Unique lifting property

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^{Partial: -}

• 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

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!

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

Warning:That grade doesn't exist!

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!
• 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:
            1. [ilmath]p^{-1}(U)\eq\bigcup_{\alpha\in I}V_\alpha[/ilmath]
            2. The [ilmath](V_\alpha)_\alpha[/ilmath] are open (given by being in [ilmath]\mathcal{H} [/ilmath])
            3. The [ilmath](V_\alpha)_\alpha[/ilmath] are pairwise disjoint
            4. 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]
        • 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:
                    1. that [ilmath]W[/ilmath] is open in [ilmath](Y,\mathcal{ K })[/ilmath] as by definition of a topology the intersection of open sets is open.
                    2. [ilmath]W\neq\emptyset[/ilmath] ([ilmath]W[/ilmath] is non-empty) as - at least - [ilmath]s\in W[/ilmath]
                  • 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 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]
                • Thus [ilmath]W\subseteq Y[/ilmath], that is [ilmath]Y[/ilmath] contains a neighbourhood of [ilmath]s[/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.

Grade: A

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We still need to show [ilmath]T[/ilmath] is open