\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*{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}
 먼저 $f_{\sharp}(\{\alpha\})$ 가 잘 정의되는지 살펴보자.
 즉 두 $\alpha \sim \alpha'$ 에 대해 $f \circ \alpha \sim f \circ
 \alpha'$ 임을 보이자. $\alpha$ 와  $\alpha'$ 사이의 homotopy F에
 대해 f $ \circ$ F는 $f \circ \alpha \sim f \circ
 \alpha'$ 사이에 homotopy를 준다. 왜냐하면

  $\hspace{2em}\,\,\,\,(f \circ F)$(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($x_0) =
 y_0 ,\,\,\,\,\,\,\, \forall s \in I$.

다음으로 $f_{\sharp}$이 homomorphism임을 보이자.

$f_{\sharp}(\{\alpha\}\{\beta\})=f_{\sharp}(\{\alpha *
\beta\})=\{f \circ (\alpha * \beta)\}$이고

$$(\alpha * \beta)(t)=\left\{\begin{array}{1} \alpha(2t),\,\,\,\,\,\,
0\leq t \leq \frac{1}{2}
 \\ \beta(2t-1),\,\,\,\,\,\,\frac{1}{2}\leq t \leq 1 \end{array} \right.$$

이므로 $f\circ(\alpha*\beta)$는 정확히 $(f\circ \alpha)*(f\circ
\beta)$가 된다. 따라서

$f_{\sharp}(\{\alpha\}\{\beta\})=\{(f\circ \alpha)*(f\circ
\beta)\}=\{f\circ \alpha\}\{f \circ \beta
\}=(f_{\sharp}\{\alpha\})(f_{\sharp}\{\beta\})$.

\end{proof}

\begin{thm}
1.  $f : (X,x_0)\rightarrow (Y,y_0)$, $g : (Y,y_0) \rightarrow
(Z,z_0)$에 대해


$\hspace{4em}\,\,\,\,(g \circ f)_{\sharp} = g_{\sharp}\circ
f_{\sharp}이다.$


 $\hspace{3em}$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}

위와 같은 성질을 \textit{Functorial property} 라고 한다.\\

{\bf Remark.} The previous theorem $\Rightarrow$ If f has an
inverse $f^{-1}$ then $(f^{-1})_{\sharp}=(f_{\sharp})^{-1}$.

\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}
\end{document}