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.

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:

1. It is surjective
2. 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

[ilmath]\xymatrix{ X \ar[d]_q \ar[dr]^f & \\ Y \ar@{.>}[r]_{\bar{f}} & Z }[/ilmath]
Passing to the quotient

This is very similar to the quotient of a function.

• 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:

1. 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.
2. 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:

1. Identification map
2. 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

1. Quotient topology, given a topological space [ilmath](X,\mathcal{J})[/ilmath] and a set [ilmath]Y[/ilmath], as usual.
2. 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
3. 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]