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\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} $: $\pi(X,x_0) \rightarrow
\pi(Y,y_0)$.
\end{thm}
\begin{proof}
F(x,t)를 f 와 g 사이의 homotopy 라 하면,
G(x,t)=F($\alpha(x),t)$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

 $g\circ f = 1_X$ relative to $x_0, f\circ g = 1_Y$ 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{defn}
\textit{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}

따라서 위 따름정리 2는  $\,\,\pi_1$ 이  homotopy ( base point
$x_0$ 를 보존하는 ) equivalent 한 공간들에 대해 같다는 것을
보여준다.

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

$f(x)=\frac{x}{|x|}\,\, ,  g = inclusion$ 으로 주면 $f \circ g =
1_{Y}$ 이고

$g \circ f $ 는 $F(x,t)=(1-t)x + t\frac{x}{|x|}$ 에 의해 $1_{X}$
와 homotopic하다.\\\\{\bf 숙제 4.}

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}