Difference between revisions of "Quotient topology"
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| + | <math>\require{AMScd}</math> | ||
| + | <math>\begin{equation}\begin{CD} | ||
| + | S^{{\mathcal{W}}_\Lambda}\otimes T @>j>> T\\ | ||
| + | @VVV @VV{\End P}V\\ | ||
| + | (S\otimes T)/I @= (Z\otimes T)/J | ||
| + | \end{CD}\end{equation}</math> | ||
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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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====Stronger than continuity==== | ====Stronger than continuity==== | ||
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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| + | See [[Motivation for quotient topology]] for a discussion on where this goes. | ||
==Definition== | ==Definition== | ||
Revision as of 09:22, 7 April 2015
[math]\require{AMScd}[/math] [math]\begin{equation}\begin{CD} S^{{\mathcal{W}}_\Lambda}\otimes T @>j>> T\\ @VVV @VV{\End P}V\\ (S\otimes T)/I @= (Z\otimes T)/J \end{CD}\end{equation}[/math]
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]
Notes
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
See Motivation for quotient topology for a discussion on where this goes.
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]
TODO: Munkres page 138
References
- ↑ Topology - Second Edition - James R Munkres
