Difference between revisions of "Disjoint union topology"

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(Created page with "{{Stub page|grade=A|msg=Grade A until it is more presentable.}} ==Definition== Suppose {{M|\big((X_\alpha,\mathcal{J}_\alpha)\big)_{\alpha\in I} }} be an indexed family of t...")
 
m (Definition: Added note about canonical injections being topological embeddings - still needs refactoring)
 
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* The [[topology]] where {{M|U\in\mathcal{P}(\coprod_{\alpha\in I}X_\alpha)}} is considered [[open set|open]] ''if and only if'' {{M|1=\forall \alpha\in I[X_\alpha\cap U\in\mathcal{J}_\alpha]}}<ref group="Note">There's a very nasty abuse of notation going on here. First, note a set {{M|U}} is going to be a bunch of points of the form {{M|(x,\gamma)}} for various {{M|x}}s and {{M|\gamma}}s ({{M|\in I}}). There is no "canonical projection" FROM the product to the spaces, as this would not be a [[function]]!</ref> - be sure to notice the abuse of notation going on here.
 
* The [[topology]] where {{M|U\in\mathcal{P}(\coprod_{\alpha\in I}X_\alpha)}} is considered [[open set|open]] ''if and only if'' {{M|1=\forall \alpha\in I[X_\alpha\cap U\in\mathcal{J}_\alpha]}}<ref group="Note">There's a very nasty abuse of notation going on here. First, note a set {{M|U}} is going to be a bunch of points of the form {{M|(x,\gamma)}} for various {{M|x}}s and {{M|\gamma}}s ({{M|\in I}}). There is no "canonical projection" FROM the product to the spaces, as this would not be a [[function]]!</ref> - be sure to notice the abuse of notation going on here.
 
{{Todo|Flesh out notes, mention subspace {{M|X_\alpha\times\{\alpha\} }} and such}}
 
{{Todo|Flesh out notes, mention subspace {{M|X_\alpha\times\{\alpha\} }} and such}}
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'''Claim 1: ''' this is indeed a topology
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{{Todo|Define the [[canonical injections of the disjoint union topology]] here}}
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Note that [[the canonical injections of the disjoint union topology are topological embeddings]]
  
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==[[Characteristic property of the disjoint union topology|Characteristic property]]==
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{{:Characteristic property of the disjoint union topology/Statement}}
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==Proof of claims==
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{{Requires proof|msg=Actually surprisingly easy to prove, done on paper. page 1, 7/8/2016, Intro to top manifolds notes. Filed}}
 
==Notes==
 
==Notes==
 
<references group="Note"/>
 
<references group="Note"/>

Latest revision as of 23:47, 25 September 2016

Stub grade: A
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Grade A until it is more presentable.

Definition

Suppose [ilmath]\big((X_\alpha,\mathcal{J}_\alpha)\big)_{\alpha\in I} [/ilmath] be an indexed family of topological spaces that are non-empty[1], the disjoint union topology is a topological space:

  • with underlying set [ilmath]\coprod_{\alpha\in I}X_\alpha[/ilmath], this is the disjoint union of sets, recall [ilmath](x,\beta)\in\coprod_{\alpha\in I}X_\alpha\iff \beta\in I\wedge x\in X_\beta[/ilmath] and
  • The topology where [ilmath]U\in\mathcal{P}(\coprod_{\alpha\in I}X_\alpha)[/ilmath] is considered open if and only if [ilmath]\forall \alpha\in I[X_\alpha\cap U\in\mathcal{J}_\alpha][/ilmath][Note 1] - be sure to notice the abuse of notation going on here.

TODO: Flesh out notes, mention subspace [ilmath]X_\alpha\times\{\alpha\} [/ilmath] and such


Claim 1: this is indeed a topology


TODO: Define the canonical injections of the disjoint union topology here


Note that the canonical injections of the disjoint union topology are topological embeddings

Characteristic property

Let [ilmath]\big((X_\alpha,\mathcal{J}_\alpha)\big)_{\alpha\in I} [/ilmath] be a collection of topological spaces and let [ilmath](Y,\mathcal{ K })[/ilmath] be another topological space]]. We denote by [ilmath]\coprod_{\alpha\in I}X_\alpha[/ilmath] the disjoint unions of the underlying sets of the members of the family, and by [ilmath]\mathcal{J} [/ilmath] the disjoint union on it (so [ilmath](\coprod_{\alpha\in I}X_\alpha,\mathcal{J})[/ilmath] is the disjoint union topological construct of the [ilmath]\big((X_\alpha,\mathcal{J}_\alpha)\big)_{\alpha\in I} [/ilmath] family) and lastly, let [ilmath]f:\coprod_{\alpha\in I}X_\alpha\rightarrow Y[/ilmath] be a map (not necessarily continuous) then:
we state the characteristic property of disjoint union topology as follows:


TODO: rewrite and rephrase this


  • [ilmath]f:\coprod_{\alpha\in I}X_\alpha\rightarrow Y[/ilmath] is continuous if and only if [ilmath]\forall\alpha\in I\big[f\big\vert_{X_\alpha^*}:i_\alpha({X_\alpha})\rightarrow Y\text{ is continuous}\big][/ilmath]

Where (for [ilmath]\beta\in I[/ilmath]) we have [ilmath]i_\beta:X_\beta\rightarrow\coprod_{\alpha\in I}X_\alpha[/ilmath] given by [ilmath]i_\beta:x\mapsto(\beta,x)[/ilmath] are the canonical injections

Proof of claims

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Actually surprisingly easy to prove, done on paper. page 1, 7/8/2016, Intro to top manifolds notes. Filed

Notes

  1. There's a very nasty abuse of notation going on here. First, note a set [ilmath]U[/ilmath] is going to be a bunch of points of the form [ilmath](x,\gamma)[/ilmath] for various [ilmath]x[/ilmath]s and [ilmath]\gamma[/ilmath]s ([ilmath]\in I[/ilmath]). There is no "canonical projection" FROM the product to the spaces, as this would not be a function!

References

  1. Introduction to Topological Manifolds - John M. Lee



TODO: Investigate the need to be non-empty, I suspect it's because the union "collapses" in this case, and the space wouldn't be a part of union