\documentclass[12pt ]{article}
\setlength{\textwidth}{14 true cm} \setlength{\textheight}{20 true
cm}

\usepackage{hangul}
\usepackage{amscd,amsmath}
\usepackage{amsfonts}
\usepackage{amssymb,theorem}
\usepackage{longtable}
\newcommand{\Aff}{\mbox{\it Aff}}
\newcommand{\aff}{\mbox{\it aff}}


\newcommand{\alp}{\alpha}
\newcommand{\bet}{\beta}
\newcommand{\del}{\delta}
\newcommand{\gam}{\gamma}
\newcommand{\vep}{\varepsilon}
\newcommand{\eps}{\epsilon}
\newcommand{\lam}{\lambda}
\newcommand{\kap}{\kappa}
\newcommand{\sig}{\sigma}
\newcommand{\ome}{\omega}
\newcommand{\Gam}{\Gamma}
\newcommand{\Ome}{\Omega}
\newcommand{\Sig}{\Sigma}
\newcommand{\Del}{\Delta}
\newcommand{\Lam}{\Lambda}


\newtheorem{thm}{정리}
\newtheorem{cor}[thm]{따름정리}
\newtheorem{lem}[thm]{보조정리}
\newtheorem{prop}[thm]{명제}
\newtheorem{cl}{Claim}

{\theorembodyfont{\rm}
\newtheorem{ex}{예}
\newtheorem{que}{질문}
\newtheorem{notation}{Notation}[section]
\newtheorem{defn}{정의}
\newtheorem{rem}{주}
\newtheorem{note}{Note}
}
\renewcommand{\thenote}{}
\renewcommand{\therem}{}

\newenvironment{proof}{{\bf 증명}}{\hfill\framebox[2mm]{}}
\newenvironment{proof1}{{\bf 정리증명}}{\hfill\framebox[2mm]{}}

\begin{document}
 \parindent=0cm
\section*{Base point change}

For $x_0,x_1 \in X$, compare $\pi_1(X,x_0)$ with  $\pi_1(X,x_1)$.
Choose a path $\rho$ from $x_0$ to $x_1$ and \textit{define
$\phi_{\rho} : \pi_1(X,x_0)\rightarrow \pi_1(X,x_1)$ by
$\{\alpha\} \mapsto \{\rho^{-1}*\alpha*\rho\}$.} Then
\textit{$\phi_{\rho}$ is an isomorphism.}\\
\begin{proof}

(1) $\phi_{\rho}$가 homomorphism임을 보이자 :

$\,\,\hspace{1em}\phi_{\rho}\{\alpha * \beta\}=\{\rho^{-1} *
\alpha * \beta* \rho\}$

$ \hspace{6em}=\{ \rho^{-1} * \alpha * \rho * \rho^{-1}* \beta *
\rho\}$

$ \hspace{6em}=\{ \rho^{-1} * \alpha * \rho \}\{\rho^{-1}* \beta *
\rho\}$

$ \hspace{6em}=\phi_{\rho}\{\alpha\}\phi_{\rho}\{\beta\}$.

(2) Define $\phi_{\rho^{-1}}$ for $\rho^{-1}$ similarly and show
$\phi_{\rho^{-1}}=(\phi_{\rho})^{-1}$ :

$\,\,(\phi_{\rho^{-1}}\circ \phi_{\rho})\{\alpha\}=
\phi_{\rho^{-1}}\{\rho^{-1} * \alpha * \rho\}$

$ \hspace{7em}=\{\rho * \rho^{-1}
* \alpha * \rho * \rho^{-1}\}$

$\hspace{7em}=\{\alpha\}$.

여기서 $\rho * \rho^{-1} \sim e \sim  \rho^{-1} * \rho $ 임을
사용하였다.

 Similarly $\phi_{\rho}\circ \phi_{\rho^{-1}}= id.$
\end{proof}\\

{\bf Remark.} If X is path connected, then we denote its
fundamental group simply by $\pi_1$(X)  if the choice of base
point is not
important.\\

만일 path $\rho$가 주어지면 그에 따라 isomorphism $\phi_{\rho}$ 가
얻어진다. 이 때  또다른 path $\sigma$가 주어지는 경우  이에 따른
$\phi_{\rho}$ 와 $\phi_{\sigma}$ 와의 관계를 살펴보자. 먼저 $\rho
\simeq \sigma$ 인 경우 $\phi_{\rho} \cong
\phi_{\sigma}$ 이다. $\rho \nsim \sigma$인 경우는 어떻게 될까?\\
{\bf 숙제 5. } $\hspace{7em}\,\,\,\phi_{\rho}$

$\hspace{7em}\pi_1(X,x_0)\rightarrow \pi_1(X,x_1)$

$\hspace{9em}\phi_{\sigma}\searrow \hspace{3em}\uparrow \mu$

$\hspace{13em}\pi_1(X,x_1)$

에서 $\mu=\phi_{\rho}\circ\phi_{\sigma^{-1}}$ 로 두면 이는
isomorphism 이 되고, 이것은 어떤 loop에 의한  conjugation (an
inner automorphism of $\pi(X,x_1))$이 됨을 보여라.
\end{document}
