Difference between revisions of "Notes:Quotient topology"

Revision as of 17:07, 21 April 2016

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.

According to John M. Lee

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

• Let X and Z be 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