Difference between revisions of "Quotient topology"
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===Theorems=== | ===Theorems=== | ||
{{Begin Theorem}} | {{Begin Theorem}} | ||
| − | Theorem: The quotient topology, {{M|\mathcal{Q} }} is the largest topology such that the quotient map, {{M|p}} is continuous | + | Theorem: The quotient topology, {{M|\mathcal{Q} }} is the largest topology such that the quotient map, {{M|p}}, is continuous |
{{Begin Proof}} | {{Begin Proof}} | ||
| + | For a map {{M|p:X\rightarrow Y}} where {{M|(X,\mathcal{J})}} is a [[Topological space]] we will show that the topology on {{M|Y}} given by: | ||
| + | * <math>\mathcal{Q}=\{V\in\mathcal{P}|p^{-1}(V)\in\mathcal{J}\}</math> | ||
| + | is the largest topology on {{M|Y}} we can have ''such that'' {{M|p}} is [[Continuous map|continuous]] | ||
| + | |||
| + | '''Proof method:''' suppose there's a larger topology, reach a contradiction. | ||
| + | |||
| + | Suppose that {{M|\mathcal{K} }} is any topology on {{M|Y}} and that {{M|p:(X,\mathcal{J})\rightarrow(Y,\mathcal{K})}} is continuous. | ||
| + | |||
| + | Suppose that {{M|\mathcal{K}\ne\mathcal{Q} }} | ||
| + | |||
| + | |||
| + | Let {{M|V\in\mathcal{K} }} such that {{M|V\notin \mathcal{Q} }} | ||
| + | |||
| + | By continuity of {{M|p}}, {{M|p^{-1}(V)\in\mathcal{J} }} | ||
| + | |||
| + | This contradicts that {{M|V\notin\mathcal{Q} }} as {{M|\mathcal{Q} }} contains all subsets of {{M|Y}} whose inverse image (preimage) is open in {{M|X}} | ||
| + | |||
| + | |||
| + | Thus any topology on {{M|Y}} where {{M|p}} is continuous is contained in the quotient topology | ||
{{End Proof}} | {{End Proof}} | ||
{{End Theorem}} | {{End Theorem}} | ||
Revision as of 12:48, 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
TODO: Now we can explore the characteristic property (with [ilmath]\text{Id}:\tfrac{X}{\sim}\rightarrow\tfrac{X}{\sim} [/ilmath] ) for now
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
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
