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\begin{document}
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\section*{Base point change}
Let $X$ be path-connected. 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 define $\phi_{\rho} : \pi_1(X,x_0) \rightarrow
\pi_1(X,x_1)$ by $[\alpha] \mapsto [\overline{\rho}*\alpha*\rho]$.
Then {$\phi_{\rho}$ is an isomorphism.}\\

\begin{proof}

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

$\,\,\hspace{1em}\phi_{\rho}[\alpha * \beta]=[\overline{\rho}
* \alpha * \beta* \rho]$

$ \hspace{5.5em}=[ \overline{\rho} * \alpha * \rho *
\overline{\rho}* \beta * \rho]$

$ \hspace{5.5em}=[ \overline{\rho} * \alpha * \rho
][\overline{\rho}* \beta * \rho]$

$ \hspace{5.5em}=\phi_{\rho}[\alpha]\phi_{\rho}[\beta]$.\\


(2) Define $\phi_{\overline{\rho}} : \pi_1(X,x_1) \rightarrow
\pi_1(X,x_0)$ by $[\bet] \mapsto [\rho * \bet * \overline{\rho}]$.


Then,$(\phi_{\overline{\rho}} \circ \phi_{\rho})[\alpha]=
\phi_{\overline{\rho}}[\overline{\rho} * \alpha * \rho] =[\rho
* \overline{\rho}
* \alpha * \rho * \overline{\rho}] = [\alpha]$.\\
따라서 $\phi_{\overline{\rho}} \circ \phi_{\rho} = id$.\\



Similarly, $\phi_{\rho}\circ \phi_{\overline{\rho}}= id$.


\end{proof}\\


한편, 또다른 path $\sigma$가 주어졌을때  만일 $\rho \simeq \sigma$
이라면  $\phi_{\rho}$ = $\phi_{\sigma}$ 이다. \\



{\bf 숙제 2. }\\
일반적으로 homotopic 하지 않은 두 path ${\rho}$, ${\sigma}$에
대해\\
$\hspace*{11em}\,\,\,\phi_{\rho}$

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

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

$\hspace{12em}\pi_1(X,x_1)$

에서 $\mu=\phi_{\rho}\circ\phi_{\sigma^{-1}}$ 로 두면 이는
isomorphism 이 되고, 이것은 어떤 loop에 의한  conjugation (an
inner automorphism of $\pi_1(X,x_1))$이 됨을 보여라.

\end{document}
