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\begin{document}
 \parindent=0cm
  \section*{Unique Path Lifting  }


\begin{thm}\textit{
(Unique path lifting property)\\
Let $p: \widetilde{X} \rightarrow X$ be a covering map and let
$\alp : I \rightarrow X$ be a path with $\alp(0)=x_0 \in X $ and  $
p(\widetilde{x_0})=x_0$. Then $\alp$ has a unique path lifting
$\widetilde{\alp}:I\rightarrow X$ with
$\widetilde{\alp}(0)=\widetilde{x_0}$ i.e., $p\circ
\widetilde{\alp}(t)=\alp (t).\,\,\forall \,\,t \in I.$
 }\end{thm}

 \begin{proof}
\,\,\textit{(Existence)}\\ For each t, $\alp(t)\in X$ has an open
neighborhood $U_t$ which is evenly covered by
$\displaystyle{\coprod_{a \in A}}V_{t,a}$ .  Since I=[0,1]
is compact, we can choose a Lebesgue number $\epsilon > 0$ for a
cover $\{\alp^{-1}(U_t)| t
\in I \}$ of I. Choose  partition of $\,\,$ I,\\
$\,\,0=t_0<t_1<...<t_{n+1}=1\,\,$ so that
$t_{i+1}-t_{i}<\epsilon,\,\,\,\,i=1,...n$.

 Then note that $\alp[t_i,t_{i+1}] \subset U_t$ for some t and we
 lift $\alp|_{[t_i,t_{i+1}]}$ inductively :

 Suppose $\alp|_{[t_0,t_{i}]}$ is already lifted (note that the
 initial point $x_0$ is lifted to $\widetilde{x_0}$). Then $\alp[t_i,t_{i+1}]\subset
 U_t$ for some t and $p^{-1}(U_t)=\displaystyle{\coprod_{a \in
 A}}V_{t,a}$ and there exists a unique $a \in A$ such that
 $\tilde{\alp}(t_i)\in V_{t,a}\,.$\\And since $p|_{V_{t,a}}:V_{t,a}\rightarrow
 U_t\,\,$is homeomorphism we can lift $\alp|_{[t_i,t_{i+1}]}$ using $(p|_{V_{t,a}})^{-1}$ and the proof is
 completed.\\

\textit{(Uniqueness)}\\ Suppose $p\circ \tilde{\alp_i}=\alp$ and
$\tilde{\alp_i}(0)=\widetilde{x_0}\,\,$  $i=1, 2$. Then we show that
$J=\{t\in I | \widetilde{\alp_1}(t)=\widetilde{\alp_2}(t)\}$ is
open and closed non-empty set : \\1 .   (J is  non-empty) :
$\,\,x_0\in I$.\\2 . (J is open) : $t\in J$에 대해 $\alp(t)\in
X$는 evenly cover되는 U를 가지고 $p^{-1}(U)=\coprod V_{a}$ 에서
$\widetilde{\alp_i}(t)$ 를 포함하는 $V_{a}$는 유일하다. 그리고
$V_{a}$ 에서는 p가 homeomorphism이므로
$\widetilde{\alp_1}=(p|_{V_{a}})^{-1}\circ
\alp=\widetilde{\alp_2}$ on $(t-\eps,t+\eps)$ 이다. 따라서
$\exists \,\eps > 0\,\,such\,\,\, that\,\, (t-\eps,t+\eps)\in
J$.\\3 .  (J is closed) : $J^c =\{ t\in I | \widetilde{\alp_1}(t)
\neq \widetilde{\alp_2}(t) \}$이 open 임을
보이자.\\$\widetilde{\alp_1}(t) \neq \widetilde{\alp_2}(t)$ for
some $t \in I$ 라면,  $\,\,\alp(t)$ 에 대해 evenly cover되는 U 가
존재해서 $p^{-1}(U)=\coprod V_{a}$ 이다. 그리고 각
$\widetilde{\alp_i}(t)$와 만나는 $V_{a_1},V_{a_2}$ 가
유일하게 존재하고, $V_{a_1} \neq V_{a_2}$ 이다. 즉
$\widetilde{\alp_1}(t-\eps,t+\eps)\subset V_{a_1}\,\,,\,\,
\widetilde{\alp_2}(t-\eps,t+\eps)\subset V_{a_2}$ 를  만족하는
$\eps \,>\, 0$ 이 존재한다. 따라서 $J^c$ 는 open이다.\\ I는
connected 이므로 위 1,2,3, 에 의해 J = I 이다. 따라서  I 내부
전체에서 $\widetilde{\alp_1}=\widetilde{\alp_2}$ 이다.
\end{proof}\\


{\bf Remark.} \textit{(Uniqueness of lifting)} Y가 connected 이고
$f : (Y,y_0)\rightarrow (X,x_0)$ 가 lifting $\tilde{f} :
(Y,y_0)\rightarrow (\widetilde{X},\widetilde{x_0}) $를 가지면,
이는 unique하다.
\\(증명) Y의 connectedness 를 이용, 위  정리의 증명에서 $I$ 대신 $Y$ 를 써서 똑같이  하면 된다.\\

\begin{thm}
\textit{(Lifting of homotopy of paths.)\\Let $p :
(\widetilde{X},\widetilde{x_0})\rightarrow (X,x_0)$ be a covering
space. And  $\alp : I \rightarrow X$ with $\alp(0) = x_0 $ and $F
: \alp \simeq \beta $ be a homotopy between $\alp$ and $ \beta$.
Then $ \,\,\exists!\,\,\widetilde{F} :
I\times I \rightarrow \widetilde{X}$ such that $p\circ
\widetilde{F}=F$ and $\widetilde{F}(0,0)=\widetilde{x_0}$.\\
In particular, $\widetilde{F}$ gives a homotopy between
$\widetilde{\alp} = F_0$ and
$\widetilde{\beta}=F_1$.\\Furthermore (1)if F keeps initial point
$x_0$ fixed, i.e., $F(0,u)=x_0 \,\,\forall u\in I$, then}

\textit{$\hspace{6em}\,\, \widetilde{F}$ keeps initial point
$\widetilde{x_0}$ fixed,\\\hspace{4em}and \hspace{3.2em}(2)if F
keeps end points $\alp(0)=\beta(0)$ and $\alp(1)=\beta(1)$ fixed,
then }

\textit{$\hspace{6em}\widetilde{F}$ keeps end points
$\widetilde{\alp(0)}=\widetilde{\beta(0)}$ and
$\widetilde{\alp(1)}=\widetilde{\beta(1)}$ fixed.}
\end{thm}
\begin{proof}
이 증명 역시 존재성만 보이면, 유일성은 $I^2$의 connectedness에
의해 보장된다.\\For each $(t,u)$,$\,\, F(t,u)$ has an open
neighborhood $\,U_{(t,u)}\,$ which is evenly covered by
$p^{-1}(U_{(t,u)})=\displaystyle{\coprod_{a \in
A}}V_{(t,u),\,a}\,$. Choose a Lebesgue  number $\eps > 0$ for a
cover $\{\,F^{-1}(U_{(t,u)})\,\,\,|\,\,\,(t,u)\in I\times I\,\,\}$
for compact $I\times I.$  Choose a partition\\
$0=t_0<t_1<\cdot\cdot\cdot<t_{n+1}=1$ with $t_{i+1}-t_i <
\frac{\eps}{2}\\ 0=u_0<u_1<\cdot\cdot\cdot<u_{n+1}=1$ with
$u_{i+1}-u_i < \frac{\eps}{2}\,\,\,\,\,\,$ so that \\each
$[t_i,t_{i+1}]\times[u_i,u_{i+1}] \subset F^{-1}(U_{(t,u)})$ for
some $(t,u)$.

As in Theorem 1, $F$ is defined inductively starting from
$[t_0,t_{1}]\times[u_0,u_{1}]$ so that $F(0,0)=\widetilde{x_0}\in
V_{(0,0),\,\alp} $ using the homeomorphism $p|_{V_{(0.0),\,a}}
: V_{(0,0),\,a}\rightarrow U_{(0,0)}$. Then lift
$F|_{[t_1,t_2]\times[u_0,u_1]}\,\,,\, \cdot\cdot\cdot\, ,\,\,
F|_{[t_n,t_{n+1}]\times[u_0,u_1]} $ successively as before to
obtain a lifting of $F|_{[0,1]\times[u_0,u_1]}$ .

Now lift $F|_{[0,1]\times[u_1,u_2]}$ using the already lifted
portion as above, and lift  $F|_{I\times[u_2,u_3]}$ ,
$\cdot\cdot\cdot$ etc., finally to get a lifting $\widetilde{F} :
[0,1]\times[0,1]\rightarrow \widetilde{X}$. \\ 위 정리의 (1)과
(2)는 lifting의 uniqueness에 의해서 constant map의 lifting은 constant map일 수 밖에 없으므로
 성립한다.
\end{proof}\\

{\bf Exercise.}  $\widetilde{F}$ 가 연속임을 보여라.\\
(Hint) $X=A\cup B$, $A$ and $B$ both closed(or open) in X 라면

$f:X\rightarrow Y$ 에서 \textit{$f|_A$ and $f|_B$ 가 연속이면 f는
연속}임을 보인후 이를 이용하라.\\


\begin{cor}\textit{
$\alp \sim \beta \Rightarrow \widetilde{\alp}\sim
\widetilde{\beta}$.}
\\여기서 $\alp,\beta$는 X의 path들이고 $\widetilde{\alp}$,$\widetilde{\beta}$는 같은 initial point를 가지는 lifting 들이다.
\end{cor}

\end{document}