# Quotient topology

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## Definition

There are a few definitions of the quotient topology however they do not conflict. This page might change shape while things are put in place

### Quotient topology via an equivalence-relation definition

Given a topological space, [ilmath](X,\mathcal{J})[/ilmath] and an equivalence relation on [ilmath]X[/ilmath], [ilmath]\sim[/ilmath][Note 1], the quotient topology on [ilmath]\frac{X}{\sim} [/ilmath], [ilmath]\mathcal{K} [/ilmath] is defined as:

• The set [ilmath]\mathcal{K}\subseteq\mathcal{P}(\frac{X}{\sim})[/ilmath] such that:
• [ilmath]\forall U\in\mathcal{P}(\frac{X}{\sim})[U\in\mathcal{K}\iff \pi^{-1}(U)\in\mathcal{J}][/ilmath] or equivalently
• [ilmath]\mathcal{K}=\{U\in\mathcal{P}(\frac{X}{\sim})\ \vert\ \pi^{-1}(U)\in\mathcal{J}\}[/ilmath]

In words:

• The topology on [ilmath]\frac{X}{\sim} [/ilmath] consists of all those sets whose pre-image (under [ilmath]\pi[/ilmath]) are open in [ilmath]X[/ilmath]
Claim 1: [ilmath]\mathcal{K} [/ilmath] is indeed a topology on [ilmath]\frac{X}{\sim} [/ilmath]

### Quotient topology via a mapping to a set definition

Let [ilmath](X,\mathcal{J})[/ilmath] be a topological space and let [ilmath]h:X\rightarrow Y[/ilmath] be a surjective map onto a set [ilmath]Y[/ilmath], then the quotient topology, [ilmath]\mathcal{K}\subseteq\mathcal{P}(Y)[/ilmath] is a topology we define on [ilmath]Y[/ilmath] as follows:

• [ilmath]\forall U\in\mathcal{P}(Y)[Y\in\mathcal{K}\iff h^{-1}(U)\in\mathcal{J}][/ilmath] or equivalently:
• [ilmath]\mathcal{K}=\{U\in\mathcal{P}(Y)\ \vert\ h^{-1}(U)\in\mathcal{J}\}[/ilmath]

The quotient topology on [ilmath]Y[/ilmath] consists of all those subsets of [ilmath]Y[/ilmath] whose pre-image (under [ilmath]h[/ilmath]) is open in [ilmath]X[/ilmath]

Claim 2: these definitions are equivalent

## Immediate theorems

The next two theorems demonstrate the purpose, the job if you will, of the quotient topology. The second (passing to the quotient) is the most important.

### Universal property of the quotient topology

 In this commutative diagram[ilmath]f[/ilmath] is continuous[ilmath]\iff[/ilmath][ilmath]f\circ q[/ilmath] is continuous [ilmath]\xymatrix{ X \ar[d]_{q} \ar[dr]^{f\circ q} & \\ Y \ar[r]_f & Z }[/ilmath]
Let [ilmath](X,\mathcal{ J })[/ilmath] and [ilmath](Y,\mathcal{ K })[/ilmath] be topological spaces and let [ilmath]q:X\rightarrow Y[/ilmath] be a quotient map. Then[1]:
• For any topological space, [ilmath](Z,\mathcal{ H })[/ilmath] a map, [ilmath]f:Y\rightarrow Z[/ilmath] is continuous if and only if the composite map, [ilmath]f\circ q[/ilmath], is continuous

### Passing to the quotient

 [ilmath]f[/ilmath] descends to the quotient [ilmath]\xymatrix{ X \ar[d]_\pi \ar[dr]^f & \\ \frac{X}{\sim} \ar@{.>}[r]^{\overline{f} }& Y}[/ilmath]

Suppose that [ilmath](X,\mathcal{ J })[/ilmath] is a topological space and [ilmath]\sim[/ilmath] is an equivalence relation, let [ilmath](\frac{X}{\sim},\mathcal{ Q })[/ilmath] be the resulting quotient topology and [ilmath]\pi:X\rightarrow\frac{X}{\sim} [/ilmath] the resulting quotient map, then:

• Let [ilmath](Y,\mathcal{ K })[/ilmath] be any topological space and let [ilmath]f:X\rightarrow Y[/ilmath] be a continuous map that is constant on the fibres of [ilmath]\pi[/ilmath][Note 2] then:
• there exists a unique continuous map, [ilmath]\bar{f}:\frac{X}{\sim}\rightarrow Y[/ilmath] such that [ilmath]f=\overline{f}\circ\pi[/ilmath]

We may then say [ilmath]f[/ilmath] descends to the quotient or passes to the quotient

Note: this is an instance of passing-to-the-quotient for functions

## Notes

1. Recall that for an equivalence relation there is a natural map that sends each [ilmath]x\in X[/ilmath] to [ilmath][x][/ilmath] (the equivalence class containing [ilmath]x[/ilmath]) which we denote here as [ilmath]\pi:X\rightarrow\frac{X}{\sim} [/ilmath]. Recall also that [ilmath]\frac{X}{\sim} [/ilmath] denotes the set of all equivalence classes of [ilmath]\sim[/ilmath].
2. That means that:

# OLD PAGE

Note: Motivation for quotient topology may be useful

## Definition of the Quotient topology

$\begin{xy} \xymatrix{ X \ar[r]^p \ar[dr]_f & Q \ar@{.>}[d]^{\tilde{f}}\\ & Y } \end{xy}$

## (OLD)Definition of Quotient topology

If $(X,\mathcal{J})$ is a topological space, $A$ is a set, and $p:(X,\mathcal{J})\rightarrow A$ is a surjective map then there exists exactly one topology $\mathcal{J}_Q$ relative to which $p$ is a quotient map. This is the quotient topology induced by $p$

The quotient topology is actually a topology

TODO: Easy enough

## Quotient map

Let [ilmath](X,\mathcal{J})[/ilmath] and [ilmath](Y,\mathcal{K})[/ilmath] be topological spaces and let [ilmath]p:X\rightarrow Y[/ilmath] be a surjective map.

[ilmath]p[/ilmath] is a quotient map[1] if we have $U\in\mathcal{K}\iff p^{-1}(U)\in\mathcal{J}$

That is to say $\mathcal{K}=\{V\in\mathcal{P}(Y)|p^{-1}(V)\in\mathcal{J}\}$

Also known as:

• Identification map

### Stronger than continuity

If we had [ilmath]\mathcal{K}=\{\emptyset,Y\}[/ilmath] then [ilmath]p[/ilmath] is automatically continuous (as it is surjective), the point is that [ilmath]\mathcal{K} [/ilmath] is the largest topology we can define on [ilmath]Y[/ilmath] such that [ilmath]p[/ilmath] is continuous

### Theorems

Theorem: The quotient topology, [ilmath]\mathcal{Q} [/ilmath] is the largest topology such that the quotient map, [ilmath]p[/ilmath], is continuous. That is to say any other topology such on [ilmath]Y[/ilmath] such that [ilmath]p[/ilmath] is continuous is contained in the quotient topology

For a map [ilmath]p:X\rightarrow Y[/ilmath] where [ilmath](X,\mathcal{J})[/ilmath] is a Topological space we will show that the topology on [ilmath]Y[/ilmath] given by:

• $\mathcal{Q}=\{V\in\mathcal{P}|p^{-1}(V)\in\mathcal{J}\}$

is the largest topology on [ilmath]Y[/ilmath] we can have such that [ilmath]p[/ilmath] is continuous

Proof method: suppose there's a larger topology, reach a contradiction.

Suppose that [ilmath]\mathcal{K} [/ilmath] is any topology on [ilmath]Y[/ilmath] and that [ilmath]p:(X,\mathcal{J})\rightarrow(Y,\mathcal{K})[/ilmath] is continuous.

Suppose that [ilmath]\mathcal{K}\ne\mathcal{Q} [/ilmath]

Let [ilmath]V\in\mathcal{K} [/ilmath] such that [ilmath]V\notin \mathcal{Q} [/ilmath]

By continuity of [ilmath]p[/ilmath], [ilmath]p^{-1}(V)\in\mathcal{J} [/ilmath]

This contradicts that [ilmath]V\notin\mathcal{Q} [/ilmath] as [ilmath]\mathcal{Q} [/ilmath] contains all subsets of [ilmath]Y[/ilmath] whose inverse image (preimage) is open in [ilmath]X[/ilmath]

Thus any topology on [ilmath]Y[/ilmath] where [ilmath]p[/ilmath] is continuous is contained in the quotient topology

This theorem hints at the Characteristic property of the quotient topology

## Quotient space

Given a Topological space [ilmath](X,\mathcal{J})[/ilmath] and an Equivalence relation [ilmath]\sim[/ilmath], then the map: $q:(X,\mathcal{J})\rightarrow(\tfrac{X}{\sim},\mathcal{Q})$ with $q:p\mapsto[p]$ (which is a quotient map) is continuous (as above)

The topological space [ilmath](\tfrac{X}{\sim},\mathcal{Q})[/ilmath] is the quotient space[2] where [ilmath]\mathcal{Q} [/ilmath] is the topology induced by the quotient

Also known as:

• Identification space