Notes:Covering spaces
From Maths
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
- 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:
- Covering map
- [ilmath]p:E\rightarrow X[/ilmath] is a covering map if:
- [ilmath]p[/ilmath] is surjective
- [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]"
- [ilmath]p:E\rightarrow X[/ilmath] is a covering map if:
Lift of a continuous map
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
- 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]
- Let [ilmath]g,h:Y\rightarrow E[/ilmath] be two lifts of [ilmath]f[/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
- 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.
- 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!
- 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:
- 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]
- 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]
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]
- Let:
- [math]S:\eq\{y\in Y\ \vert\ g(y)\eq h(y)\} [/math]
- [math]T:\eq\{y\in Y\ \vert\ g(y)\neq h(y)\} [/math]
- Note that [ilmath]Y\eq S\cup T[/ilmath] and [ilmath]S\cap T\eq\emptyset[/ilmath] (these are easily shown)
- Note also that [ilmath]Y-T\eq S[/ilmath] and [ilmath]Y-S\eq T[/ilmath]
- If we were to show that [ilmath]S[/ilmath] and [ilmath]T[/ilmath] are open it follows that [ilmath]S[/ilmath] and [ilmath]T[/ilmath] are closed too.
- As [ilmath]Y[/ilmath] is a connected topological space the only sets with this property (of being open and closed) are [ilmath]Y[/ilmath] and [ilmath]\emptyset[/ilmath]
- Notice that by hypothesis, [ilmath]S[/ilmath] isn't empty
- We see that [ilmath]S\eq Y[/ilmath] and [ilmath]T\eq\emptyset[/ilmath]
- If we were to show that [ilmath]S[/ilmath] and [ilmath]T[/ilmath] are open it follows that [ilmath]S[/ilmath] and [ilmath]T[/ilmath] are closed too.
- Let:
- So if we prove [ilmath]S[/ilmath], [ilmath]T[/ilmath] are both open then the result follows.