Difference between revisions of "Quotient topology"

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'''Note:''' [[Motivation for quotient topology]] may be useful
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==Definition==
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If <math>(X,\mathcal{J})</math> is a [[Topological space|topological space]], <math>A</math> is a set, and <math>p:(X,\mathcal{J})\rightarrow A</math> is a [[Surjection|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>
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{{Begin Theorem}}
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The quotient topology is actually a topology
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{{Begin Proof}}
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{{Todo|Easy enough}}
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{{End Proof}}
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{{End Theorem}}
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==Quotient map==
 
==Quotient map==
 
Let {{M|(X,\mathcal{J})}} and {{M|(Y,\mathcal{K})}} be [[Topological space|topological spaces]] and let {{M|p:X\rightarrow Y}} be a [[Surjection|surjective]] map.
 
Let {{M|(X,\mathcal{J})}} and {{M|(Y,\mathcal{K})}} be [[Topological space|topological spaces]] and let {{M|p:X\rightarrow Y}} be a [[Surjection|surjective]] map.
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{{M|p}} is a quotient map<ref>Topology - Second Edition - James R Munkres</ref> if we have <math>U\in\mathcal{K}\iff p^{-1}(U)\in\mathcal{J}</math>
 
{{M|p}} is a quotient map<ref>Topology - Second Edition - James R Munkres</ref> if we have <math>U\in\mathcal{K}\iff p^{-1}(U)\in\mathcal{J}</math>
  
===Notes===
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===Stronger than continuity===
====Stronger than continuity====
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If we had {{M|1=\mathcal{K}=\{\emptyset,Y\} }} then {{M|p}} is automatically continuous (as it is surjective), the point is that {{M|\mathcal{K} }} is the [[Topology#Finer.2C_Larger.2C_Stronger|largest topology]] we can define on {{M|Y}} such that {{M|p}} is continuous
 
If we had {{M|1=\mathcal{K}=\{\emptyset,Y\} }} then {{M|p}} is automatically continuous (as it is surjective), the point is that {{M|\mathcal{K} }} is the [[Topology#Finer.2C_Larger.2C_Stronger|largest topology]] we can define on {{M|Y}} such that {{M|p}} is continuous
  
 
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{{Todo|Now we can explore the characteristic property (with {{M|\text{Id}:\tfrac{X}{\sim}\rightarrow\tfrac{X}{\sim} }} ) for now}}
See [[Motivation for quotient topology]] for a discussion on where this goes.
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==Definition==
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If <math>(X,\mathcal{J})</math> is a [[Topological space|topological space]], <math>A</math> is a set, and <math>p:(X,\mathcal{J})\rightarrow A</math> is a [[Surjection|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>
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{{Todo|Munkres page 138}}
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==References==
 
==References==

Revision as of 11:13, 7 April 2015

Note: Motivation for quotient topology may be useful


Definition

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




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

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



TODO: Now we can explore the characteristic property (with [ilmath]\text{Id}:\tfrac{X}{\sim}\rightarrow\tfrac{X}{\sim} [/ilmath] ) for now



References

  1. Topology - Second Edition - James R Munkres