Unique lifting property
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
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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 nonempty)
 [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 YT[/ilmath] (and [ilmath]T\eq YS[/ilmath]) in some form. We need [ilmath]S\eq YT[/ilmath] and [ilmath]T\eq YS[/ilmath] at least!
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 YS[/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 nonempty 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\mkern15mu\cdot\mkern8mu}}{\mathop{\bigcup\mkern13mu\cdot\mkern5mu}}{\mathop{\bigcup\mkern13mu\cdot\mkern5mu}}{\mathop{\bigcup\mkern13mu\cdot\mkern5mu}} }[/math][math]\newcommand{\udot}{\cup\mkern12.5mu\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]
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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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