\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 $\alpha \sim \alpha'$에 대해 $f \circ \alpha \sim f \circ \alpha'$ 임을 보이자. $\alpha$ 와 $\alpha'$ 사이의 homotopy를 $F$라 하면,\\ $\hspace*{2em}(f \circ F)$(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($x_0) = y_0 \hspace{3em} \forall s \in I$\\ 이므로 $f \circ F$는 $f \circ \alpha $와 $f \circ \alpha'$ 사이에 homotopy를 준다.\\ 다음으로 $f_{\sharp}$이 homomorphism임을 보이자.\\ $f_{\sharp}([\alpha][\beta])=f_{\sharp}([\alpha * \beta])=[f \circ (\alpha * \beta)]$이고,\\ \begin{displaymath} (\alpha * \beta)(t)=\left\{\begin{array}{cl} \alpha(2t)& 0\leq t \leq \frac{1}{2} \\ \beta(2t-1) & \frac{1}{2}\leq t \leq 1 \end{array} \right. \end{displaymath} 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}(Functorial Property) \hspace*{1em}$1) \ f : (X,x_0)\rightarrow (Y,y_0)$, $g : (Y,y_0) \rightarrow (Z,z_0)$\\ $\hspace*{2em}\Rightarrow (g \circ f)_{\sharp} =g_{\sharp}\circ f_{\sharp}$.\\ \hspace*{1em}$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}\\ 따라서 $\pi_1$은 Category of topological space with base point에서 Category of group으로 가는 Functor이다.\\ {\bf Applications} \\ 1. If $f$ has an inverse $f^{-1}$ then $(f^{-1})_{\sharp}=(f_{\sharp})^{-1}$ by functorial property.\\ $\because f_\sharp \circ (f^{-1})_\sharp=(f\circ f^{-1})_\sharp=id_\sharp=id$. \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} 나중에 엄밀한 증명을 하겠지만, 예를 들어 $\mathbb{R}^2$와 $\mathbb{R}^2-\{0\}$의 fundamental group은 각각 $0$과 $\mathbb{Z}$이므로 이 두 공간은 homeomorphic하지 않다.\\ 2. Brouwer Fixed Point Theorem \begin{thm} 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} $f$가 fixed point를 갖지 않는다고 가정하자. \\ $g(x)$를 $f(x)$에서 출발하여 $x$를 지나는 반직선과 $\partial D^2$의 교점으로 정의하면, 함수 $g : D^2 \rightarrow \partial D^2$가 정의된다. 즉, \begin{displaymath} g(x) = f(x) + t(x-f(x)) \ \ where \ t>0 \ and\ \|f(x) + t(x-f(x)) \|=1. \end{displaymath} 이때, $g$는 연속이고, $\partial D^2=S^1$에서 $g(x)=x$이므로, 다음 diagram이 commute한다. \\ \hspace*{7.5em} $D^2\hspace{1ex}\overset{g}{\longrightarrow}\hspace{1ex}\partial D^2$\\ \hspace*{8.0em} $i\nwarrow \hspace{3ex} \nearrow id$\\ \hspace*{9.0em} $\hspace{0.5ex}\partial D^2$\\ 이에 대응하는 fundamental group들을 생각해 보면,\\ \hspace*{4.0em} $0=\pi_1(D^2,1)\overset{g_\sharp}{\longrightarrow}\pi_1(S^1,1)=\mathbb{Z}$\\ \hspace*{7.0em} $i_\sharp\nwarrow \hspace{6ex} \nearrow id_\sharp$\\ \hspace*{8.0em} $\pi_1(S^1,1)=\mathbb{Z}$\\ 이 되고, $0 = g_{\sharp}\circ i_{\sharp}=(g \circ i)_{\sharp}=id_{\sharp}=id : \mathbb{Z} \rightarrow \mathbb{Z}$ 이므로 이는 모순이다. \end{proof} \end{document}