Quotient Topology

Open Map

A function

f : X \to Y

is said to be an open map if for every open set

U \in 𝒯_{X}

the set

f (U)

is open in

Y

The Projection Maps Are Open

The project maps:

π_{X} : X \times Y \to X

and

π_{Y} : X \times Y \to Y

are open

Let $M$ be an open set in $𝒯_{X \times Y}$ therefore it is of the form $M = ⋃ 𝒜$ where $𝒜 \subseteq {U \times V : U \in 𝒯_{X}, V \in 𝒯_{Y}}$ .

But then $π_{X} (M) = π_{X} (⋃ 𝒜)$ = $⋃ {π_{X} (U_{A} \times V_{A}) : U_{A} \times V_{A} \in 𝒜} = ⋃ {U_{A} : U_{A} \times V_{A} \in 𝒜}$ so that $π_{X} (M)$ is a unions of open sets of $𝒯_{X}$ and therefore is open in $X$ . Symetrically wealso have that $π_{Y}$ is open.

Quotient Map

Given an

(X, \sim)

, then we define the quotient map of

\sim

as a function

π : X \to X ∖ \sim

(where we are mapping into the space modded out by the quotient)

π (x) = [x]

Quotient Topology by Universal Properties

Given a quotient map

π : X \to X ∖ \sim

then there exists a unique topology on

X ∖ \sim

such that

$π$ is continuous
If $f : X ∖ \sim \to Z$ is a function and $f \circ h$ is continuous, then so is $f$

Specifically it is given by the set

{V \subseteq X ∖ \sim : π^{- 1} (V) \in 𝒯_{X}}

We denote this topology by

𝒯_{X ∖ \sim}

First we show that it is a topology and then we show that it is unique.

Direction of a Vector Space

Suppose that

V

is a vector space then we define the direction of $V$ by considering the equivalence relation induced by

{(v_{1}, v_{2}) : v_{1} = α v_{2}, α \in ℝ^{> 0}}

which we denote by

D

then we define

dir (V) = V ∖ D

The Direction of the Reals

What is the quotient topology of

dir (ℝ)

Before starting we can see that $dir (ℝ) = {[- 1], [0], [1]}$ , now let's figure out the topology, we know that it is made up of sets whose inverse images under $π : ℝ \to dir (ℝ)$ is open.

We note that $π^{- 1} ([- 1]) = (- \infty, 0)$ , that $π^{- 1} ([0]) = {0}$ and that $π^{- 1} ([1]) = (0, \infty)$ . Any other subset of $dir (ℝ)$ is a union of these sets, moreover since the inverse image allows unions to pass through, it shows that the inverse image of every subset is also open, thus is the discrete topology.

Direction of R2

What is the quotient topology of

dir (ℝ^{2})

We note that $dir (ℝ^{2}) = {[e^{2 π i θ}] : θ \in [0, 1)} \cup {[0]}$ . Now let's start figuring out the topology, so we have to find those subsets of $dir (ℝ^{2})$ such that their inverse images are open in $ℝ^{2}$

Now we need to find all subsets whose inverse image under $π$ is open, one way of doing this is to consider single points, and then try to construct open inverse images by combining the single points, in this case a single point on the unit circle gets mapped to a ray that doesn't pas through the origin.which on it's own is not open in $ℝ^{2}$ so then we have to give it some area by considering an an open shell sector.