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\begin{document}
 \parindent=0cm
\section*{Homotopy invariance(preliminary version)}
\begin{thm}
$f,g:(X,x_0) \rightarrow (Y,y_0)$  and  $f \simeq g$ relative to
$x_0$.\\
$ \Rightarrow f_{\sharp}=g_{\sharp} $ : $\pi_1(X,x_0) \rightarrow
\pi_1(Y,y_0)$.
\end{thm}

\begin{proof}
\begin{figure}[htb]

\centerline{\includegraphics*[scale=0.4,clip=true]{homo1.eps}}

\end{figure}
$F$(t,s)를 $f$ 와 $g$ 사이의 homotopy 라 하면,
$G$(t,s)=$F(\alpha$(t),s)는 $f \circ \alpha$ 와 $g \circ \alpha$
사이의 homotopy를 준다. 따라서

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

\begin{cor}
$f:(X,x_0) \rightarrow (Y,y_0), g:(Y,y_0)\rightarrow (X,x_0)$ and
$g\circ f \simeq 1_X$ relative to $x_0$ , $f \circ g \simeq 1_Y$
relative to $y_0$.

$\Rightarrow f_{\sharp}=(g_{\sharp})^{-1} : \pi_1 (X,x_0)
\rightarrow \pi_1 (Y,y_0)$ is an isomorphism.
\end{cor}

\begin{proof}
$g \circ f \simeq 1_X$ relative to $x_0$ 이므로  $g_{\sharp} \circ
f_{\sharp} = (g \circ f)_{\sharp} = (id_X)_{\sharp} =
id$이다.\\

같은 방법으로 $f_{\sharp} \circ g_{\sharp} = (f \circ g)_{\sharp}
= (id_Y)_{\sharp} = id$이다.

\end{proof}

\begin{defn}
$X$ is homotopy equivalent to $Y$ (or $X$ has the same homotopy type as $Y$), denoted by $X \simeq Y$,\\
if  $\,\,\,\,\exists f$ : $X \rightarrow Y$ and $g$ : $Y
\rightarrow X$ such that $f\circ g \simeq 1_Y$ and $g\circ f
\simeq 1_X$.
\\In this case $f$ is called a homotopy equivalence.
\end{defn}

{\bf Example.} $\textbf{R}^2 \setminus \{0\} \simeq S^1.$\\

$\textbf{R}^2 \setminus \{0\}$를 $X$, $S^1$를 $Y$라고 하자.\\
$f(x)=\frac{x}{|x|}\,\, , g = inclusion$ 으로 주면 $f \circ g =
1_{Y}$ 인 것은 자명하다.

또한 $F(x,t)=(1-t)x + t\frac{x}{|x|}$ 이라고 두면 \\

$\hspace{2em}$  $F(x,0) = x$  $\hspace{2em}$  $F(x,1) =
\frac{x}{|x|} =f(x)$\\

따라서 $g \circ f \simeq id_X$ 이고 $X$는 $Y$와 same homotopy type을 갖는다.\\

{\bf 숙제 1.}

1. $\textbf{R}^n \setminus \{0\} \simeq S^{n-1}.$\\
2. $\simeq$ is an equivalence relation.\\
3. M\"{o}bius band 와 annulus는 homotopy type 이 같은가?\\
4. $T^2 \setminus \{point\} \simeq$ figure eight.\\


\end{document}