\documentclass[12pt ]{article}
\setlength{\textwidth}{14 true cm} \setlength{\textheight}{20 true
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\newcommand{\Aff}{\mbox{\it Aff}}
\newcommand{\aff}{\mbox{\it aff}}


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\newcommand{\bet}{\beta}
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\newcommand{\eps}{\epsilon}
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\newcommand{\kap}{\kappa}
\newcommand{\sig}{\sigma}
\newcommand{\ome}{\omega}
\newcommand{\Gam}{\Gamma}
\newcommand{\Ome}{\Omega}
\newcommand{\Sig}{\Sigma}
\newcommand{\Del}{\Delta}
\newcommand{\Lam}{\Lambda}
\newcommand{\dis}{\displaystyle}
\newcommand{\disi}{\displaystyle{\sum_{i=1}^n}}
\newcommand{\disj}{\displaystyle{\sum_{j=1}^m}}
\newcommand{\pa}{\partial}
\newcommand{\cb}{\mathbb{C}}
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\newcommand{\rb}{\mathbb{R}}
\newcommand{\zb}{\mathbb{Z}}
\newcommand{\tx}{(\widetilde{X}, \widetilde{x}_0)}
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\newcommand{\txxx}{(\widetilde{X}_2, \widetilde{x}_2)}
\newcommand{\x}{(X,x_0)}
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\newcommand{\yt}{\widetilde{y}}
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\newtheorem{lem}[thm]{º¸Á¶Á¤¸®}
\newtheorem{prop}[thm]{¸íÁ¦}
\newtheorem{cl}{Claim}


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\newtheorem{que}{Áú¹®}
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\newtheorem{defn}{Á¤ÀÇ}
\newtheorem{rem}{ÁÖ}
\newtheorem{note}{Note}
}
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\begin{document}
\parindent=0cm
\psset{unit=2cm}
\section*{IV.2 Functorial property and Homotopy invariance}

\textbf{(1) Functorial property of $\pi_{n}$}\\

\framebox{\hspace*{1em}\parbox[b]{14cm}{ $f : (X,x_0 ) \rightarrow
(Y,y_0 ) \Rightarrow f_{*} = \pi_{n} (f)
: \pi_{n} (X,x_0 ) \rightarrow \pi_{n} (Y, y_0 )\\
\hspace*{18em} [\alp ]\hspace{1em} \mapsto \hspace{1em}[f\circ \alp ]$}}\\
%functorial property
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\begin{center}
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    \rput(-1.4,-0.3){$x_0$}%
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    \rput(0,0){$\overset{f}{\longrightarrow}$}

    \psframe[linewidth=0.5mm](-1.1,-1.2)(-0.7,-0.8)%
    \rput(-0.9,-0.6){$\alp$}%
    \rput(0.6,-0.6){$f\circ \alp$}
    \psline{->}(-1,-0.7)(-1,-0.4)%
    \psline{->}(-0.7,-0.7)(0.8,-0.4)
    %\rput(-1,0){$\alp$}

    }
}
\end{pspicture}
\end{center}
1. $f_{*}$ is a homomorphism : $f\circ (\alp *\bet ) = (f\circ
\alp) * (f\circ \bet)$\\

2. id$_*$ = id : clear\\
$\hspace*{1em} (g\circ f)_{*} = g_{*} \circ f_{*}: \hspace*{1em}
(X,x_0 ) \overset{f}{\rightarrow}(Y,y_0 )
\overset{g}{\rightarrow} (Z,z_0 )\\
\hspace*{7.5em}\Rightarrow \pi_{n} (X,x_0 )
\overset{f_{*}}{\rightarrow} \pi_{n}(Y,y_0 )
\overset{g_{*}}{\rightarrow} \pi_{n}(Z,z_0 )\\
\hspace*{11em} [\alp]\hspace{1em}\mapsto [f\circ \alp ] \mapsto
[g\circ f\circ
\alp]$\\

3. $f , g : (X,x_0 ) \rightarrow (Y,y_0 )$ and $ f
\underset{H}{\simeq} g$(rel $x_0)\\
\Rightarrow f_{*} = g_{*} : $  The proof is the same as in $\pi_1$ -case, i.e.,\\
$\hspace*{5em} f\circ\alp = H_0 \circ \alp \simeq H_1 \circ \alp =
g\circ \alp$\\

%functorial property
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%    \rput(-1,0.5){$X$}%
%    \rput(-1.4,-0.3){$x_0$}%
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%
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%    \rput(0,0.4){$g$}

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%    \rput(0,-0.4){$f$}

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\newpage
\textbf{Change of Base point}\\

Let $x_0 ,x_1 \in X$ and $\rho$ be a path from $x_0 $ to $x_1$.\\
Given $\alp : (I^{n} , \partial I^{n} ) \rightarrow (X,x_0 )$,
define $\Phi : I^{n} \times I \rightarrow X$ as an extension of a
map $\phi : J=I^{n} \times 0 \cup \partial I^{n} \times I
\rightarrow X$ defined by $\phi|_{I^{n} \times 0} = \alp$ and
$\phi|_{\partial I^{n} \times \{t\}}=\rho(t)$.\\
(Note that $J$ is a strong deformation retract of $I^{n+1}$ and
hence any map defined on $J$ has an extension on $I^{n+1}$.)
%functorial property
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} }


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Define $\phi_{\rho} : \pi_{n} (X,x_0 ) \rightarrow\pi_{n} (X,x_1
)\\ \hspace*{7em}[\alp]\hspace{0.7em} \mapsto [\Phi |_{I^{n}\times\{1\}}]$\\

1. independent of choice of $\Phi$:\\

%independent of choice of Phi (1)
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\begin{center}
\begin{pspicture}(4,0.9)%
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\rput(-1,0.5){Let} \rput(5,0.5){be two extensions.}

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\psline{->}(1,0)(1,0.5)%
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\rput(0.5,-0.2){$\alp$}%

\rput(0.5,0.5){$\Phi$}%

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\pspolygon[fillstyle=solid,fillcolor=white,linewidth=0.4mm](0,0)(1,0)(1,1)(0,1)(0,0)%
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\psline{->}(1,0)(1,0.5)%
\rput(1.3,0.5){$\rho$}%
\rput(0.5,-0.2){$\alp$}%

\rput(0.5,0.5){$\Psi$}%
}}
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Define a homotopy $H$ between $\Phi$ and $\Psi$ as an extension of a map :\\
$I^{n}\times I\times 0 \cup \partial (I^{n}\times I)\times I
\rightarrow X$ given by the following pictures.\\

%independent of choice of Phi(2)
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\rput(0.5,-0.2){$\rho$}%
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\rput(-0.2,0.2){$\alp$}%

\rput(0.65,0.15){$\Phi$}%

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\pspolygon[fillstyle=solid,fillcolor=gray,linewidth=0.4mm](0,0)(1,0)(1.3,0.3)(0.3,0.3)(0,0)%
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\rput(0.65,0.15){$\Psi$}%

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\psline(0,0.4)(1,0.4)%
\psline(0,0.6)(1,0.6)%
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%\rput(0.65,0.15){$\Phi$}%

\rput(-0.9,0.8){$\longrightarrow$}%
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\psline(1,0)(1,1)%
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Then $H|_{I^{n}\times I\times\{1\}}$ gives a homotopy between $\Phi|_{I^{n}\times 1}$ and $\Psi|_{I^{n}\times 1}$.\\
$\therefore \Phi |_{I^{n}\times 1} \sim \Psi |_{I^{n}\times 1}$\\

2. $\phi_{\rho}$ is well-defined i.e., $\alp \underset{H}{\sim}
\alp ' \Rightarrow \Phi |_{I^{n}\times 1} \sim \Phi '
|_{I^{n}\times 1}$\\

%well-defined(1)
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\rput(0.5,0.5){$\Phi '$}%
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%well-defined(2)
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\rput(0.65,0.15){$\Phi$}%

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\psline(1,0)(1,1)%
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\psline(0.3,0.5)(1.3,0.5)%
\psline(0.3,0.7)(1.3,0.7)%
\psline(0.3,0.9)(1.3,0.9)%
\psline(0.3,1.1)(1.3,1.1)%
\psline{->}(0.5,-0.1)(0.5,0.5)%
\rput(0.5,-0.2){$\rho$}%
\psline(0.3,1)(0.3,1.3)

 }}
\end{pspicture}
\end{center}

3. $\phi_{\rho}$ depends only on the homopoty class of $\rho$(rel
$\partial$) i.e., $\rho \underset{H}{\sim}\rho ' \Rightarrow
\phi_{\rho} = \phi_{\rho '}$, \\so that we can write as
$\phi_{[\rho]}$.\\

%homotopy(1)
\psset{unit=1.5cm}
\begin{center}
\begin{pspicture}(4,0.9)%
%\psgrid[gridwidth=0.1pt,subgridwidth=0.1pt,gridcolor=red,subgridcolor=green]

\pspolygon[fillstyle=solid,fillcolor=white,linewidth=0.4mm](0,0)(1,0)(1,1)(0,1)(0,0)%
\psline{->}(0,0)(0,0.5)%
\rput(-0.3,0.5){$\rho$}%
\psline{->}(1,0)(1,0.5)%
\rput(1.3,0.5){$\rho$}%
\rput(0.5,-0.2){$\alp$}%

\rput(0.5,0.5){$\Phi$}%

\rput(2,0){\rput(0,0){
\pspolygon[fillstyle=solid,fillcolor=white,linewidth=0.4mm](0,0)(1,0)(1,1)(0,1)(0,0)%
\psline{->}(0,0)(0,0.5)%
\rput(-0.3,0.5){$\rho '$}%
\psline{->}(1,0)(1,0.5)%
\rput(1.3,0.5){$\rho '$}%
\rput(0.5,-0.2){$\alp $}%

\rput(0.5,0.5){$\Phi '$}%
}}
\end{pspicture}
\end{center}

%well-defined(2)
\psset{unit=2cm}
\begin{center}
\begin{pspicture}(5,1.7)%
%\psgrid[gridwidth=0.1pt,subgridwidth=0.1pt,gridcolor=red,subgridcolor=green]

\pspolygon[fillstyle=solid,fillcolor=gray,linewidth=0.4mm](0,0)(1,0)(1.3,0.3)(0.3,0.3)(0,0)%
\psline{->}(0,0)(0.5,0)%
\rput(0.5,-0.2){$\rho$}%
\psline{->}(0.3,0.3)(0.8,0.3)%
\rput(0.8,0.5){$\rho$}%
\rput(-0.2,0.2){$\alp$}%

\rput(0.65,0.15){$\Phi$}%

\psline(0,0)(0,1)%
\psline(0.3,0.3)(0.3,1.3)%
\psline(1,0)(1,1)%
\psline(1.3,0.3)(1.3,1.3)%

\rput(0,1){\rput(0,0){
\pspolygon[fillstyle=solid,fillcolor=gray,linewidth=0.4mm](0,0)(1,0)(1.3,0.3)(0.3,0.3)(0,0)%
\psline{->}(0,0)(0.5,0)%
\rput(0.5,-0.2){$\rho '$}%
\psline{->}(0.3,0.3)(0.8,0.3)%
\rput(0.8,0.5){$\rho '$}%
\rput(-0.2,0.2){$\alp $}%

\rput(0.65,0.15){$\Phi '$}%

}}

\rput(3,0){\rput(0,0){
\pspolygon[fillstyle=solid,fillcolor=gray,linewidth=0.4mm](0,0)(1,0)(1.3,0.3)(0.3,0.3)(0,0)%
\psline{->}(0,0)(0.5,0)%
%\rput(0.5,-0.2){$\rho$}%
\psline{->}(0.3,0.3)(0.8,0.3)%
%\rput(0.8,0.5){$\rho$}%
%\rput(-0.2,0.2){$\alp$}%
\psline(0,0.2)(1,0.2)%
\psline(0,0.4)(1,0.4)%
\psline(0,0.6)(1,0.6)%
\psline(0,0.8)(1,0.8)%

%\rput(0.65,0.15){$\Phi$}%

\rput(-0.9,0.8){$\longrightarrow$}%
\psline(0,0)(0,1)%

\psline(1,0)(1,1)%
\psline(1.3,0.3)(1.3,1.3)%

\psline[linecolor=blue](0,0)(0.3,0.3)%

\rput(0,1){\rput(0,0){
\pspolygon[fillstyle=solid,fillcolor=gray,linewidth=0.4mm](0,0)(1,0)(1.3,0.3)(0.3,0.3)(0,0)%
\psline{->}(0,0)(0.5,0)%
%\rput(0.5,-0.2){$\rho$}%
\psline{->}(0.3,0.3)(0.8,0.3)%
%\rput(0.8,0.5){$\rho$}%
%\rput(-0.2,0.2){$\alp$}%

%\rput(0.65,0.15){$\Phi$}%


}}


\rput(0.5,-0.2){$\rho$}%

\pspolygon[fillstyle=solid,fillcolor=blue,linewidth=0.4mm](0,0)(0,1)(1,1)(1,0)(0,0)%

\pspolygon[fillstyle=solid,fillcolor=blue,linewidth=0.4mm](1,0.3)(1.3,0.3)(1.3,1.3)(1,1)(1,0.3)%
\rput(0.5,0.6){$H$}%

\psline(0.5,1)(0.5,1.4)(0.7,1.3)%
\rput(0.5,1.6){$\rho '$}

 }}
\end{pspicture}
\end{center}

4. $\phi_{[\rho]}$ is a homomorphism i.e., $\phi_{[\rho]} [\alp
*\bet ] = \phi_{[\rho]} [\alp]\phi_{[\rho]}[\bet]$.\\

%homomorphism
\psset{unit=2cm}
\begin{center}
\begin{pspicture}(3,1.3)%
%\psgrid[gridwidth=0.1pt,subgridwidth=0.1pt,gridcolor=red,subgridcolor=green]
\rput(-0.2,0.3){$(t_2 , \cdots , t_{n} )$}%
\psline(0,0.2)(0.7,0.2)
\psline(0.5,0)(1.5,0)(1.5,1)(0.5,1)(0.5,0)%
\psline(1,0)(1,1)(1.3,1.3)%
\psline(1.5,0)(1.8,0.3)(1.8,1.3)(1.5,1)%
\psline(1.8,1.3)(0.8,1.3)(0.5,1)%
\psline[linestyle=dashed](0.8,1.3)(0.8,0.3)(1.8,0.3)%
\psline{->}(0.5,0)(1.2,0.7)%
\rput(1.95,0){$t_1 $}%
\psline[linewidth=0.4mm]{->}(0.5,0)(1.8,0)%

\rput(0.7,0.6){$\Phi _{\alp}$}%
\rput(1.34,0.6){$\Phi _{\bet}$}%

\rput(0.7,-0.2){$\alp$}%
\rput(1.3,-0.2){$\bet$}%

\psline[linecolor=blue,linewidth=0.4mm](0.5,1)(1.5,1)
\rput(0,1.2){$\phi _{\rho} (\alp)$}%
\psline[linecolor=blue](0.2,1.2)(0.8,1)
\rput(2.3,1.2){$\phi _{\rho} (\bet)$}%
\psline[linecolor=blue](2.1,1.2)(1.3,1)%

\psline[linecolor=blue](1,1)(2,0.8)%
\rput(3,0.8){$\phi _{\rho} (\alp *\bet ) = (\phi _{\rho} \alp
)(\phi _{\rho} \bet)$}

\end{pspicture}
\end{center}
\newpage
5. $\phi_{\rho * \rho '} = \phi_{\rho '} \cdot\phi_{\rho }$ where
$\rho(1) = \rho'(0)$:\\

%product
\psset{unit=3cm}
\begin{center}
\begin{pspicture}(3,1.3)%
%\psgrid[gridwidth=0.1pt,subgridwidth=0.1pt,gridcolor=red,subgridcolor=green]
\psframe(0,0)(1,0.5) \psframe(0,0.7)(1,1.2) %
\rput(0.5,0.3){$\Phi$} \rput(0.5,1){$\Phi '$}%
\rput(-0.1,0.2){$\rho$} \rput(1.1,0.2){$\rho$}%
\rput(-0.1,0.9){$\rho '$} \rput(1.1,0.9){$\rho '$}%
\psframe(1.5,0.1)(2.5,1.1)%
\rput(2,0.4){$\Phi$} \rput(2,0.9){$\Phi '$}%
\rput(1.4,0.3){$\rho$} \rput(2.6,0.3){$\rho$}%
\rput(1.4,0.9){$\rho '$} \rput(2.6,0.9){$\rho '$}%
\psline(1.5,0.6)(2.5,0.6)%

\rput(0.5,-0.1){$\alp$} \rput(0.5,0.6){$\phi_ {\rho} (\alp)$}
\rput(0.5,1.3){$\phi_{\rho '}\phi_ {\rho} (\alp)$}
\rput(2,-0.1){$\alp$} \rput(2,1.2){$\phi_{\rho *\rho '}(\alp)$}
\end{pspicture}
\end{center}


6. $\phi_{\rho}$ is an isomorphism and $\phi_{\overline{\rho}}=
\phi_{\rho}^{-1}:\\
\phi_{\rho}\cdot\phi_{\overline{\rho}} \overset{5}{=}
\phi_{\overline{\rho}*\rho} \overset{3}{=} \phi_{x_1} =
$id$_{\pi_{n}(X,x_1 )}$ and similarly
$\phi_{\overline{\rho}}\cdot\phi_{\rho} =$ id$_{\pi_{n}(X,x_0
)}$\\

\textbf{Remark.} If $\rho$ is a loop, then $\phi_{\rho} : \pi_{n}
(X, x_0 ) \rightarrow \pi_{n} (X, x_0 ).$ Hence, we have a right
action of
$\pi_{1} (X,x_0 )$ on $\pi_{n}(X,x_0 ).$\\

\textbf{¼÷Á¦ 12.} (1) $ X$ is n-simple.\\
$\hspace*{5em}\overset{def}{\Leftrightarrow}\exists x_0 \in X$
s.t. $\pi_1 (X,x_0 )$ action on $ \pi_{n} (X,x_0 )$ is trivial.\\
$\hspace*{5em}\Leftrightarrow \forall x \in X, \pi_1 (X,x )$
action on $ \pi_{n}
(X,x )$ is trivial.\\
$\hspace*{4.5em}$(2) $X$ is 1-simple. $\Leftrightarrow \pi_1 (X)$
is abelian.\\
\newpage
\textbf{Homotopy invariance}\\

1. $f \overset{H}{\simeq} g : X\rightarrow Y\hspace{1em}
\Rightarrow\hspace{1em}\pi_{n} (X,x)
\hspace{1em}\overset{g_{*}}{\longrightarrow}
\hspace{1em} \pi_{n}(Y,g(x))\\
\hspace*{13em} f_{*} \searrow \hspace{1em}\curvearrowright
\hspace{1em}\nearrow \phi_{\rho} : \cong\hspace{3em} $where $\rho(t) =H(x,t)\\
\hspace*{15em}\pi_{n} (Y,f(x))$\\

\begin{proof}

%functorial property
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\begin{center}
\begin{pspicture}(4,1)%
%\psgrid[gridwidth=0.1pt,subgridwidth=0.1pt,gridcolor=red,subgridcolor=green]
\psframe[linecolor=blue,linewidth=0.4mm](0,0.3)(0.4,0.7)%
\rput(0.2,0.2){$I^{n}$}%
\rput(-0.1,0.5){$I$}%
\rput(0.8,0.6){$\alp\times$ id}%
\psline{->}(0.5,0.5)(1.1,0.5)%
\pscurve[linecolor=blue]{->}(0.3,0.75)(0.8,0.8)(1.2,0.75)%
\pscurve[linecolor=blue]{->}(0.3,0.15)(0.8,0.1)(1.2,0.15)%

\rput(2.5,0.5){

    \psellipse(1,0)(0.7,0.5)
    \rput(1,0.5){$Y$}%

    \pscurve{->}(-0.3,0.3)(0,0.32)(0.6,0.3)
    \rput(0,0.4){$g$}

    \rput(0,0){$H$}

    \psline{->}(-0.3,-0.3)(0,-0.32)(0.6,-0.3)
    \rput(0,-0.4){$f$}

    %X
    \rput(-0.9,0.5){$X\times I$}%

    \psset{unit=1.3cm}{
    \rput(-1,0.6){
    \psellipse(-1,0)(0.9,0.5)%
    \psdot(-1.3,-0.3)%
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    \pscurve[linecolor=blue,linestyle=dashed](-0.9,-0.27)(-0.8,0)(-0.9,0.2)}}

    \psset{unit=1.3cm}{
    \rput(-1,-0.6){
    \psellipse(-1,0)(0.9,0.5)%
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    \pscurve[linecolor=blue,linestyle=dashed](-0.9,-0.27)(-0.8,0)(-0.9,0.2)}}

    \psline[linecolor=blue](-2.3,-0.9)(-2.3,0.3)%
    \psline[linecolor=blue](-1.5,-0.6)(-1.5,0.6)%
    \psline(-2.9,-0.6)(-2.9,0.6)%
    \psline(-1.1,-0.6)(-1.1,0.6)%


    %Y

    \psset{unit=1cm}{
    \rput(4,0.6){
    \psellipse(-1,0)(0.9,0.5)%
    \psdot(-1.3,-0.3)%
    %\rput(-1.6,-0.3){$g(x)$}
    \pscurve[linecolor=blue](-1.3,-0.3)(-0.9,-0.27)(-0.5,-0.1)(-0.5,0.1)(-0.9,0.2)(-1.3,-0.3)
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    \pscurve[linecolor=blue](2.7,0.3)(2.6,0)(2.7,-0.8)%
    \pscurve[linecolor=blue](3.5,0.6)(3.4,0)(3.5,-0.6)%
    \rput(2.3,0){$\rho(t)$}
    \psline(2.1,-0.6)(2.1,0.6)%
    \psline(3.9,-0.6)(3.9,0.6)%

    \psset{unit=1cm}{
    \rput(4,-0.6){
    \psellipse(-1,0)(0.9,0.5)%
    \psdot(-1.3,-0.3)%
    %\rput(-1.6,-0.3){$f(x)$}
    \pscurve[linecolor=blue](-1.3,-0.3)(-0.9,-0.27)(-0.5,-0.1)(-0.5,0.1)(-0.9,0.2)(-1.3,-0.3)
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    \pscurve[linecolor=blue,linestyle=dashed](-0.9,-0.27)(-0.8,0)(-0.9,0.2)}}
}
\end{pspicture}
\end{center}

\begin{floatingfigure}[l]{2cm}
\psset{unit=2cm}
\begin{pspicture}(1.5,0.7)%
\rput(0,0.6){$\Rightarrow$}
\psframe[linewidth=0.4mm](0.5,0.2)(1.1,0.8)%
\psline{->}(0.5,0.2)(0.5,0.5)%
\psline{->}(1.1,0.2)(1.1,0.5)%
\rput(0.2,0.5){$\rho$}%
\rput(1.4,0.5){$\rho$}%
\rput(0.8,0){$f\circ\alp$}%
\rput(0.8,1){$g\circ\alp$}%

\end{pspicture}
\end{floatingfigure}

$\hspace{3em}\Rightarrow\hspace{2em} [g\circ\alp ]=\phi_{\rho}[f\circ\alp]\\
\hspace*{7em}\parallel\hspace{3em}\parallel\\
\hspace*{6em}g_{*}[\alp]\hspace{2em}\phi_{\rho}\circ f_{*}[\alp]$

\end{proof}\\

2. $f : X\rightarrow Y$ is a homotopy equivalence. \\ $\Rightarrow
f_{*} : \pi_{n}(X,x)\rightarrow \pi_{n} (Y,y)$ is an
isomorphism.\\

\begin{proof}
exactly same as $\pi_1$-case:\\
\begin{floatingfigure}[l]{5.5cm}
\psset{unit=2cm}
\begin{pspicture}(3,1)%
\psellipse(0.7,0.3)(0.5,0.3)%
\psellipse(2.3,0.3)(0.5,0.3)%
\psline{->}(1.3,0.4)(1.7,0.4)%
\psline{<-}(1.3,0.2)(1.7,0.2)%
\rput(1.5,0.6){$f$}%
\rput(1.5,0.1){$g$}%
\psdot(0.9,0.5) \psdot(0.4,0.2)\psdot(2.3,0.3)
\psline(0.9,0.5)(0.4,0.2)
\rput(0.6,0.4){$\rho$}\rput(1.0,0.4){$x$}\rput(0.8,0.2){$g\circ
f(x)$}\rput(2.5,0.3){$f(x)$}
%\rput(0.8,0){$f\cdot\alp$}%
%\rput(0.8,1){$g\cdot\alp$}%
\end{pspicture}\end{floatingfigure}

$g\circ f\simeq $id$_{X}$ and $f\circ g\simeq $ id$_{Y}$\\
$g_* \circ f_* = (g\circ f)_* = \phi_{\rho}\circ id_* : \cong
\hspace{0.5em}\Rightarrow g_*$ is onto.\\
¸¶Âù°¡Áö·Î $f_* \circ g_* : \cong \hspace{0.5em}\Rightarrow g_*$ is 1-1.\\
$\therefore f_*$ is $\cong$. \\(µÎ¹ø° ÁÙÀÇ $f_*$¿Í ¼¼¹ø° ÁÙÀÇ
$f_*$´Â ´Ù¸£´Ù.)\end{proof}

\end{document}