Difference between revisions of "Notes:Covering spaces"

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(Saving work, easy stuff, worth doing proof)
 
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#* Let {{M|p:E\rightarrow X}} be a covering map and let {{M|F:[0,1]\times[0,1]\rightarrow X}} be a [[homotopy]] (a continuous map in this case), then by the covering map: {{M|\exists e_0\in E[p(e_0)\eq F(0,0)]}}
 
#* Let {{M|p:E\rightarrow X}} be a covering map and let {{M|F:[0,1]\times[0,1]\rightarrow X}} be a [[homotopy]] (a continuous map in this case), then by the covering map: {{M|\exists e_0\in E[p(e_0)\eq F(0,0)]}}
 
#** Then we claim there exists a unique lift of {{M|F}}, say {{M|G:[0,1]\times[0,1]\rightarrow E}} such that {{M|G(0,0)\eq e_0}}
 
#** Then we claim there exists a unique lift of {{M|F}}, say {{M|G:[0,1]\times[0,1]\rightarrow E}} such that {{M|G(0,0)\eq e_0}}
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==Proof==
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===Uniqueness of lifts===
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* We want to show {{M|\big(\exists y_0\in Y[g(y_0)\eq h(y_0)]\big)\implies g\eq h}}, to show {{M|g\eq h}} we must show: {{M|\forall y\in Y[g(y)\eq h(y)]}}

Revision as of 17:21, 25 February 2017

Gamelin & Greene

  • Evenly covered
    • Let [ilmath]p:E\rightarrow X[/ilmath] be continuous map, we say [ilmath]U[/ilmath] open in [ilmath]X[/ilmath] is evenly covered by [ilmath]p[/ilmath] if:
      • [ilmath]p^{-1}(U)[/ilmath] equals a union of disjoint open sets of [ilmath]E[/ilmath] such that each one of these disjoint open sets is homeomorphic onto [ilmath]U[/ilmath] if you restrict [ilmath]p[/ilmath] to it
  • Covering map
    • [ilmath]p:E\rightarrow X[/ilmath] is a covering map if:
      1. [ilmath]p[/ilmath] is surjective
      2. [ilmath]\forall x\in X\exists U\in\mathcal{O}(x,X)[U\text{ evenly covered by }p][/ilmath]
    • We say "[ilmath]E[/ilmath] is a covering space of [ilmath]X[/ilmath]"

Lift of a continuous map

[ilmath]\xymatrix{ & & E \ar[d]^p \\ Y \ar@{.>}[rru]^g \ar[rr]^f & & X }[/ilmath]

Let [ilmath]p:E\rightarrow X[/ilmath] be a covering map, let [ilmath](Y,\mathcal{ K })[/ilmath] be a topological space and let [ilmath]f:Y\rightarrow X[/ilmath] be a continuous map

  • If there is a map: [ilmath]g:Y\rightarrow E[/ilmath] such that [ilmath]p\circ g\eq f[/ilmath] then [ilmath]g[/ilmath] is called a lift of [ilmath]f[/ilmath]

i.e. if the diagram on the right commutes

Claims

  1. Uniqueness of lifts
    • Let [ilmath]p:E\rightarrow X[/ilmath] be a covering map, let [ilmath]Y[/ilmath] be a connected topological space and let [ilmath]f:Y\rightarrow X[/ilmath] be a continuous map.
      • Let [ilmath]g,h:Y\rightarrow E[/ilmath] be two lifts of [ilmath]f[/ilmath]
        • if [ilmath]\exists y\in Y[g(y)\eq h(y)][/ilmath] then [ilmath]g\eq y[/ilmath]
    • Take some time to think about this, it's basically saying if they're lifts over the same part then they agree, the diagram above commuting has large implications
  2. Path lifting theorem
    • Let [ilmath]p:E\rightarrow X[/ilmath] be a covering map, let [ilmath]\gamma\in[/ilmath][ilmath]C([0,1],X)[/ilmath] (a path), then as [ilmath]p[/ilmath] is a covering map [ilmath]\exists e_0\in E[\gamma(0)\eq p(e_0)][/ilmath], we claim:
      • There is a unique path (or "lift" of [ilmath]\gamma[/ilmath]): [ilmath]\alpha:[0,1]\rightarrow E[/ilmath] such that [ilmath]\alpha(0)\eq e_0[/ilmath] and such that [ilmath]p\circ\alpha\eq\gamma[/ilmath]
    • Caveat:Note that if [ilmath]\gamma[/ilmath] was a loop that [ilmath]\alpha[/ilmath] need not be a loop!
  3. Path-homotopy lifting theorem (named by Alec)
    • Let [ilmath]p:E\rightarrow X[/ilmath] be a covering map and let [ilmath]F:[0,1]\times[0,1]\rightarrow X[/ilmath] be a homotopy (a continuous map in this case), then by the covering map: [ilmath]\exists e_0\in E[p(e_0)\eq F(0,0)][/ilmath]
      • Then we claim there exists a unique lift of [ilmath]F[/ilmath], say [ilmath]G:[0,1]\times[0,1]\rightarrow E[/ilmath] such that [ilmath]G(0,0)\eq e_0[/ilmath]

Proof

Uniqueness of lifts

  • We want to show [ilmath]\big(\exists y_0\in Y[g(y_0)\eq h(y_0)]\big)\implies g\eq h[/ilmath], to show [ilmath]g\eq h[/ilmath] we must show: [ilmath]\forall y\in Y[g(y)\eq h(y)][/ilmath]
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