Difference between revisions of "Quotient topology"
m (→Theorems) |
m |
||
| Line 26: | Line 26: | ||
===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 | ||
| − | |||
| − | |||
===Theorems=== | ===Theorems=== | ||
| Line 54: | Line 52: | ||
{{End Proof}} | {{End Proof}} | ||
{{End Theorem}} | {{End Theorem}} | ||
| + | This theorem hints at the [[Characteristic property of the quotient topology]] | ||
==Quotient space== | ==Quotient space== | ||
Revision as of 12:54, 7 April 2015
Note: Motivation for quotient topology may be useful
Contents
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
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]
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
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[2] where [ilmath]\mathcal{Q} [/ilmath] is the topology induced by the quotient
Also known as:
- Identification space
