Difference between revisions of "The canonical injections of the disjoint union topology are topological embeddings"
(Created page with "{{Stub page|grade=A|msg=Important for progress!}} __TOC__ ==Statement== Let {{M|\big((X_\alpha,\mathcal{J}_\alpha)\big)_{\alpha\in I} }} be a collection of topological space...") |
(Added proof notes, there are still some ambiguities involved) |
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*# {{M|i_\beta}} being [[injective]] and | *# {{M|i_\beta}} being [[injective]] and | ||
*# {{M|i_\beta}} being a [[homeomorphism]] between {{M|X_\beta}} and {{M|i_\beta(X_\beta)}} | *# {{M|i_\beta}} being a [[homeomorphism]] between {{M|X_\beta}} and {{M|i_\beta(X_\beta)}} | ||
− | {{Requires proof|grade=A|msg=Done on paper, it isn't hard. I want to save my work now though}} | + | {{Requires proof|grade=A|msg=Done on paper, it isn't hard. I want to save my work now though '''NOTES ARE BELOW'''}} |
+ | I only cover part 3 here. | ||
+ | |||
+ | We have shown {{M|i_\beta:X_\beta\rightarrow\coprod_{\alpha\in I}X_\alpha}} is [[continuous]] and [[injective]]. It only remains to show that it is a [[homeomorphism]] onto (in the [[surjective]] sense of the word "onto") its {{link|image|map}}. | ||
+ | * First note that [[every injection yields a bijection onto its image]] | ||
+ | ** Thus we get a map {{M|\overline{i_\beta}:X_\beta\rightarrow i_\beta(\coprod_{\alpha\in I}X_\alpha)}} given by {{M|1=\overline{i_\beta}:x\mapsto i_\beta(x)}} (note that this means {{M|\overline{i_\beta}:x\mapsto(\beta,x)}}) which is a [[bijection]] | ||
+ | * Next note that [[every bijection yields an inverse function]], so now we have {{M|(\overline{i_\beta})^{-1}:i_\beta(\coprod_{\alpha\in I}X_\alpha)\rightarrow X_\beta}} | ||
+ | * We only really need to show that {{M|(\overline{i_\beta})^{-1}:i_\beta(\coprod_{\alpha\in I}X_\alpha)\rightarrow X_\beta}} is [[continuous]] | ||
+ | * Let {{M|U\in\mathcal{J}_\beta}} (so {{M|U}} is [[open set|open]] in {{M|X_\beta}}) be given | ||
+ | ** Then we must show that {{M|((\overline{i_\beta})^{-1})^{-1}(U)\in\mathcal{J} }} in order for {{M|(\overline{i_\beta})^{-1}:i_\beta(\coprod_{\alpha\in I}X_\alpha)\rightarrow X_\beta}} to be continuous | ||
+ | ** But! {{M|1=((\overline{i_\beta})^{-1})^{-1}(U)=\overline{i_\beta}(U)}} | ||
+ | ** Recall we defined a set to be open in {{M|\coprod_{\alpha\in I}X_\alpha}} if its intersection with (the image of) each {{M|X_\alpha}} is open in {{M|X_\alpha}} {{Caution|Terminology is a bit fuzzy here. I need to fix that}} | ||
+ | *** Let {{M|\gamma\in I}} be given | ||
+ | **** If {{M|\gamma\ne\beta}} then | ||
+ | ***** {{M|1=\overline{i_\beta}(U)\cap X_\gamma^*=\emptyset}} and by definition, {{M|\emptyset\in\mathcal{J}_\gamma}} | ||
+ | ****** {{Caution|This is where the notation gets weird. The image of the emptyset is the empty set, not sets of the form {{M|(\beta,x)}} ....}} | ||
+ | **** If {{M|1=\gamma=\beta}} then | ||
+ | ***** {{M|1=\overline{i_\beta}(U)\cap X_\beta^*=U}} {{Caution|or the image of {{M|U}} }} - which is open as {{M|U\in\mathcal{J}_\beta}}! | ||
==References== | ==References== | ||
<references/> | <references/> | ||
{{Theorem Of|Topology}} | {{Theorem Of|Topology}} |
Latest revision as of 12:34, 26 September 2016
Contents
Statement
Let [ilmath]\big((X_\alpha,\mathcal{J}_\alpha)\big)_{\alpha\in I} [/ilmath] be a collection of topological spaces and let [ilmath](\coprod_{\alpha\in I}X_\alpha,\mathcal{J})[/ilmath] be the disjoint union space of that family. With this construction we get some canonical injections:
- For each [ilmath]\beta\in I[/ilmath] we get a map (called a canonical injection) [ilmath]i_\beta:X_\beta\rightarrow\coprod_{\alpha\in I}X_\alpha[/ilmath] given by [ilmath]i_\beta:x\mapsto(\beta,x)[/ilmath]
We claim that each [ilmath]i_\beta[/ilmath] is a topological embedding[1] (that means [ilmath]i_\beta[/ilmath] is injective and continuous and a homeomorphism between [ilmath]X_\beta[/ilmath] and [ilmath]i_\beta(X_\beta)[/ilmath] (its image))
Proof
Let [ilmath]\beta\in I[/ilmath] be given.
- The proof that [ilmath]i_\beta:X_\beta\rightarrow\coprod_{\alpha\in I}X_\alpha[/ilmath] by [ilmath]i_\beta:x\mapsto(\beta,x)[/ilmath] consists of three parts:
- Continuity of [ilmath]i_\beta[/ilmath] - covered on the canonical injections of the disjoint union topology page so not shown on this pag
- [ilmath]i_\beta[/ilmath] being injective and
- [ilmath]i_\beta[/ilmath] being a homeomorphism between [ilmath]X_\beta[/ilmath] and [ilmath]i_\beta(X_\beta)[/ilmath]
The message provided is:
I only cover part 3 here.
We have shown [ilmath]i_\beta:X_\beta\rightarrow\coprod_{\alpha\in I}X_\alpha[/ilmath] is continuous and injective. It only remains to show that it is a homeomorphism onto (in the surjective sense of the word "onto") its image.
- First note that every injection yields a bijection onto its image
- Thus we get a map [ilmath]\overline{i_\beta}:X_\beta\rightarrow i_\beta(\coprod_{\alpha\in I}X_\alpha)[/ilmath] given by [ilmath]\overline{i_\beta}:x\mapsto i_\beta(x)[/ilmath] (note that this means [ilmath]\overline{i_\beta}:x\mapsto(\beta,x)[/ilmath]) which is a bijection
- Next note that every bijection yields an inverse function, so now we have [ilmath](\overline{i_\beta})^{-1}:i_\beta(\coprod_{\alpha\in I}X_\alpha)\rightarrow X_\beta[/ilmath]
- We only really need to show that [ilmath](\overline{i_\beta})^{-1}:i_\beta(\coprod_{\alpha\in I}X_\alpha)\rightarrow X_\beta[/ilmath] is continuous
- Let [ilmath]U\in\mathcal{J}_\beta[/ilmath] (so [ilmath]U[/ilmath] is open in [ilmath]X_\beta[/ilmath]) be given
- Then we must show that [ilmath]((\overline{i_\beta})^{-1})^{-1}(U)\in\mathcal{J} [/ilmath] in order for [ilmath](\overline{i_\beta})^{-1}:i_\beta(\coprod_{\alpha\in I}X_\alpha)\rightarrow X_\beta[/ilmath] to be continuous
- But! [ilmath]((\overline{i_\beta})^{-1})^{-1}(U)=\overline{i_\beta}(U)[/ilmath]
- Recall we defined a set to be open in [ilmath]\coprod_{\alpha\in I}X_\alpha[/ilmath] if its intersection with (the image of) each [ilmath]X_\alpha[/ilmath] is open in [ilmath]X_\alpha[/ilmath] Caution:Terminology is a bit fuzzy here. I need to fix that
- Let [ilmath]\gamma\in I[/ilmath] be given
- If [ilmath]\gamma\ne\beta[/ilmath] then
- [ilmath]\overline{i_\beta}(U)\cap X_\gamma^*=\emptyset[/ilmath] and by definition, [ilmath]\emptyset\in\mathcal{J}_\gamma[/ilmath]
- Caution:This is where the notation gets weird. The image of the emptyset is the empty set, not sets of the form [ilmath](\beta,x)[/ilmath] ....
- [ilmath]\overline{i_\beta}(U)\cap X_\gamma^*=\emptyset[/ilmath] and by definition, [ilmath]\emptyset\in\mathcal{J}_\gamma[/ilmath]
- If [ilmath]\gamma=\beta[/ilmath] then
- [ilmath]\overline{i_\beta}(U)\cap X_\beta^*=U[/ilmath] Caution:or the image of [ilmath]U[/ilmath] - which is open as [ilmath]U\in\mathcal{J}_\beta[/ilmath]!
- If [ilmath]\gamma\ne\beta[/ilmath] then
- Let [ilmath]\gamma\in I[/ilmath] be given
References