# 7.03 Path-lifting: uniqueness

Below the video you will find accompanying notes and some pre-class questions.

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# Notes

*(0.00)* In the last section, we saw that if you have a path
\(\gamma\) in \(X\) starting at \(x=\gamma(0)\), a covering space
\(p\colon Y\to X\) and a point \(y\in p^{-1}(x)\) then there exists a
lifted path \(\tilde{\gamma}\) in \(Y\) with
\(\tilde{\gamma}(0)=y\). In this section, we will see that this lift
is unique.

## Uniqueness of lifts

*(1.00)* We will prove the following, more general, result.

*all*\(t\in T\) if and only if they are equal for

*some*\(t\in T\).

*(2.35)*For path lifting, \(T=[0,1]\) and \(F=\gamma\); the lemma will imply that two lifts which agree at \(0\in[0,1]\) agree everywhere.

*(2.49)*Recall that \(\tilde{F}\) is a lift if \(p\circ\tilde{F}=F\).

*(3.14)*We need to prove that if \(\tilde{F}_1(t)=\tilde{F}_2(t)\) for some \(t\in T\) then \(\tilde{F}_1(t)=\tilde{F}_2(t)\) for all \(t\in T\).

Let \(S=\{t\in T\ :\ \tilde{F}_1(t)=\tilde{F}_2(t)\}\). We need to prove that \(S\) is open and that \(T\setminus S\) is open (i.e. \(S\) is closed). Since \(T\) is connected, this implies that either \(S=\emptyset\) or \(T\setminus S=\emptyset\), so if \(S\) is nonempty then \(S=T\) and \(\tilde{F}_1\equiv\tilde{F}_2\).

*(5.22)* Given a point \(t\in T\) and let \(x=F(t)\). Since
\(p\colon Y\to X\) is a covering space, there is an elementary
neighourhood \(W\subset X\) of \(x\). In \(Y\) there are elementary
sheets \(V_1\) and \(V_2\), projecting to \(W\), containing
\(\tilde{F}_1(t)\) and \(\tilde{F}_2(t)\) respectively.

*(7.32)* Since \(\tilde{F}_1,\tilde{F}_2\) are continuous, there
exist open sets \(U_1,U_2\subset T\) such that \(t\in U_i\) and
\(\tilde{F}_1(U_i)\subset V_i\) for \(i=1,2\). Let \(U=U_1\cap
U_2\).

*(8.45)* We want to show that \(T\setminus S\) is open, so suppose
that \(t\in T\setminus S\). Then I claim that \(U\subset T\setminus
S\), which will prove that every point in \(T\setminus S\) has an
open neighbourhood contained in \(T\setminus S\), which implies that
\(T\setminus S\) is open.

*(9.40)* The claim can be proved as follows. Since \(t\in T\setminus
S\), \(\tilde{F}_1(t)\neq\tilde{F}_2(t)\), so \(V_1\) and \(V_2\)
are disjoint. Therefore \(\tilde{F}_1(t')\neq\tilde{F}_2(t')\) for
all \(t'\in U\) (because \(t'\in U\) implies \(\tilde{F}_1(t')\in
V_1\) and \(\tilde{F}_2(t')\in V_2\), and \(V_1\cap
V_2=\emptyset\)).

*(11.25)* We also want to show that \(S\) is open, so suppose that
\(t\in S\). Then \(\tilde{F}_1(t)=\tilde{F}_2(t)\), so
\(V_1=V_2=:V\). There exists a local inverse \(q\colon W\to V\) to
\(p\), i.e. \(p\circ q=id_W\), \(q\circ p=id_V\).

*(12.53)* If \(t'\in U\) then \(\tilde{F}_1(t')\in V\) and
\(\tilde{F}_2(t')\in V\), so
\(q(F(t'))=q(p(\tilde{F}_1(t')))=\tilde{F}_1(t')\) and
\(q(F(t'))=q(p(\tilde{F}_2(t')))=\tilde{F}_2(t')\), so we deduce
that \(\tilde{F}_1(t')=\tilde{F}_2(t')\). Therefore \(\tilde{F}_1\)
and \(\tilde{F}_2\) agree on \(U\), which proves that \(S\) is open.

This proves the lemma.

# Pre-class questions

- There is a condition on covering spaces which I'm using in this
section but which I forgot to mention in earlier sections. Can you
spot what it is?

# Navigation

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