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\begin{document}
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  \section*{Contractible space and Brouwer fixed point}


  \begin{defn}\textit{
  A space X is contractible to $x_0 \in X$  if  \\$id_X \simeq c,$
  where c : X $ \rightarrow \{x_0\} \subset X $is a constant
  map.}\\
  \end{defn}

  {\bf Remark}\textit{  contractible $\Rightarrow$ path connected.}\\

{\bf 예} 1. $\textbf{R}^n$ is contractible.

      $\hspace{2em}$  F(x)=tx  로 주면 이는 id 와  {0} 간에 homotopy 를 준다.

  $\hspace{1em}$ 2. $D^n$ is contractible. 역시 F(x)=tx 로 주면 된다.

   $\hspace{1em}$ 3. Any space which is homeomorphic to $D^n$.

   $\hspace{1em}$ 4. A "tree" is contractible.(그림)

   $\hspace{1em}$ 5. $S^1$ is not contractible.\\

   {\bf 숙제 6.} X$\cong$Y  and X is contractible $\Rightarrow$ Y
   is also contractible.

   $\hspace{3em}$Can you show that $S^1$ is not
   contractible.($\pi_1$을 쓰지 않고 직접적으로

   $\hspace{3em}$ 보일 수 있나?)\\

{\bf Remark.}

1. \textit{X is contractible to $x_0 \in X \Rightarrow$ X is
contractible to any other point of X }:

 X가 contractible to $x_0$
이면  path connected 이므로  $\forall\,\,x_1 \in$ 에 대해
$x_0,x_1$ 사이에  path $\rho$가 존재한다. F 를 $id_X$ 와 $c_{x_0}$
간의 homotopy라 할  때 아래와 같이 정의된 H는 $id_X$와 $c_{x_0}$
사이에 원하는 homotopy를 준다.


$$H(x,t)=\left \{\begin{array}{1}F(x,2t)\,\,\,\,\,\,  0\leq t \leq \frac{1}{2}\\ \rho(2t-1)\,\,\,\,\,\,  \frac{1}{2} \leq t \leq 1.\end{array}
\right.$$\\


2.\textit{ X is contractible $\Leftrightarrow$ X $\simeq$
\{point\} :}

($\Rightarrow$ 증명) $\{x_0\} \hookrightarrow X \rightarrow
\{x_0\}$에서

$\hspace{7em}i \hspace{2em}c_{x_0}$

$c_{x_0} \circ i = id_{x_0}$ 이고 X 가 contractible 이므로 $ i
\circ c_{x_0} \simeq id_{X} $이다. 따라서 $X \simeq \{x_0\}$ 이다.

($\Leftarrow$ 증명) $X \simeq \{x_0\}$ 이므로 homotopy
equivalence f: $X \rightarrow \{x_0\},g : \{x_0\} \rightarrow X$
가 존재한다. 이 때 f는 constant map $c_{x_0}$ 가 되고 따라서 $g
\circ f$ 역시 constant map이 된다. 그런데 $g \circ f \simeq
id_{X}$ 이므로 X 는 contractible하다.

\begin{thm}
X is contrantible $\Rightarrow$  $\pi_1(X) = 0$.
\end{thm}
\begin{proof}
$X \simeq \{x_0\}$ 이므로 $\pi_1(X) \cong \pi_1(\{point\})=0$.
\end{proof}\\

{\bf Fact.} $\pi_1(S^1) \cong \textbf{Z}$.

 따라서 $S^1$ 은 contractible하지 않다.
 $\textbf{R}^2\backslash\{0\}$ 역시 마찬가지이다.

 \begin{thm}
 \textit{(Brouwer fixed point theorem)\\
 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}
Suppose not. Then $x \neq f(x), \forall x \in D^2$.

Define a function $g : D^2 \rightarrow \partial D^2$ as follows :

Let $g(x)$ be the point of intersection of the half line from
$f(x)$ to $x$ with $\partial D^2$.

i.e., $g(x) = f(x) + t(x-f(x))$ where $t$ is the unique solution
of $\|f(x) + t(x-f(x)) \|=1$. Then $g$ is continuous and $g$ is
$id$ on $\partial D^2=S^1\,\,\,$ i.e.,

$\hspace{2em}i \hspace{3em}g$

 $\,\,\,S^1 \hookrightarrow D^2 \rightarrow S^1$   and   $g \circ i =
 id$ 이므로 대응하는 fundamental group들을 생각해 보면,\\

 $\hspace{5em}i_{\sharp}\hspace{7em}g_{\sharp}$

 $\pi_1(S^1,1)\hspace{1em}\rightarrow\hspace{1em} \pi_1(D^2,1)\hspace{1em}\rightarrow\hspace{1em}\pi_1(S^1,1)$

$\hspace{2em} \textbf{Z}\hspace{6em}0 \hspace{7em}\textbf{Z}$\\

이 되고 이는 funtorial property에 의해 모순이다. 즉

$0 = g_{\sharp}\circ i_{\sharp}=(g \circ
i)_{\sharp}=id_{\sharp}=id : \textbf{Z} \rightarrow \textbf{Z}$
이므로 이는 모순이 된다.\end{proof}


\end{document}