Difference between revisions of "Quotient topology"

Revision as of 12:48, 7 April 2015 (view source) Alec (Talk | contribs) m (→‎Theorems) ← Older edit		Revision as of 12:54, 7 April 2015 (view source) Alec (Talk | contribs) m Newer edit →
Line 26:		Line 26:
	===Stronger than continuity===		===Stronger than continuity===
	If we had {{M|1=\mathcal{K}=\{\emptyset,Y\} }} then {{M|p}} is automatically continuous (as it is surjective), the point is that {{M|\mathcal{K} }} is the [[Topology#Finer.2C_Larger.2C_Stronger|largest topology]] we can define on {{M|Y}} such that {{M|p}} is continuous		If we had {{M|1=\mathcal{K}=\{\emptyset,Y\} }} then {{M|p}} is automatically continuous (as it is surjective), the point is that {{M|\mathcal{K} }} is the [[Topology#Finer.2C_Larger.2C_Stronger|largest topology]] we can define on {{M|Y}} such that {{M|p}} is continuous
−
−	{{Todo|Now we can explore the characteristic property (with {{M|\text{Id}:\tfrac{X}{\sim}\rightarrow\tfrac{X}{\sim} }} ) for now}}
	===Theorems===		===Theorems===
Line 54:		Line 52:
	{{End Proof}}		{{End Proof}}
	{{End Theorem}}		{{End Theorem}}
		+	This theorem hints at the [[Characteristic property of the quotient topology]]
	==Quotient space==		==Quotient space==

---

Revision as of 12:54, 7 April 2015

Note: Motivation for quotient topology may be useful

Definition of Quotient topology

If

$$(X,\mathcal{J})$$ is a topological space, $$A$$ is a set, and $$p:(X,\mathcal{J})\rightarrow A$$ is a surjective map then there exists exactly one topology $$\mathcal{J}_Q$$ relative to which $$p$$ is a quotient map. This is the quotient topology induced by $$p$$

Quotient map

Let $(X,\mathcal{J})$ and $(Y,\mathcal{K})$ be topological spaces and let $p:X\rightarrow Y$ be a surjective map.

$p$ is a quotient map^[1] if we have

$$U\in\mathcal{K}\iff p^{-1}(U)\in\mathcal{J}$$

That is to say

$$\mathcal{K}=\{V\in\mathcal{P}(Y)|p^{-1}(V)\in\mathcal{J}\}$$

Also known as:

• Identification map

Stronger than continuity

If we had $\mathcal{K}=\{\emptyset,Y\}$ then $p$ is automatically continuous (as it is surjective), the point is that $\mathcal{K}$ is the largest topology we can define on $Y$ such that $p$ is continuous

Theorems

This theorem hints at the Characteristic property of the quotient topology

Quotient space

Given a Topological space $(X,\mathcal{J})$ and an Equivalence relation $\sim$, then the map:

$$q:(X,\mathcal{J})\rightarrow(\tfrac{X}{\sim},\mathcal{Q})$$ with $$q:p\mapsto[p]$$ (which is a quotient map) is continuous (as above)

The topological space $(\tfrac{X}{\sim},\mathcal{Q})$ is the quotient space^[2] where $\mathcal{Q}$ is the topology induced by the quotient

Also known as:

• Identification space

References

1. Jump up ↑ Topology - Second Edition - James R Munkres
2. Jump up ↑ Introduction to topological manifolds - John M Lee - Second edition