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 $\sim$ denote an equivalence relation, let $(X,\mathcal{J})$ be a topological space. We get a map, $\pi:X\rightarrow\frac{X}{\sim}$ that takes $\pi:x\rightarrow[x]$

• The quotient topology on $\frac{X}{\sim}$ is the finest such that $\pi$ is continuous

Let $\mathcal{K}$ denote a topology on $\frac{X}{\sim}$, then we may define $\mathcal{K}$ as:

• $\mathcal{K}:=\{U\in\mathcal{P}(\frac{X}{\sim})\ \vert\ \pi^{-1}(U)\in\mathcal{J} \}$, that is:
  • $U\in\mathcal{P}(\frac{X}{\sim})$ is open if $\pi^{-1}(U)$ is open in $X$ - 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 "$U\in\mathcal{P}(\frac{X}{\sim})$ is open in $\frac{X}{\sim}$ if and only if $\pi^{-1}(U)\in\mathcal{J}$, not sure why they stress it so.

Quotient map

A map between two topological spaces $(X,\mathcal{J})$ and $(Y,\mathcal{K})$ is a quotient map if:

1. It is surjective
2. The topology on $Y$ ($\mathcal{K}$) is the quotient topology that'd be induced on $Y$ by the map $q$

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:

• $x\sim y\iff q(x)=q(y)$ and that's where this comes from

Passing to the quotient

• Let X and Z be topological spaces,
• let $q:X\rightarrow Y$ be a quotient map,
• let $f:X\rightarrow Z$ be any continuous mapping such that $q(x)=q(y)\implies f(x)=f(y)$

Then

• There exists a unique continuous map, $\bar{f}:Y\rightarrow Z$ such that $f=\bar{f}\circ q$

Munkres