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\begin{document}
 \parindent=0cm
  \section*{$p_{\sharp}\pi_1(\widetilde{X},\widetilde{x})$}


\begin{thm}

Let $p: (\widetilde{X}, \widetilde{x_0}) \rightarrow (X,x_0)$ be a
covering.\\Then $p_\sharp : \pi_1(\widetilde{X}, \widetilde{x_0})
\rightarrow \pi_1(X,x_0)$ is a monomorphism.

\end{thm}

\begin{proof}

$p_\sharp\{\widetilde{\alp}\} = \{p \circ \widetilde{\alp}\} =
\{\alp\} = 1$�̶�� �����ϸ� $\alp \sim x_0$�� �����ϴ� homotopy
$F$�� �����Ѵ�.($x_0$�� constant loop)

�������� 3���κ��� $\widetilde{\alp} \sim \widetilde{x_0}$��
��´�. ���⿡�� $\widetilde{x_0}$�� constant loop�̸鼭 ���ÿ�
$x_0$�� lifting�̴�.

$\therefore \{\widetilde{\alp}\}=\{\widetilde{x_0}\}=1$
\end{proof}\\

{\bf Remark.}

1. $p_{\sharp}\pi_1(\widetilde{X},\widetilde{x_0})$ ��
$\pi_1(X,x_0)$ �� subgroup���� �� �� �ִ�.

2. $T^2$�� $S^2$�� covering�� �� �� ���� :

$\pi_1(T^2)=\mathbf{Z}\bigoplus \mathbf{Z}$�̰� ���� ��������
�ʾ����� $\pi_1(S^2)=0$ ���� �̿��ϸ�, $T^2$�� $S^2$��  covering��
�� �� ������ �� �� �ִ�.


\begin{thm}
Let $p:\widetilde{X}\rightarrow X$ be a covering with
$p(\widetilde{x_0})=p(\widetilde{x_0}')=x_0$. Then $
p_{\sharp}\pi_1(\widetilde{X},\widetilde{x}_0)$ and
$p_{\sharp}\pi_1(\widetilde{X},\widetilde{x_0}')$ are conjugate
in $\pi_1(X,x_0)$, i.e., $\exists \{\tau\}\in \pi_1(X,x_0)$ such
that
$\{\tau\}p_{\sharp}\pi_1(\widetilde{X},\widetilde{x_0})\{\tau\}^{-1}=p_{\sharp}\pi_1(\widetilde{X},\widetilde{x_0}')$

\end{thm}

\begin{proof}
���� $\widetilde{x_0}'$�� $\widetilde{x_0}$������ path $\sigma$��
���� ��, �װ��� X�� ���� $\tau:\,\,=p\circ \sigma$�� ������ ����.
 �׸��� ������  �Ϲ�����  diagram�� ������ ����,

$\hspace{9.3em}\phi_{\rho}$

$\hspace{3em}\pi_1(X,x_0)\hspace{2em}\rightarrow\hspace{2em}\pi_1(X,x_1)$

$\hspace{4em}f_{\sharp}\downarrow\hspace{8.5em}\downarrow
f_{\sharp}\hspace{5em}$ key diagram ($*$)

$\hspace{2.3em}\pi_1(Y,f(x_0))\hspace{1.5em}\rightarrow\hspace{1.5em}\pi_1(Y,f(x_1))$

$\hspace{9.3em}\phi_{f \circ \rho}$

 �� diagram��  commute���� ������.

$\{\alp\} \in \pi_1(X,x_0)$�� ����
$f_{\sharp}(\phi_{\rho}(\{\alp\})=\phi_{f \circ \rho
}(f_{\sharp}(\{\alp\}))$ ���� ������ �ϴµ�

$f_{\sharp}(\phi_{\rho}(\{\alp\})=f_{\sharp}(\{\rho * \alp *
\overline{\rho}\})=\{f \circ (\rho * \alp *
\overline{\rho})\}=\{(f \circ \rho)*(f \circ \alp )*(f\circ
\overline{\rho})\}$

$\phi_{f \circ \rho }(f_{\sharp}(\{\alp\}))=\phi_{f\circ
\rho}\{f\circ \alp\}=\{(f \circ \rho)*(f \circ \alp
)*(\overline{f\circ \rho})\} $

�̰�, $f\circ \overline{\rho}=\overline{f\circ \rho} $ �̹Ƿ�, ��
diagram��  commute�Ѵ�.

���� ���� ������  case�� �´� ���� diagram �� commute�Ѵ�.

$\hspace{9.3em}\phi_{\sigma}$

$\hspace{3em}\pi_1(\widetilde{X},\widetilde{x_0})\hspace{2em}\rightarrow\hspace{2em}\pi_1(\widetilde{X},\widetilde{x_0}')$

$\hspace{4em}p_{\sharp}\downarrow\hspace{8.5em}\downarrow
p_{\sharp}\hspace{5em}$  diagram ($*$)'

$\hspace{3em}\pi_1(X,x_0)\hspace{2em}\rightarrow\hspace{2em}\pi_1(X,x_0)$

$\hspace{9.3em}\phi_{\tau}$

���� �� ������ �����Ǿ���.

\end{proof}\\

{\bf Remark.}

1 .\textit{ If
$\,\,\,p_{\sharp}\pi_1(\widetilde{X},\widetilde{x_0})$ is a
normal subgroup of $\,\,\,\,\pi_1(X,x_0)\,\,$, then\\
$p_{\sharp}\pi_1(\widetilde{X},\widetilde{x_0})=p_{\sharp}\pi_1(\widetilde{X},\widetilde{x_0}')$
for all $\widetilde{x_0}' \in p^{-1}(x_0)$.}

��, normality�� $\widetilde{X}$ ��  base point �� ���ÿ�
�����ϴ�. ���� ��  diagram $(*)'$���� $X$ �� base point $x_0$��
���ÿ��� �����ϴ�. �� ��� $p:\widetilde{X}\rightarrow X$ ��
\textit{regular} covering (or \textit{normal} covering) �̶��
�θ���.

2 . �� �����κ��� \textit{The conjugacy class of a subgroup
$\,\,p_{\sharp}\pi_1(\widetilde{X},\widetilde{x_0})$��
$\,\{p_{\sharp}\pi_1(\widetilde{X},\widetilde{x_0}')\,\,\,|\,\,\,\widetilde{x_0}'
\in p^{-1}(x_0)\}$�� ����.}\\




  \end{document}
