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\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}