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\begin{document}
 \parindent=0cm
 \section*{Functorial Property}
 \begin{thm}
f : $(X,x_0)\rightarrow (Y,y_0)$  induces  a  homomorphism

$\hspace{3em}f_{\sharp}$ : $\pi_1(X,x_0) \rightarrow \pi_1
(Y,y_0)$ given by $ [\alpha] \mapsto [f \circ \alpha]$.
\end{thm}
\begin{proof}
 ���� $f_{\sharp}([\alpha])$�� �� ���ǵǴ��� ���캸��.
 �� $\alpha \sim \alpha'$�� ���� $f \circ \alpha \sim f \circ
 \alpha'$ ���� ������. $\alpha$ ��  $\alpha'$ ������ homotopy�� $F$��
 �ϸ�,\\
$\hspace*{2em}(f \circ F)$(t,0) = f(F(t,0)) = f($\alpha(t))=
(f\circ \alpha)(t)$\\
$\hspace*{2em}(f \circ F)$(t,1) = f(F(t,1)) = f($\alpha'(t))=
(f\circ \alpha')(t)$\\
$\hspace*{2em}(f \circ F)$(0,s) = f(F(0,s)) = f($x_0) = y_0
\hspace{3em} \forall s \in I$\\
�̹Ƿ� $f \circ F$�� $f \circ \alpha $�� $f \circ
 \alpha'$ ���̿� homotopy�� �ش�.\\
�������� $f_{\sharp}$�� homomorphism���� ������.\\
$f_{\sharp}([\alpha][\beta])=f_{\sharp}([\alpha * \beta])=[f \circ
(\alpha * \beta)]$�̰�,\\
\begin{displaymath}
(\alpha * \beta)(t)=\left\{\begin{array}{cl} \alpha(2t)& 0\leq t
\leq \frac{1}{2}
 \\ \beta(2t-1) & \frac{1}{2}\leq t \leq 1 \end{array}
 \right.
\end{displaymath}
�̹Ƿ� $f\circ(\alpha*\beta)$�� ��Ȯ�� $(f\circ \alpha)*(f\circ
\beta)$�� �ȴ�. ����,\\
$f_{\sharp}([\alpha][\beta])=[(f\circ \alpha)*(f\circ
\beta)]=[f\circ \alpha][f \circ \beta
]=f_{\sharp}([\alpha])f_{\sharp}([\beta])$.
\end{proof}

\begin{thm}(Functorial Property)

\hspace*{1em}$1) \ f : (X,x_0)\rightarrow (Y,y_0)$, $g : (Y,y_0)
\rightarrow
(Z,z_0)$\\
$\hspace*{2em}\Rightarrow (g \circ f)_{\sharp} =g_{\sharp}\circ f_{\sharp}$.\\
\hspace*{1em}$2) \  id_{\sharp} = id $.
\end{thm}
\begin{proof}

$(g\circ f)_{\sharp}[\alpha] = [(g \circ f)\circ \alpha] = [g\circ
(f \circ \alpha)] = g_{\sharp}[f \circ
\alpha]=g_{\sharp}(f_{\sharp}[\alpha])= (g_{\sharp}\circ
f_{\sharp})[\alpha].$

$id_{\sharp}[\alpha]=[id \circ \alpha]=[\alpha]$.
\end{proof}\\

���� $\pi_1$�� Category of topological space with base point����
Category of group���� ���� Functor�̴�.\\

{\bf Applications} \\
1. If $f$ has an inverse $f^{-1}$ then
$(f^{-1})_{\sharp}=(f_{\sharp})^{-1}$ by functorial property.\\
$\because f_\sharp \circ (f^{-1})_\sharp=(f\circ
f^{-1})_\sharp=id_\sharp=id$.

\begin{cor}
$f:(X,x_0)\rightarrow (Y,y_0)$ is a homeomorphism.

$\Rightarrow f_{\sharp}:\pi(X,x_0)\rightarrow \pi(Y,y_0)$ is an
isomorphism.
\end{cor}

���߿� ������ ������ �ϰ�����, ���� ��� $\mathbb{R}^2$��
$\mathbb{R}^2-\{0\}$�� fundamental group�� ���� $0$��
$\mathbb{Z}$�̹Ƿ� �� �� ������ homeomorphic���� �ʴ�.\\

2. Brouwer Fixed Point Theorem
\begin{thm} Let f : $D^2 \rightarrow D^2$ be a map. Then f has a fixed point,
  i.e., $\exists x \in D^2$ such that f(x)=x.
 \end{thm}
\begin{proof}
$f$�� fixed point�� ���� �ʴ´ٰ� ��������. \\
$g(x)$�� $f(x)$���� ����Ͽ�
 $x$�� ������ �������� $\partial D^2$�� �������� �����ϸ�,
�Լ� $g : D^2 \rightarrow \partial D^2$�� ���ǵȴ�. ��,
\begin{displaymath}
g(x) = f(x) + t(x-f(x)) \ \ where \ t>0 \ and\  \|f(x) + t(x-f(x))
\|=1.
\end{displaymath}
�̶�, $g$�� �����̰�, $\partial D^2=S^1$���� $g(x)=x$�̹Ƿ�, ���� diagram�� commute�Ѵ�. \\

\hspace*{7.5em} $D^2\hspace{1ex}\overset{g}{\longrightarrow}\hspace{1ex}\partial D^2$\\
\hspace*{8.0em} $i\nwarrow \hspace{3ex} \nearrow id$\\
\hspace*{9.0em} $\hspace{0.5ex}\partial D^2$\\

�̿� �����ϴ� fundamental group���� ������ ����,\\

\hspace*{4.0em} $0=\pi_1(D^2,1)\overset{g_\sharp}{\longrightarrow}\pi_1(S^1,1)=\mathbb{Z}$\\
\hspace*{7.0em} $i_\sharp\nwarrow \hspace{6ex} \nearrow id_\sharp$\\
\hspace*{8.0em} $\pi_1(S^1,1)=\mathbb{Z}$\\

�� �ǰ�, $0 = g_{\sharp}\circ i_{\sharp}=(g \circ
i)_{\sharp}=id_{\sharp}=id : \mathbb{Z} \rightarrow \mathbb{Z}$
�̹Ƿ� �̴� ����̴�.
\end{proof}
\end{document}
