Difference between revisions of "Unique lifting property"
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** In words: that for all points in {{M|S}} there is a neighbourhood to that point contained entirely in {{M|S}} | ** In words: that for all points in {{M|S}} there is a neighbourhood to that point contained entirely in {{M|S}} | ||
Thus {{M|S}} is [[open set|open]], as required. | Thus {{M|S}} is [[open set|open]], as required. | ||
| + | ===Proof that {{M|T}} is an [[open set]]=== | ||
| + | {{Extra Maths}} | ||
| + | * Let {{M|t\in T}} be given - {{Caution|The result requires that {{M|T}} is empty, but we can still do this}}<ref group="Note">{{XXX|Investigate}}</ref> | ||
| + | ** Define {{M|r_1:\eq g(t)}}, note {{M|r_1\in E}} | ||
| + | *** Define {{M|r_2:\eq h(t)}}, note {{M|r_2\in E}} | ||
| + | **** Define {{M|z:\eq p(r_1):\eq p(g(t))\eq f(t)}} by {{M|g}} being a lifting, note {{M|z\in X}} | ||
| + | ***** Note that {{M|z:\eq p(g(t))\eq f(t)\eq p(h(t))\eq: p(r_2)}} by {{M|h}} being a lifting, so we have | ||
| + | ****** {{M|p(r_1)\eq p(r_2)}} | ||
| + | ***** As {{M|p}} is a {{link|covering map|topology}} we see that: | ||
| + | ****** {{M|\exists U\in\mathcal{J}[z\in U\wedge U\text{ is evenly covered by }p]}} | ||
| + | ******* By evenly covered: {{M|\exists(V_\alpha)_{\alpha\in I}\subseteq\mathcal{H}\big[\bigudot_{\alpha\in I}V_\alpha\eq p^{-1}(U)\wedge \big(\forall\beta\in I[V_\beta\cong_{p\vert_{V_\beta}^\text{Im} } U]\big)\big]}} (note that {{M|\bigudot}} emphasises the [[union]] is of [[pairwise disjoint]] sets) | ||
| + | ***** Define {{M|(V_\alpha)_{\alpha\in I} \subseteq H}} to be the arbitrary family of sets [[opens set|open]] in {{M|E}} which are the {{link|sheets of the covering|topology}} of {{M|U}}. | ||
| + | ****** Then: | ||
| + | ******# {{M|\exists\beta\in I[r_1\in V_\beta]}} | ||
| + | ******# {{M|\exists\gamma\in I[r_2\in V_\gamma]}} | ||
| + | ****** '''Lemma: ''' {{M|\beta\neq\gamma}} | ||
| + | ******* Suppose that {{M|\beta\eq\gamma}} (and thus use {{M|V_\beta}} as the notation and discard {{M|\gamma}}) | ||
| + | ******** Then {{M|r_1\in V_\beta}} and {{M|r_2\in V_\beta}} | ||
| + | ******** Define {{M|q:\eq p\vert^\text{Im}_{V_\beta}:V_\beta\rightarrow U}}, so {{M|V_\beta}}[[homeomorphic|{{M|\cong}}]]{{M|{}_q U}}<ref group="Note">This is what {{link|evenly covered|topology}} means, each sheet of the covering is [[homeomorphic]] to {{M|U}} via the [[restriction onto its image]] of the {{link|covering map|topology}}</ref> | ||
| + | ********* Note that as it is a [[homeomorphism]] it is a [[bijection]] and as it is a bijection it is an [[injection]], so: | ||
| + | ********** {{M|\forall a,b\in V_\beta[a\neq b\implies q(a)\neq q(b)]}}<ref group="Note">As the [[injection]] page describes, this is equivalent to and is sometimes given as the following form: | ||
| + | * {{M|\forall a,b\in V_\beta[q(a)\eq q(b)\implies a\eq b]}} | ||
| + | This is just the contrapositive of what we have</ref> | ||
| + | ********* So {{M|r_1\neq r_2\implies q(r_1)\neq q(r_2)}} | ||
| + | ********** But {{M|q(r_1)\eq p(r_1)\eq p(r_2)\eq q(r_2)}} (from above) - so {{M|q(r_1)\eq q(r_2)}} | ||
| + | *********** This is a contradiction, so we cannot have {{M|\beta\eq\gamma}} | ||
| + | ******* Thus {{M|\beta\neq\gamma}} | ||
| + | ****** By the lemma and the property of the {{link|covering sheets|topology}} (that they're [[pairwise disjoint]]) we see: | ||
| + | ******* {{M|V_\beta\cap V_\gamma\eq\emptyset}} (they're disjoint) | ||
| + | ****** Define {{M|q_\beta:\eq p\vert^\text{Im}_{V_\beta}:V_\beta\rightarrow U}} so that {{M|V_\beta\cong_{q_\beta} U}} | ||
| + | ******* Define {{M|q_\gamma:\eq p\vert^\text{Im}_{V_\gamma}:V_\gamma\rightarrow U}} so that {{M|V_\gamma\cong_{q_\gamma} U}} | ||
| + | ******** Now {{M|V_\beta\cong_{q_\beta} U {}_{q_\gamma}\cong V_\gamma}} | ||
| + | ******** Define {{M|W_\beta:\eq g^{-1}(V_\beta)}} - by [[continuity]] of {{M|g}} and the fact that {{M|V_\beta}} is open we see {{M|W_\beta\in\mathcal{K} }}, {{ie}} that {{M|W_\beta}} is [[open set|open]] in {{M|Y}} | ||
| + | ********* Define {{M|W_\gamma:\eq h^{-1}(V_\gamma)}} - by [[continuity]] of {{M|h}} and the fact that {{M|V_\gamma}} is open we see {{M|W_\gamma\in\mathcal{K} }}, {{ie}} that {{M|W_\gamma}} is [[open set|open]] in {{M|Y}} | ||
| + | ********** Note that {{M|t\in g^{-1}(r_1)}} and {{M|r_1\in V_\beta}}, so {{M|t\in W_\beta}} | ||
| + | ********** Note also that {{M|t\in h^{-1}(r_2)}} and {{M|r_2\in V_\gamma}}, so {{M|t\in W_\gamma}} | ||
| + | ********** Define {{M|W:\eq W_\beta\cap W_\alpha}}, as we're in a [[topology]] the [[intersection]] of (finitely) many open sets is again open, so we see {{M|W\in\mathcal{K} }} - {{ie}} {{M|W}} is open in {{M|Y}}. | ||
| + | *********** Notice that {{M|t\in W}} as for an [[intersection]] we see {{M|t\in W_\beta\cap W_\gamma\iff(t\in W_\beta\wedge t\in W_\gamma)}} | ||
| + | *********** Notice also that {{M|g(W)\subseteq V_\beta}} and {{M|h(W)\subseteq V_\gamma}} | ||
| + | *********** As {{M|V_\beta\cap V_\gamma\eq\emptyset}} we see {{M|h(W)\cap g(W)\eq\emptyset}} - {{XXX|Make a statement page with this, I've used it before!}} | ||
| + | ************ Because {{M|h(W)\cap g(W)\eq\emptyset}} they ''cannot'' agree anywhere, thus {{M|W\subseteq T}} | ||
| + | ************* Since {{M|W}} is [[open set|open]] and contains {{M|t}} we have found an [[open neighbourhood]] contained within {{M|T}} for each point in {{M|T}} | ||
| − | {{Requires proof|We still need to show {{M|T}} is open|grade=A}} | + | {{Requires proof|We still need to show {{M|T}} is open |
| + | * Update: almost there! [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 15:45, 2 March 2017 (UTC) | ||
| + | * Tidy up and we're done [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 16:23, 2 March 2017 (UTC)|grade=A}} | ||
==Notes== | ==Notes== | ||
<references group="Note"/> | <references group="Note"/> | ||
Latest revision as of 16:23, 2 March 2017
Contents
Statement
Let [ilmath](X,\mathcal{ J })[/ilmath] be a topological space, suppose that [ilmath](E,\mathcal{ H })[/ilmath] is a covering space (with covering map [ilmath]p:E\rightarrow X[/ilmath]). Suppose [ilmath](Y,\mathcal{ K })[/ilmath] is a connected topological space and [ilmath]f:Y\rightarrow X[/ilmath] is a continuous map, then[1]Partial:[2] - [Note 1]
- given two lifts of [ilmath]f[/ilmath] through [ilmath]p[/ilmath], say [ilmath]g,h:Y\rightarrow E[/ilmath] we have:
- [ilmath]\big(\exists y\in Y[g(y)\eq h(y)\big)\implies \big(\ \underbrace{\forall y\in Y[g(y)\eq h(y)]}_{\text{i.e. that }g\eq h}\ \big)[/ilmath]
- In words: if there exists a point on which [ilmath]g[/ilmath] and [ilmath]h[/ilmath] agree then [ilmath]g[/ilmath] and [ilmath]h[/ilmath] are equal as functions
- [ilmath]\big(\exists y\in Y[g(y)\eq h(y)\big)\implies \big(\ \underbrace{\forall y\in Y[g(y)\eq h(y)]}_{\text{i.e. that }g\eq h}\ \big)[/ilmath]
Bonus corollary
Recall that a logical implication is logically equivalent to the contrapositive, that is
- [ilmath](A\implies B)\iff(\neg B\implies\neg A)[/ilmath]
So, should the above claim be true, we also get:
- [ilmath]\big(\exists y\in Y[g(y)\neq h(y)]\big)\implies\big(\forall y\in Y[g(y)\neq h(y)]\big)[/ilmath]
- In words: if there exists a [ilmath]y_0\in Y[/ilmath] such that [ilmath]g[/ilmath] and [ilmath]h[/ilmath] disagree at [ilmath]y_0[/ilmath] then they disagree everywhere.
- Caveat:This does not mean [ilmath]g(Y)\cap h(Y)\eq\emptyset[/ilmath] necessarily!
- In words: if there exists a [ilmath]y_0\in Y[/ilmath] such that [ilmath]g[/ilmath] and [ilmath]h[/ilmath] disagree at [ilmath]y_0[/ilmath] then they disagree everywhere.
Proof
Grade: C
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Warning:What follows is VERY messy. I was distracted when writing it and kept switching between it and something else. However the written proof was BARELY legible so this is better than nothing. Hopefully
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TODO: It'd be good to put some reasoning here. Like given a [ilmath]z\in Y[/ilmath] to show [ilmath]g[/ilmath] and [ilmath]h[/ilmath] agree there we really need to work in terms of open neighbourhoods to [ilmath]z[/ilmath], which we can get from somewhere they're known to agree on. However this only works if it's connected, as if [ilmath]z[/ilmath] is in another connected components "stuff just doesn't work" <--make this more formal
Let us make the following definitions:
- [ilmath]S:\eq\{y\in Y\ \vert\ h(y)\eq g(y)\} [/ilmath] (when we introduce the hypothesis, by that hypothesis this will be non-empty)
- [ilmath]T:\eq\{y\in Y\ \vert\ h(y)\neq g(y)\} [/ilmath]
Lemma 1: [ilmath]S[/ilmath] and [ilmath]T[/ilmath] are disjoint.
- Suppose [ilmath]S\cap T\neq\emptyset[/ilmath], clearly this [ilmath]iff\exists z\in S\cap T[/ilmath]
- Suppose there is such a [ilmath]z\in S\cap T[/ilmath], by definition of intersection that means [ilmath]\exists z\in Y[z\in S\wedge z\in T][/ilmath]
- [ilmath]z\in S\iff g(z)\eq h(z)[/ilmath] and [ilmath]z\in T\iff g(z)\neq h(z)[/ilmath]
- We see [ilmath]g(z)\eq h(z)\neq g(z)[/ilmath] so [ilmath]g(z)\neq g(z)[/ilmath] - a contradiction!
- [ilmath]z\in S\iff g(z)\eq h(z)[/ilmath] and [ilmath]z\in T\iff g(z)\neq h(z)[/ilmath]
- Suppose there is such a [ilmath]z\in S\cap T[/ilmath], by definition of intersection that means [ilmath]\exists z\in Y[z\in S\wedge z\in T][/ilmath]
- We see there cannot be any elements in [ilmath]S\cap T[/ilmath] for if there was we have a contradiction. As shown
- So [ilmath]S\cap T\eq\emptyset[/ilmath] - i.e. [ilmath]S[/ilmath] and [ilmath]T[/ilmath] are disjoint
Lemma 2: [ilmath]S\cup T\eq Y[/ilmath] AND [ilmath]S\eq Y-T[/ilmath] (and [ilmath]T\eq Y-S[/ilmath]) in some form. We need [ilmath]S\eq Y-T[/ilmath] and [ilmath]T\eq Y-S[/ilmath] at least!
TODO: Todo
As [ilmath]Y[/ilmath] is a connected topological space we see that the only sets that are both open and closed are [ilmath]Y[/ilmath] itself and [ilmath]\emptyset[/ilmath], if the result holds (which we very much hope it does) then [ilmath]S\eq Y[/ilmath] and [ilmath]T\eq\emptyset[/ilmath], so if we show [ilmath]S[/ilmath] is open. As [ilmath]T\eq Y-S[/ilmath] we would see [ilmath]T[/ilmath] is closed as a result.
If we show [ilmath]T[/ilmath] is open too, then [ilmath]S[/ilmath] would be closed, thus showing they're both open and closed!
So we must have [ilmath]S\eq Y[/ilmath] (as [ilmath]S[/ilmath] is non-empty by hypothesis) and [ilmath]T\eq\emptyset[/ilmath], but if [ilmath]S\eq Y[/ilmath] then they agree everywhere!
- TODO: Notes:Covering spaces shows it better
Proof that [ilmath]S[/ilmath] is open
- Let [ilmath]s\in S[/ilmath] be given. There is at least one to give by hypothesis. We must find a neighbourhood of [ilmath]s[/ilmath] that is fully contained in [ilmath]S[/ilmath], recall that a set is open if and only if it is a neighbourhood to all of its points, this is what we hope to show.
- Define [ilmath]r:\eq h(s)\eq g(s)[/ilmath] and notice [ilmath]r\in E[/ilmath]
- Define [ilmath]z:\eq p(r)[/ilmath] and notice [ilmath]p(r)\in X[/ilmath]
- As [ilmath]z\in X[/ilmath] and [ilmath]p[/ilmath] is a covering map there exists an open neighbourhood, [ilmath]U\in\mathcal{J} [/ilmath] of [ilmath]z[/ilmath] such that [ilmath]U[/ilmath] is evenly covered by [ilmath]p[/ilmath]
- Choose [ilmath]U[/ilmath] to be such an open neighbourhood
- By choice of [ilmath]U[/ilmath] we see [ilmath]\exists(V_\alpha)_{\alpha\in I}\subseteq\mathcal{H} [/ilmath]
- such that:
- [ilmath]p^{-1}(U)\eq\bigcup_{\alpha\in I}V_\alpha[/ilmath]
- The [ilmath](V_\alpha)_\alpha[/ilmath] are open (given by being in [ilmath]\mathcal{H} [/ilmath])
- The [ilmath](V_\alpha)_\alpha[/ilmath] are pairwise disjoint
- For each [ilmath]V_\beta\in(V_\alpha)_\alpha[/ilmath] we have [ilmath]V_\beta\cong_{p\vert^\text{Im}_{V_\beta} } U[/ilmath], that is the covering map restricted to its image on [ilmath]V_\beta[/ilmath] is a homeomorphism onto [ilmath]U[/ilmath]
- such that:
- Choose [ilmath](V_\alpha)_{\alpha\in I}\subseteq\mathcal{H} [/ilmath] to be this family of sheets of the covering
- As [ilmath]z\in U[/ilmath] we see [ilmath]p^{-1}(z)\subseteq p^{-1}(U)[/ilmath]
- As [ilmath]z:\eq p(r)[/ilmath] we see that [ilmath]r\in p^{-1}(z) [/ilmath] so [ilmath]r\in p^{-1}(U)[/ilmath]
- As [ilmath]p^{-1}(U)\eq\bigcup_{\alpha\in I}V_\alpha[/ilmath] we see [ilmath]r\in \bigcup_{\alpha\in I}V_\alpha[/ilmath], by definition of union:
- [ilmath]\big(r\in \bigcup_{\alpha\in I}V_\alpha\big)\iff\big(\exists\beta\in I[r\in V_\beta\big)[/ilmath]
- So [ilmath]\exists\beta\in I[r\in V_\beta][/ilmath]
- Define [ilmath]V:\eq V_\beta[/ilmath] where [ilmath]V_\beta[/ilmath] is the element of [ilmath](V_\alpha)_{\alpha\in I} [/ilmath] with [ilmath]r\in V_\beta[/ilmath] as discussed above
- As [ilmath]r:\eq h(s)[/ilmath] and [ilmath]h(s)\eq g(s)[/ilmath] we see that:
- [ilmath]r\in V\iff[h(s)\in V\wedge g(s)\in V][/ilmath]
- So [ilmath]s\in h^{-1}(V)[/ilmath] and [ilmath]s\in g^{-1}(V)[/ilmath]
- Notice, by continuity and [ilmath]V[/ilmath] being open in [ilmath](E,\mathcal{ H })[/ilmath] that [ilmath]h^{-1}(V)[/ilmath] and [ilmath]g^{-1}(V)[/ilmath] are both open in [ilmath](Y,\mathcal{ K })[/ilmath].
- Thus [ilmath]s\in h^{-1}(V)\cap g^{-1}(V)[/ilmath] (as by definition of intersection [ilmath](s\in A\cap B)\iff(s\in A\wedge s\in B)[/ilmath] - we have the RHS, so we have the left.
- Define [ilmath]W:\eq h^{-1}(V)\cap g^{-1}(V)[/ilmath], so [ilmath]W\subseteq Y[/ilmath]
- notice:
- Define [ilmath]q:\eq p\vert_V^\text{Im}:V\rightarrow U[/ilmath] be the homeomorphism of the restriction of [ilmath]p[/ilmath] to [ilmath]V[/ilmath] which is onto [ilmath]U[/ilmath].
- This means that [ilmath]q[/ilmath] is injective, i.e.:
- [ilmath]\forall v,w\in V[q(v)\eq q(w)\implies v\eq w][/ilmath]
- Notice also that [ilmath]h(W)\subseteq V[/ilmath] and [ilmath]g(W)\subseteq V[/ilmath]
- Let [ilmath]w\in W[/ilmath] be given
- By definition of being lifts: [ilmath]f(w)\eq p(h(w))\eq p(g(w))[/ilmath]
- As [ilmath]h(w)\in V[/ilmath] and [ilmath]g(w)\in V[/ilmath] we see that [ilmath]p(h(w))\eq q(w)[/ilmath] and [ilmath]p(g(w))\eq q(g(w))[/ilmath]
- So [ilmath]f(w)\eq q(h(w))\eq q(g(w))[/ilmath]
- But [ilmath]q[/ilmath] is injective, so [ilmath]q(h(w))\eq q(g(w))\implies g(w)\eq h(w)[/ilmath]
- So [ilmath]f(w)\eq q(h(w))\eq q(g(w))[/ilmath]
- so we have [ilmath]g(w)\eq h(w)[/ilmath]
- so [ilmath]w\in S[/ilmath] (by definition of [ilmath]S[/ilmath])
- Since [ilmath]w\in W[/ilmath] was arbitrary we have shown [ilmath]\forall w\in W[w\in S][/ilmath]
- This means that [ilmath]q[/ilmath] is injective, i.e.:
- Thus [ilmath]W\subseteq Y[/ilmath], that is [ilmath]Y[/ilmath] contains a neighbourhood of [ilmath]s[/ilmath]
- Define [ilmath]W:\eq h^{-1}(V)\cap g^{-1}(V)[/ilmath], so [ilmath]W\subseteq Y[/ilmath]
- As [ilmath]r:\eq h(s)[/ilmath] and [ilmath]h(s)\eq g(s)[/ilmath] we see that:
- Define [ilmath]V:\eq V_\beta[/ilmath] where [ilmath]V_\beta[/ilmath] is the element of [ilmath](V_\alpha)_{\alpha\in I} [/ilmath] with [ilmath]r\in V_\beta[/ilmath] as discussed above
- By choice of [ilmath]U[/ilmath] we see [ilmath]\exists(V_\alpha)_{\alpha\in I}\subseteq\mathcal{H} [/ilmath]
- Define [ilmath]z:\eq p(r)[/ilmath] and notice [ilmath]p(r)\in X[/ilmath]
- Define [ilmath]r:\eq h(s)\eq g(s)[/ilmath] and notice [ilmath]r\in E[/ilmath]
- Since [ilmath]s\in S[/ilmath] was arbitrary, we have shown [ilmath]\forall s\in S\exists W\in\mathcal{K}[x\in W\wedge W\subseteq Y][/ilmath]
- In words: that for all points in [ilmath]S[/ilmath] there is a neighbourhood to that point contained entirely in [ilmath]S[/ilmath]
Thus [ilmath]S[/ilmath] is open, as required.
Proof that [ilmath]T[/ilmath] is an open set
[math]\newcommand{\bigudot}{ \mathchoice{\mathop{\bigcup\mkern-15mu\cdot\mkern8mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}} }[/math][math]\newcommand{\udot}{\cup\mkern-12.5mu\cdot\mkern6.25mu\!}[/math][math]\require{AMScd}\newcommand{\d}[1][]{\mathrm{d}^{#1} }[/math]
- Let [ilmath]t\in T[/ilmath] be given - Caution:The result requires that [ilmath]T[/ilmath] is empty, but we can still do this[Note 2]
- Define [ilmath]r_1:\eq g(t)[/ilmath], note [ilmath]r_1\in E[/ilmath]
- Define [ilmath]r_2:\eq h(t)[/ilmath], note [ilmath]r_2\in E[/ilmath]
- Define [ilmath]z:\eq p(r_1):\eq p(g(t))\eq f(t)[/ilmath] by [ilmath]g[/ilmath] being a lifting, note [ilmath]z\in X[/ilmath]
- Note that [ilmath]z:\eq p(g(t))\eq f(t)\eq p(h(t))\eq: p(r_2)[/ilmath] by [ilmath]h[/ilmath] being a lifting, so we have
- [ilmath]p(r_1)\eq p(r_2)[/ilmath]
- As [ilmath]p[/ilmath] is a covering map we see that:
- [ilmath]\exists U\in\mathcal{J}[z\in U\wedge U\text{ is evenly covered by }p][/ilmath]
- By evenly covered: [ilmath]\exists(V_\alpha)_{\alpha\in I}\subseteq\mathcal{H}\big[\bigudot_{\alpha\in I}V_\alpha\eq p^{-1}(U)\wedge \big(\forall\beta\in I[V_\beta\cong_{p\vert_{V_\beta}^\text{Im} } U]\big)\big][/ilmath] (note that [ilmath]\bigudot[/ilmath] emphasises the union is of pairwise disjoint sets)
- [ilmath]\exists U\in\mathcal{J}[z\in U\wedge U\text{ is evenly covered by }p][/ilmath]
- Define [ilmath](V_\alpha)_{\alpha\in I} \subseteq H[/ilmath] to be the arbitrary family of sets open in [ilmath]E[/ilmath] which are the sheets of the covering of [ilmath]U[/ilmath].
- Then:
- [ilmath]\exists\beta\in I[r_1\in V_\beta][/ilmath]
- [ilmath]\exists\gamma\in I[r_2\in V_\gamma][/ilmath]
- Lemma: [ilmath]\beta\neq\gamma[/ilmath]
- Suppose that [ilmath]\beta\eq\gamma[/ilmath] (and thus use [ilmath]V_\beta[/ilmath] as the notation and discard [ilmath]\gamma[/ilmath])
- Then [ilmath]r_1\in V_\beta[/ilmath] and [ilmath]r_2\in V_\beta[/ilmath]
- Define [ilmath]q:\eq p\vert^\text{Im}_{V_\beta}:V_\beta\rightarrow U[/ilmath], so [ilmath]V_\beta[/ilmath][ilmath]\cong[/ilmath][ilmath]{}_q U[/ilmath][Note 3]
- Note that as it is a homeomorphism it is a bijection and as it is a bijection it is an injection, so:
- [ilmath]\forall a,b\in V_\beta[a\neq b\implies q(a)\neq q(b)][/ilmath][Note 4]
- So [ilmath]r_1\neq r_2\implies q(r_1)\neq q(r_2)[/ilmath]
- But [ilmath]q(r_1)\eq p(r_1)\eq p(r_2)\eq q(r_2)[/ilmath] (from above) - so [ilmath]q(r_1)\eq q(r_2)[/ilmath]
- This is a contradiction, so we cannot have [ilmath]\beta\eq\gamma[/ilmath]
- But [ilmath]q(r_1)\eq p(r_1)\eq p(r_2)\eq q(r_2)[/ilmath] (from above) - so [ilmath]q(r_1)\eq q(r_2)[/ilmath]
- Note that as it is a homeomorphism it is a bijection and as it is a bijection it is an injection, so:
- Thus [ilmath]\beta\neq\gamma[/ilmath]
- Suppose that [ilmath]\beta\eq\gamma[/ilmath] (and thus use [ilmath]V_\beta[/ilmath] as the notation and discard [ilmath]\gamma[/ilmath])
- By the lemma and the property of the covering sheets (that they're pairwise disjoint) we see:
- [ilmath]V_\beta\cap V_\gamma\eq\emptyset[/ilmath] (they're disjoint)
- Define [ilmath]q_\beta:\eq p\vert^\text{Im}_{V_\beta}:V_\beta\rightarrow U[/ilmath] so that [ilmath]V_\beta\cong_{q_\beta} U[/ilmath]
- Define [ilmath]q_\gamma:\eq p\vert^\text{Im}_{V_\gamma}:V_\gamma\rightarrow U[/ilmath] so that [ilmath]V_\gamma\cong_{q_\gamma} U[/ilmath]
- Now [ilmath]V_\beta\cong_{q_\beta} U {}_{q_\gamma}\cong V_\gamma[/ilmath]
- Define [ilmath]W_\beta:\eq g^{-1}(V_\beta)[/ilmath] - by continuity of [ilmath]g[/ilmath] and the fact that [ilmath]V_\beta[/ilmath] is open we see [ilmath]W_\beta\in\mathcal{K} [/ilmath], i.e. that [ilmath]W_\beta[/ilmath] is open in [ilmath]Y[/ilmath]
- Define [ilmath]W_\gamma:\eq h^{-1}(V_\gamma)[/ilmath] - by continuity of [ilmath]h[/ilmath] and the fact that [ilmath]V_\gamma[/ilmath] is open we see [ilmath]W_\gamma\in\mathcal{K} [/ilmath], i.e. that [ilmath]W_\gamma[/ilmath] is open in [ilmath]Y[/ilmath]
- Note that [ilmath]t\in g^{-1}(r_1)[/ilmath] and [ilmath]r_1\in V_\beta[/ilmath], so [ilmath]t\in W_\beta[/ilmath]
- Note also that [ilmath]t\in h^{-1}(r_2)[/ilmath] and [ilmath]r_2\in V_\gamma[/ilmath], so [ilmath]t\in W_\gamma[/ilmath]
- Define [ilmath]W:\eq W_\beta\cap W_\alpha[/ilmath], as we're in a topology the intersection of (finitely) many open sets is again open, so we see [ilmath]W\in\mathcal{K} [/ilmath] - i.e. [ilmath]W[/ilmath] is open in [ilmath]Y[/ilmath].
- Notice that [ilmath]t\in W[/ilmath] as for an intersection we see [ilmath]t\in W_\beta\cap W_\gamma\iff(t\in W_\beta\wedge t\in W_\gamma)[/ilmath]
- Notice also that [ilmath]g(W)\subseteq V_\beta[/ilmath] and [ilmath]h(W)\subseteq V_\gamma[/ilmath]
- As [ilmath]V_\beta\cap V_\gamma\eq\emptyset[/ilmath] we see [ilmath]h(W)\cap g(W)\eq\emptyset[/ilmath] - TODO: Make a statement page with this, I've used it before!
- Because [ilmath]h(W)\cap g(W)\eq\emptyset[/ilmath] they cannot agree anywhere, thus [ilmath]W\subseteq T[/ilmath]
- Since [ilmath]W[/ilmath] is open and contains [ilmath]t[/ilmath] we have found an open neighbourhood contained within [ilmath]T[/ilmath] for each point in [ilmath]T[/ilmath]
- Because [ilmath]h(W)\cap g(W)\eq\emptyset[/ilmath] they cannot agree anywhere, thus [ilmath]W\subseteq T[/ilmath]
- Define [ilmath]W_\gamma:\eq h^{-1}(V_\gamma)[/ilmath] - by continuity of [ilmath]h[/ilmath] and the fact that [ilmath]V_\gamma[/ilmath] is open we see [ilmath]W_\gamma\in\mathcal{K} [/ilmath], i.e. that [ilmath]W_\gamma[/ilmath] is open in [ilmath]Y[/ilmath]
- Define [ilmath]q_\gamma:\eq p\vert^\text{Im}_{V_\gamma}:V_\gamma\rightarrow U[/ilmath] so that [ilmath]V_\gamma\cong_{q_\gamma} U[/ilmath]
- Then:
- Note that [ilmath]z:\eq p(g(t))\eq f(t)\eq p(h(t))\eq: p(r_2)[/ilmath] by [ilmath]h[/ilmath] being a lifting, so we have
- Define [ilmath]z:\eq p(r_1):\eq p(g(t))\eq f(t)[/ilmath] by [ilmath]g[/ilmath] being a lifting, note [ilmath]z\in X[/ilmath]
- Define [ilmath]r_2:\eq h(t)[/ilmath], note [ilmath]r_2\in E[/ilmath]
- Define [ilmath]r_1:\eq g(t)[/ilmath], note [ilmath]r_1\in E[/ilmath]
Grade: A
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Notes
- ↑ Lee defines covering maps and spaces a little differently. He requires that for evenly covered that [ilmath]U[/ilmath] be homeomorphic to each sheet, and each sheet is connected and disjoint from the others. Thus [ilmath]U[/ilmath] is connected. It may not matter
- TODO: Does it?
-
- ↑ TODO: Investigate
- ↑ This is what evenly covered means, each sheet of the covering is homeomorphic to [ilmath]U[/ilmath] via the restriction onto its image of the covering map
- ↑ As the injection page describes, this is equivalent to and is sometimes given as the following form:
- [ilmath]\forall a,b\in V_\beta[q(a)\eq q(b)\implies a\eq b][/ilmath]
References
- ↑ Introduction to Topology - Theodore W. Gamelin & Robert Everist Greene
- ↑ Introduction to Topological Manifolds - John M. Lee
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