Quotient topology
The message provided is:
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 equivalencerelation 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 preimage (under [ilmath]\pi[/ilmath]) are open in [ilmath]X[/ilmath]
 Claim 1: [ilmath]\mathcal{K} [/ilmath] is indeed a topology on [ilmath]\frac{X}{\sim} [/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 preimage (under [ilmath]h[/ilmath]) is open in [ilmath]X[/ilmath]
 Claim 2: these definitions are equivalent
Immediate theorems
The next two theorems demonstrate the purpose, the job if you will, of the quotient topology. The second (passing to the quotient) is the most important.
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
Passing to the quotient
Suppose that [ilmath](X,\mathcal{ J })[/ilmath] is a topological space and [ilmath]\sim[/ilmath] is an equivalence relation, let [ilmath](\frac{X}{\sim},\mathcal{ Q })[/ilmath] be the resulting quotient topology and [ilmath]\pi:X\rightarrow\frac{X}{\sim} [/ilmath] the resulting quotient map, then:
 Let [ilmath](Y,\mathcal{ K })[/ilmath] be any topological space and let [ilmath]f:X\rightarrow Y[/ilmath] be a continuous map that is constant on the fibres of [ilmath]\pi[/ilmath]^{[Note 2]} then:
 there exists a unique continuous map, [ilmath]\bar{f}:\frac{X}{\sim}\rightarrow Y[/ilmath] such that [ilmath]f=\overline{f}\circ\pi[/ilmath]
We may then say [ilmath]f[/ilmath] descends to the quotient or passes to the quotient
 Note: this is an instance of passingtothequotient for functions
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].
 ↑
That means that:
 [ilmath]\forall x,y\in X[\pi(x)=\pi(y)\implies f(x)=f(y)][/ilmath]  as mentioned in passingtothequotient for functions, or
 [ilmath]\forall x,y\in X[f(x)\ne f(y)\implies \pi(x)\ne\pi(y)][/ilmath], also mentioned
 See : Equivalent conditions to being constant on the fibres of a map for details
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