Notes:Quotient topology
- Note to readers: the page quotient topology as it stands right now (Alec (talk) 17:07, 21 April 2016 (UTC)) is an embarrassment to me. However before I can clean it up I must unify it. I've been using it for almost 2 years now though I promise! Gosh this is embarrassing.
See Notes:Quotient topology plan for an outline of the page.
Contents
According to John M. Lee
- Addendum: for some reason I lie here, the author considers a map [ilmath]f:X\rightarrow A[/ilmath] where [ilmath]A[/ilmath] is a set and later applies the equivalence relation version to it.
Let [ilmath]\sim[/ilmath] denote an equivalence relation, let [ilmath](X,\mathcal{J})[/ilmath] be a topological space. We get a map, [ilmath]\pi:X\rightarrow\frac{X}{\sim} [/ilmath] that takes [ilmath]\pi:x\rightarrow[x][/ilmath]
- The quotient topology on [ilmath]\frac{X}{\sim} [/ilmath] is the finest such that [ilmath]\pi[/ilmath] is continuous
Let [ilmath]\mathcal{K} [/ilmath] denote a topology on [ilmath]\frac{X}{\sim} [/ilmath], then we may define [ilmath]\mathcal{K} [/ilmath] as:
- [ilmath]\mathcal{K}:=\{U\in\mathcal{P}(\frac{X}{\sim})\ \vert\ \pi^{-1}(U)\in\mathcal{J} \}[/ilmath], that is:
- [ilmath]U\in\mathcal{P}(\frac{X}{\sim})[/ilmath] is open if [ilmath]\pi^{-1}(U)[/ilmath] is open in [ilmath]X[/ilmath] - we get "only if" by going the other way. I must make a page about how definitions are "iff"s
Note: more than one book is very clear on "[ilmath]U\in\mathcal{P}(\frac{X}{\sim})[/ilmath] is open in [ilmath]\frac{X}{\sim} [/ilmath] if and only if [ilmath]\pi^{-1}(U)\in\mathcal{J} [/ilmath], not sure why they stress it so.
Quotient map
A map between two topological spaces [ilmath](X,\mathcal{J})[/ilmath] and [ilmath](Y,\mathcal{K})[/ilmath] is a quotient map if:
- It is surjective
- The topology on [ilmath]Y[/ilmath] ([ilmath]\mathcal{K} [/ilmath]) is the quotient topology that'd be induced on [ilmath]Y[/ilmath] by the map [ilmath]q[/ilmath]
Lee actually defines the quotient topology using maps first, then constructs the equiv relation version, but we can can define an equivalence relation as follows:
- [ilmath]x\sim y\iff q(x)=q(y)[/ilmath] and that's where this comes from
Passing to the quotient
Passing to the quotient |
---|
- Let X and Z be topological space|topological spaces,
- let [ilmath]q:X\rightarrow Y[/ilmath] be a quotient map,
- let [ilmath]f:X\rightarrow Z[/ilmath] be any continuous mapping such that [ilmath]q(x)=q(y)\implies f(x)=f(y)[/ilmath]
Then
- There exists a unique continuous map, [ilmath]\bar{f}:Y\rightarrow Z[/ilmath] such that [ilmath]f=\bar{f}\circ q[/ilmath]
Munkres
Munkres starts with a quotient map
- 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 surjective map
We say [ilmath]q[/ilmath] is a quotient map provided:
- [ilmath]\forall U\in\mathcal{P}(Y)[U\in\mathcal{K}\iff p^{-1}(U)\in\mathcal{J}][/ilmath]
He goes on to say:
- This condition is "stronger than continuity" (of [ilmath]q[/ilmath] presumably) probably because if we gave [ilmath]Y[/ilmath] the indiscrete topology it'd be continuous.
- He defines this in several ways, one of which is "saturation" of maps. Yeah this is just the equivalence relation version hiding (CHECK THIS THOUGH)
Quotient topology
If [ilmath](X,\mathcal{J})[/ilmath] is a topological space and [ilmath]A[/ilmath] a set and if [ilmath]p:X\rightarrow A[/ilmath] is a surjective map then:
- There is exactly on topology, [ilmath]\mathcal{K} [/ilmath] on [ilmath]A[/ilmath] relative to which [ilmath]p[/ilmath] is a quotient map (as defined above)
That topology is the quotient topology induced by [ilmath]p[/ilmath]
Quotient space
Let [ilmath](X,\mathcal{J})[/ilmath] be a topological space and let [ilmath]X^*[/ilmath] be a partition of [ilmath]X[/ilmath] into disjoint subsets whose union is [ilmath]X[/ilmath] (that is the definition of a partition). Let [ilmath]p:X\rightarrow X^*[/ilmath] be the surjective map that carries each point of [ilmath]X[/ilmath] to the element of [ilmath]X^*[/ilmath] containing that point, then:
- The quotient topology induced by [ilmath]p[/ilmath] on [ilmath]X^*[/ilmath] is called the quotient space of [ilmath]X[/ilmath]
Mond
- Note: David Mond is my tutor at university. While I do not like his style of writing (informal definitions, ambiguities in the English interpretation) he does have a great way of ordering things. That is applicable here. I also found like 9 typos in the first 8 pages; however it has many examples and many pictures, and they are lecture notes.
Mond starts with a topological space [ilmath](X,\mathcal{J})[/ilmath] and an equivalence relation, [ilmath]\sim[/ilmath]. Then:
- The quotient topology ([ilmath]\mathcal{K} [/ilmath]) on [ilmath]\frac{X}{\sim} [/ilmath] is the topology where [ilmath]\forall U\in\mathcal{P}(\frac{X}{\sim})[U\in\mathcal{K}\iff \pi^{-1}(U)\in\mathcal{J}][/ilmath]
Passing to the quotient
Mond then goes for passing to the quotient, exactly as John M. Lee has. Very weirdly worded though.
No mention of the quotient map as a concept.
Deals with equivalence relation generated by which is great.
Mendelson
Starts with what is called a quotient map above, but calls it "an identification". Then he goes straight on to "passing to the quotient" it's a very weirdly written section, but he does:
- Identification map
- Identification topology
It seems I'll have to prove these concepts are the same (having one induces the other). Provides a source for the "identification" terminology, which is useful.
Gamelin & Greene
- Quotient topology, given a topological space [ilmath](X,\mathcal{J})[/ilmath] and a set [ilmath]Y[/ilmath], as usual.
- Slightly strange, again top space [ilmath](X,\mathcal{J})[/ilmath] and an equivalence relation [ilmath]\sim[/ilmath], for [ilmath]f:\frac{X}{\sim}\rightarrow Y[/ilmath] - [ilmath]f[/ilmath] is continuous [ilmath]\iff[/ilmath] [ilmath]f\circ\pi[/ilmath] is continuous ([ilmath]\pi[/ilmath] being the canonical projection)
- This is NOT passing to the quotient
- Passing to the quotient now. Let [ilmath]f:X\rightarrow Y[/ilmath] be continuous function, and [ilmath]\sim[/ilmath] and equivalence relation on [ilmath]X[/ilmath] such that [ilmath]f[/ilmath] is constant for each [ilmath]x\in[x]\in\frac{X}{\sim} [/ilmath]. Then:
- [ilmath]\exists g:\frac{X}{\sim}\rightarrow Y[/ilmath] continuous such that [ilmath]f=g\circ\pi[/ilmath]