torus T 2)

2 I Let ∼ on I be the relation with (0, a) ∼ (1, a) for and (a, 0) ∼ (a, 1) for all a ∈ I . 2 2 I The 2-torus is the quotient space T = I / ∼. 2 2 I Let q : I → T be the quotient map q(x, y) = [x, y] where [x, y] is the equiv. class of (x, y) ∈ I 2 under ∼. 2 I Let 0 < a, b < 1. What does a nbhd of [a, b] ∈ T look like?

I Let ε = min{|a − 0|, |a − 1|, |b − 0|, |b − 1|} 2 I Then for each p ∈ Bε(a, b) ⊂ I we have [p] = {p}. 2 I Thus Bε(a, b) ⊂ I is a saturated set.

I And q|Bε(a,b) : Bε(a, b) → Bε(a, b) is an injection.

I Hence by def. of quot. top. q|Bε(a,b) is a homeomorphism. Nathan Broaddus General Topology and Knot Theory

Lecture 18 - 10/5/2012 Quotient Topology Quotient Topology

Example 181 (The n-torus T n)

n I Let ∼ on R be the relation with (x, y) ∼ (x + m, y + n) for all x, y ∈ R and n, m ∈ R n n I The n-torus is the quotient space T = R / ∼. n I What do open sets of T look like? n n I Claim: If U ⊂ R is open then q(U) is open in T n I If U ⊂ R is open then the smallest saturated set containing U is q−1(q(U))

−1 [ q (q(U)) = [x, y] (x,y)∈U [ = {(x + m, y + n)|n, m ∈ Z} (x,y)∈U [ = {(x + m, y + n)|(x, y) ∈ U} n,m∈Z

Nathan Broaddus General Topology and Knot Theory Lecture 18 - 10/5/2012 Quotient Topology Quotient Topology

Example 182

I Let ∼ on R be the relation with x ∼ y if x − y ∈ Q.

I Let Y = R/ ∼ with quotient map q : R → Y . −1 I Notice that for each y ∈ Y we have q (y) id countable (since Q is countable)

I Thus Y is uncountable (otherwise R would be a countable union of countable sets).

I If nonempty U ⊂ R is open then the smallest saturated set containg U is [ q−1(q(U)) = [x] x∈U = R

I Thus Y is an uncountable set with the trivial topology.

Nathan Broaddus General Topology and Knot Theory

Lecture 18 - 10/5/2012 Quotient Topology Quotient Topology

Deﬁnition 183 (Disjoint union and Coproduct Topology) If A and B are sets then the disjoint union of A and B is the set

A q B = A × {0} ∪ B × {1}

We write A ⊂ A q B and B ⊂ A q B even thought not strictly true.

The disjoint union of the collection of indexed sets {Aα}α∈A is a [ Aα = Aα × {α}. α∈A α∈A ` We identify Aα with Aα × {α} ⊂ α∈A Aα. ` If each Aα is a topological space then Aα is given the coproduct ` α∈A topology where B ⊂ α∈A Aα is open iﬀ B ∩ Aα is open for all α ∈ A.

Nathan Broaddus General Topology and Knot Theory Lecture 18 - 10/5/2012 Quotient Topology Quotient Topology

I Let X be a topological space and Y a set of topological spaces. Q I Let f : X → Y be a function and for each Y ∈ Y let fY : X → Y be πY ◦ f .

I Recall that Prop. 136 says that f is continuous iﬀ each fY is continuous.

I Compare to the following:

Proposition 184 (Universal property of coproduct) Let X be a set of topological spaces and Y be a topological space. A ` function f : X → Y is continuous if and only if f |X : X → Y is continuous for each X ∈ X.

Proof. Immediate from def. of open sets of ` X

Nathan Broaddus General Topology and Knot Theory

Lecture 18 - 10/5/2012 Quotient Topology Quotient Topology

Example 185

n I Let ∼ on R be the relation with x ∼ 2 x if n ∈ Z.

I Let W = R/ ∼ with quotient map q : R → W . 1 I If x 6= 0 then [x] has a nbhd homeomorphic to S .

I Only saturated open set containing [0] is R. 1 1 I Hence W =∼ S q S ∪ {[0]} where [0] is in the closure of every nonempty subset of W .

Nathan Broaddus General Topology and Knot Theory Lecture 18 - 10/5/2012 Quotient Topology Quotient Topology

Deﬁnition 186 (Graph)

1. Let X 0 be a set with the discrete topology called the vertices. 2. Let A be a set and for each

3. For each α ∈ A let Iα = I = [0, 1]. 0 4. For each α ∈ A let ϕα : ∂Iα → X where ∂Iα = {0, 1} ⊂ Iα 5. Let 0 a Y = X q Iα α∈A 6. Let ∼ on Y be the equivalence relation x ∼ y if

I x ∈ Iα and y ∈ Iβ and φα(x) = φβ (y). 0 I or x ∈ Iα and y ∈ X and φα(x) = y. 7.A graph is the quotient X 1 = Y / ∼. 8. Let q : Y → X 1 be the quotient map.

9. Let iα : Iα → Y be the inclusion 1 10. An edge of the graph X is a set of the form q ◦ iα(Iα). Nathan Broaddus General Topology and Knot Theory

Lecture 18 - 10/5/2012 Quotient Topology Quotient Topology

I Let X be a topological space and ∼ be an equivalence relation on X . I Let Z be a topological space. I What are the continuous functions f :(X / ∼) → Z? I We have that the quotient map q : X → X / ∼ is continuous so any continuous function f :(X / ∼) → Z gives continuous f ◦ q : X → Z.

I For which functions g : X → Z is there a function f :(X / ∼) → Z such that g = f ◦ q?

Proposition 187 (Universal Property of Quotient Space)

1. Let X be a topological space and ∼ be an equivalence relation on X with quotient map q : X → X / ∼ 2. Let Z be a topological space. 3. Let g : X → Z be a continuous function. There is a continuous function f :(X / ∼) → Z such that g = f ◦ q iﬀ for each [x] ∈ X / ∼ we have that g|q−1 ({[x]}) is constant.

Nathan Broaddus General Topology and Knot Theory Lecture 18 - 10/5/2012 Quotient Topology Quotient Topology

Proof of Prop. 187.

I Let X be a topological space and ∼ be an equivalence relation on X with quotient map q : X → X / ∼

I Let Z be a topological space.

I Let g : X → Z be a continuous function.

I ⇒

I Suppose there is continuous f :(X / ∼) → Z such that g = f ◦ q.

I If x0, x1 ∈ X and x0 ∼ x1 then

g(x0) = f ◦ q(x0) = f ([x0]) = f ([x1]) = f ◦ q(x1) = g(x1)

Nathan Broaddus General Topology and Knot Theory

Lecture 18 - 10/5/2012 Quotient Topology Quotient Topology

Proof of Prop. 187 (continued).

I ⇐

I Suppose that for all x0, x1 ∈ X if x0 ∼ x1 then g(x0) = g(x1)

I Deﬁne f : X / ∼→ Z to be the function f ([x]) = g(x).

I By assumption f is well-deﬁned. −1 I If V ⊂ Z is open then g (V ) is open by continuity of g.

I So we have open

g −1(V ) = (f ◦ q)−1(V ) = q−1(f −1(V ))

−1 I By def. of quotient top. f (V ) is open in X / ∼.

Nathan Broaddus General Topology and Knot Theory