# 7.03 Path-lifting: uniqueness

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

# 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.

Let $$p\colon Y\to X$$ be a covering space. Let $$T$$ be a connected space. Let $$F\colon T\to X$$ be a continuous map. Then two lifts $$\tilde{F}_1,\tilde{F}_2\colon T\to Y$$ are equal for 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

1. 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?