Difference between revisions of "Exercises:Saul - Algebraic Topology - 1/Exercise 1.2/Lemmas"
From Maths
(Saving work) |
m (Alec moved page Exercises:Saul - Algebraic Topology - 1/Exercise 1.2/Exercises:Saul - Algebraic Topology - 1/Exercise 1.2/Lemmas to Exercises:Saul - Algebraic Topology - 1/Exercise 1.2/Lemmas without leaving a redirect: Opps) |
||
| (One intermediate revision by the same user not shown) | |||
| Line 4: | Line 4: | ||
====Lemmas==== | ====Lemmas==== | ||
</noinclude> | </noinclude> | ||
| + | {{Caveat|The following listed here are for reference to someone looking at the exercises only.}} They are done from memory and have no reference (at the time of writing) - use at your own peril. | ||
| + | =====Homeomorphisms and point removal===== | ||
| + | Suppose {{Top.|X|J}} and {{Top.|Y|K}} are [[topological spaces]] and {{M|f:X\rightarrow Y}} is a [[homeomorphism]] between them (so {{M|X\cong Y}}), then for any {{M|x\in X}} we have: | ||
| + | * {{M|f\vert_{X-\{x\} }:X-\{x\}\rightarrow Y-\{f(x)\} }} is a [[homeomorphism]]<ref group="Note">This is a slight abuse of notation for a restriction, for a restriction we would have {{M|f\vert_{X-\{x\} }:X-\{x\}\rightarrow Y}} - notice the codomain has changed to {{M|Y-\{f(x)\} }}</ref> we of course consider these spaces with the [[subspace topology]] | ||
=====Point removal===== | =====Point removal===== | ||
| − | Let {{Top.|X|J}} and {{Top.|Y|K}} be [[topological spaces]] and let {{M|f:X\rightarrow Y}} be any [[map]] (possibly [[continuous]]) between them. | + | Let {{Top.|X|J}} and {{Top.|Y|K}} be [[topological spaces]] and let {{M|f:X\rightarrow Y}} be any [[map]] (possibly [[continuous]]) between them. Then |
| − | + | * {{M|f:X\rightarrow Y}} being a [[homeomorphism]] ''[[implies]]'' {{M|\forall p\in X[\mathcal{O}(p)\eq\mathcal{O}(f(p))]}} - where {{M|\mathcal{O}(p)}} is the number of path-connected components of the space {{M|X-\{p\} }} | |
| + | ** We may write {{M|\mathcal{O}(p,X)}} to mean {{M|p}} removed from the space {{M|X}}, this makes things clearer when dealing with subsets. | ||
| + | Specifically, by [[contrapositive]], if {{M|\exists p\in X[\mathcal{O}(p)\neq\mathcal{O}(f(p))]}} then {{M|f:X\rightarrow Y}} is not a homeomorphism. | ||
| + | =====Two-step point removal===== | ||
| + | This is a corollary of the two claims above. Two-step point removal means that: | ||
| + | * Suppose {{M|X\cong_f Y}}, then {{M|X-\{p\}\cong_{f\vert_{X-\{x\} } }Y-\{f(p)\} }} (by "''homeomorphisms and point removal''") | ||
| + | * Suppose {{M|X-\{p\}\cong_{f\vert_{X-\{x\} } }Y-\{f(p)\} }} are indeed homeomorphic, then {{M|\forall q\in X[\mathcal{O}(q,X-\{p\})\eq\mathcal{O}(f\vert_{X-\{x\} }(q);Y-\{f(p)\})]}} | ||
| + | We conclude: | ||
| + | * {{M|X\cong_f Y}} [[implies]] {{M|\forall q\in X[\mathcal{O}(q,X-\{p\})\eq\mathcal{O}(f\vert_{X-\{x\} }(q);Y-\{f(p)\})]}} | ||
| + | Which will be helpful for reaching contradictions. | ||
<noinclude> | <noinclude> | ||
==Notes== | ==Notes== | ||
Latest revision as of 18:46, 17 January 2017
Contents
Exercises
Exercise 1.2
Lemmas
Caveat:The following listed here are for reference to someone looking at the exercises only. They are done from memory and have no reference (at the time of writing) - use at your own peril.
Homeomorphisms and point removal
Suppose [ilmath](X,\mathcal{ J })[/ilmath] and [ilmath](Y,\mathcal{ K })[/ilmath] are topological spaces and [ilmath]f:X\rightarrow Y[/ilmath] is a homeomorphism between them (so [ilmath]X\cong Y[/ilmath]), then for any [ilmath]x\in X[/ilmath] we have:
- [ilmath]f\vert_{X-\{x\} }:X-\{x\}\rightarrow Y-\{f(x)\} [/ilmath] is a homeomorphism[Note 1] we of course consider these spaces with the subspace topology
Point removal
Let [ilmath](X,\mathcal{ J })[/ilmath] and [ilmath](Y,\mathcal{ K })[/ilmath] be topological spaces and let [ilmath]f:X\rightarrow Y[/ilmath] be any map (possibly continuous) between them. Then
- [ilmath]f:X\rightarrow Y[/ilmath] being a homeomorphism implies [ilmath]\forall p\in X[\mathcal{O}(p)\eq\mathcal{O}(f(p))][/ilmath] - where [ilmath]\mathcal{O}(p)[/ilmath] is the number of path-connected components of the space [ilmath]X-\{p\} [/ilmath]
- We may write [ilmath]\mathcal{O}(p,X)[/ilmath] to mean [ilmath]p[/ilmath] removed from the space [ilmath]X[/ilmath], this makes things clearer when dealing with subsets.
Specifically, by contrapositive, if [ilmath]\exists p\in X[\mathcal{O}(p)\neq\mathcal{O}(f(p))][/ilmath] then [ilmath]f:X\rightarrow Y[/ilmath] is not a homeomorphism.
Two-step point removal
This is a corollary of the two claims above. Two-step point removal means that:
- Suppose [ilmath]X\cong_f Y[/ilmath], then [ilmath]X-\{p\}\cong_{f\vert_{X-\{x\} } }Y-\{f(p)\} [/ilmath] (by "homeomorphisms and point removal")
- Suppose [ilmath]X-\{p\}\cong_{f\vert_{X-\{x\} } }Y-\{f(p)\} [/ilmath] are indeed homeomorphic, then [ilmath]\forall q\in X[\mathcal{O}(q,X-\{p\})\eq\mathcal{O}(f\vert_{X-\{x\} }(q);Y-\{f(p)\})][/ilmath]
We conclude:
- [ilmath]X\cong_f Y[/ilmath] implies [ilmath]\forall q\in X[\mathcal{O}(q,X-\{p\})\eq\mathcal{O}(f\vert_{X-\{x\} }(q);Y-\{f(p)\})][/ilmath]
Which will be helpful for reaching contradictions.
Notes
- ↑ This is a slight abuse of notation for a restriction, for a restriction we would have [ilmath]f\vert_{X-\{x\} }:X-\{x\}\rightarrow Y[/ilmath] - notice the codomain has changed to [ilmath]Y-\{f(x)\} [/ilmath]
References
