Difference between revisions of "Quotient topology"

Revision as of 12:34, 7 April 2015 (view source) Alec (Talk | contribs) m ← Older edit		Revision as of 12:48, 7 April 2015 (view source) Alec (Talk | contribs) m (→‎Theorems) Newer edit →
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	===Theorems===		===Theorems===
	{{Begin Theorem}}		{{Begin Theorem}}
−	Theorem: The quotient topology, {{M|\mathcal{Q} }} is the largest topology such that the quotient map, {{M|p}} is continuous	+	Theorem: The quotient topology, {{M|\mathcal{Q} }} is the largest topology such that the quotient map, {{M|p}}, is continuous
	{{Begin Proof}}		{{Begin Proof}}
		+	For a map {{M|p:X\rightarrow Y}} where {{M|(X,\mathcal{J})}} is a [[Topological space]] we will show that the topology on {{M|Y}} given by:
		+	* <math>\mathcal{Q}=\{V\in\mathcal{P}|p^{-1}(V)\in\mathcal{J}\}</math>
		+	is the largest topology on {{M|Y}} we can have ''such that'' {{M|p}} is [[Continuous map|continuous]]
		+
		+	'''Proof method:''' suppose there's a larger topology, reach a contradiction.
		+
		+	Suppose that {{M|\mathcal{K} }} is any topology on {{M|Y}} and that {{M|p:(X,\mathcal{J})\rightarrow(Y,\mathcal{K})}} is continuous.
		+
		+	Suppose that {{M|\mathcal{K}\ne\mathcal{Q} }}
		+
		+
		+	Let {{M|V\in\mathcal{K} }} such that {{M|V\notin \mathcal{Q} }}
		+
		+	By continuity of {{M|p}}, {{M|p^{-1}(V)\in\mathcal{J} }}
		+
		+	This contradicts that {{M|V\notin\mathcal{Q} }} as {{M|\mathcal{Q} }} contains all subsets of {{M|Y}} whose inverse image (preimage) is open in {{M|X}}
		+
		+
		+	Thus any topology on {{M|Y}} where {{M|p}} is continuous is contained in the quotient topology
	{{End Proof}}		{{End Proof}}
	{{End Theorem}}		{{End Theorem}}

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Revision as of 12:48, 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

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TODO: Now we can explore the characteristic property (with $\text{Id}:\tfrac{X}{\sim}\rightarrow\tfrac{X}{\sim}$ ) for now

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Theorems

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