Difference between revisions of "Quotient topology"
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− | ''' | + | {{Refactor notice|grade=A|msg=This page is an embarrassment}} |
− | {{ | + | ==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/Equivalence relation definition|Quotient topology via an equivalence-relation definition]]=== | ||
+ | {{:Quotient topology/Equivalence relation definition}} | ||
+ | ===[[Quotient topology/Mapping to a set definition|Quotient topology via a mapping to a set definition]]=== | ||
+ | {{:Quotient topology/Mapping to a set definition}} | ||
+ | : '''Claim 1: ''' these definitions are equivalent | ||
+ | ===[[Universal property of the quotient topology]]=== | ||
+ | {{:Universal property of the quotient topology/Statement}} | ||
+ | ==Proof of claims== | ||
+ | ==Notes== | ||
+ | <references group="Note"/> | ||
+ | ==References== | ||
+ | <references/> | ||
+ | {{Topology navbox|plain}} | ||
+ | {{Definition|Topology}} | ||
+ | =OLD PAGE= | ||
+ | '''Note:''' [[Motivation for quotient topology]] may be useful | ||
==Definition of the Quotient topology== | ==Definition of the Quotient topology== | ||
<math> | <math> |
Revision as of 14:20, 25 April 2016
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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][Note 1], 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
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[1]:- 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
Proof of claims
Notes
- ↑ Recall that for an equivalence relation there is a natural map that sends each [ilmath]x\in X[/ilmath] to [ilmath][x][/ilmath] (the equivalence class containing [ilmath]x[/ilmath]) which we denote here as [ilmath]\pi:X\rightarrow\frac{X}{\sim} [/ilmath]. Recall also that [ilmath]\frac{X}{\sim} [/ilmath] denotes the set of all equivalence classes of [ilmath]\sim[/ilmath].
References
|
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
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. 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
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