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], 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]

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 1: these definitions are equivalent

Universal property of the quotient topology

[ilmath]\xymatrix{ X \ar[d]_{q} \ar[dr]^{f\circ q} & \\ Y \ar[r]_f & Z }[/ilmath]
In this commutative diagram [ilmath]f[/ilmath] is continuous [ilmath]\iff[/ilmath] [ilmath]f\circ q[/ilmath] is continuous

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:

• 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

OLD PAGE

Note: Motivation for quotient topology may be useful

Definition of the Quotient topology

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

(OLD)Definition of Quotient topology

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

The quotient topology is actually a topology

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TODO: Easy enough

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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 if we have [math]U\in\mathcal{K}\iff p^{-1}(U)\in\mathcal{J}[/math]

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

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

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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:

• [math]\mathcal{Q}=\{V\in\mathcal{P}|p^{-1}(V)\in\mathcal{J}\}[/math]

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: [math]q:(X,\mathcal{J})\rightarrow(\tfrac{X}{\sim},\mathcal{Q})[/math] with [math]q:p\mapsto[p][/math] (which is a quotient map) is continuous (as above)

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

Also known as:

• Identification space

See also

• Open and closed maps