\documentclass[12pt ]{article} \setlength{\textwidth}{14 true cm} \setlength{\textheight}{20 true cm} \usepackage{graphicx} \usepackage{hangul} \usepackage{amscd,amsmath} \usepackage{amsfonts} \usepackage{amssymb,theorem} \usepackage{longtable} \usepackage{floatflt} \usepackage{texdraw, epic, pstcol ,pstricks ,pst-3d, pst-poly,pst-grad,pst-node, pst-text} \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*{Application} \textit{{\bf $1\,\,.\,\, \pi_k(S^n)$=0, if $k < n$.}} \begin{proof} simplicial approximation 정리와 $S^n\setminus \{point\} \simeq \mathbb{R}^n$이 contractible이라는 것을 이용하자. $S^k$에서 $S^n$으로 가는 $\alp$가 onto만 아니라면, $S^n$상에서 $\alp$의 image가 아닌 점 $x_0$ 를 생각할 수 있고 $S^n\setminus \{x_0\} \simeq \mathbb{R}^n$이 contractible이므로 증명이 완성된다. 먼저 simplicial map 자체는 $p$-skeleton 을 $p$-skeleton 으로 보내므로 $\alp$의 simplicial approximation $\alp'$는 onto가 될 수 없고, 따라서 $\alp'$은 한 점으로 contractible 하고 $\alp \simeq \alp '$ 이므로 $\alp$ 도 contractible하다. 이 증명에서 base point는 vertex가 되도록 simplicial complex 구조를 $S^k$와 $S^n$에 적당히 준다. 앞절 정리의 Remark에 의해 $\alp(base\,\,point)=\alp'(base\,\,point)$라 두어도 상관없고, 앞절 정리 5증명에서 homotopy $F$가 base point를 fix한다고 가정하여도 좋다. \end{proof} {\bf $2\,\,.\,\, i:K^{k+1}\hookrightarrow K$ induces an isomorphism $i_{\ast}:\pi_{k}(|K^{k+1}|)\rightarrow \pi_k(|K|)$.}\\ \begin{proof} 먼저 simplicial approximation thoerem을 이용하면 epimorphism 이 되는 것은 위 증명에서 처럼 자명하다. 이제 1-1 임을 보이자. 1-1 임을 보이기 위해 이전에 보였던 다음 note를 이용하자.\\ {\bf Note.}\textit{ $\alp:S^{k}\rightarrow X$ represent a zero element in $\pi_{k}(X)$ if and only if} \textit{$\hspace{3em}\exists\,\,extension\,\,\overline{\alp}:B^{k+1}\rightarrow X.$}\\ 어떤 $\{\alp\} \in \pi_k(|K^{k+1}|)$ 가 $i_{\ast}$에 의해 identity 로 간다면, 위의 note 에 의해 extension $\overline{\alp}$ 가 존재하고 다음 diagram 이 성립한다.여기서 처음부터 $\alp$를 simplicial이라 가정해도 상관없다.\\ $\,\,\,|K^{k+1}|\hspace{1em}\rightarrow\hspace{1em} |K|\hspace{4em}|K^{k+1}|\hspace{1em}\rightarrow\hspace{1em} |K| $ $\alp\uparrow\hspace{7.5em}\Rightarrow \hspace{3.5em}\alp\uparrow\hspace{4.5em}\uparrow \exists\,\overline{\alp}$ $\,\,\,S^{k}\hspace{11.5em}S^{k}\hspace{2.0em}\subset\hspace{1.8em} B^{k+1}$\\ 여기서 $\overline{\alp}$의 simplicial approximation $\overline{\alp}'$를 잡으면 $\,\,\,K^{k+1}\hspace{1em}\rightarrow\hspace{1em} K\hspace{4em}$ $\hspace{1em}\exists\,\overline{\alp'}\nwarrow$ $\,\,\,\hspace{4.5em}B^{k+1}$ 가 되고 이 때 앞절 정리의 remark에 의해서 $\{\overline{\alp}'|_{S^{k}}\}=\{\alp\}$ 라 두어도 된다. 그런데 $|K^{k+1}|$에서 $\{\overline{\alp}'|_{S^{k}}\}=0$ 이므로 $\{\alp\}=0$ 이다. 따라서 1-1임을 보였다.\\ \textit{{\bf 3. Edge-path group}} $v_{0} \in V(K)$에 대해서\\ $\Omega_s(K,v_0)$=the set of closed edge-paths based at $v_{0}$ $\hspace{4em}$=$\{v_0v_{i_1}\cdot\cdot\cdot v_{i_k}v_0\,\,|\,\,v_i\in V(K)$ and $\{v_0,v_{i_1}\},\cdot\cdot\cdot,\{v_{i_k},v_0\}$ : 1-simplices in $K\}$\\ 이제 $\Omega_s(K,v_0)$에 equivalence relation $\overset{s}{\sim}$를 다음 3가지 equivalence에 의해 generate된 것으로 주자. $(1)\,\,\cdot\cdot\cdot v_iv_i \cdot \cdot \,\cdot \overset{s}{\sim} \cdot \cdot \cdot v_i\cdot\cdot\,\cdot$ $(2)\,\,\cdot\cdot\cdot v_iv_jv_i\cdot\cdot\,\cdot\overset{s}{\sim}\cdot\cdot\cdot v_i\cdot\cdot\,\cdot$ $(3)\,\,\cdot\cdot\cdot v_iv_jv_k\cdot\cdot\,\cdot\overset{s}{\sim}\cdot\cdot\cdot v_iv_k\cdot\cdot\,\cdot$ if $$ is a 2-simplex in $K$\\ 이 때 $\Omega_s(K,v_0)/{\overset{s}{\sim}}:=E(K,v_0)$를 $(K,v_0)$의 edge path group이라고 한다. 이 때 group operation은 juxtaposition 이다.\\ $E(K,v_0)$ 가 group이 됨을 보이자. 먼저 $((v_0v_{i_1}\cdot\cdot\, v_0)(v_0v_{j_1}\cdot\cdot\, v_0))(v_0v_{k_1}\cdot\cdot\, v_0)=(v_0v_{i_1}\cdot\cdot\, v_0)((v_0v_{j_1}\cdot\cdot\, v_0)(v_0v_{k_1}\cdot\cdot\, v_0)) $ 는 위 세가지 equivalence 에 대해 성립하므로 결합법칙을 만족하고, $\{v_0\}\in E(K,v_0)$ 가 group operation의 identity 가 된다. 그리고$(v_0v_{i_1}\cdot\cdot\cdot v_{i_k}v_0)\in E(K,v_0)$에 대해 inverse는 $(v_0v_{i_k}\cdot\cdot\cdot v_{i_1}v_0)$가 된다. 따라서 $E(K,v_0)$ 는 group 이 된다. \begin{thm} $E(K,v_0)\cong \pi_1(|K|,v_0)$ \end{thm} \begin{proof} 두 group사이의 isomorphism을 찾기 위해 먼저 다음 함수를 생각하자. $\hspace{3em}\phi\,\,:\,\,\hspace{1em}\Omega_s(K,v_0)\hspace{1em}\rightarrow \hspace{1em}\Omega(|K|,v_0)/\sim$ $\hspace{5em}v_0v_{i_1}\cdot\cdot\cdot v_{i_{k-1}}v_0\hspace{1em}\mapsto\hspace{2em} \{\alp\}$\\ where $\alp:I\rightarrow |K|$ is a piecewise linear map representing simplicial loop corresponding to $v_{0}, v_{i_{1}}, \cdots v_{i_{k}}, v_{0}$.\\ 이 $\phi$에 의해 induce되는 $\phi_{\sharp}:\Omega_s/{\overset{s}{\sim}}\rightarrow \Omega/{\sim}$을 얻을 수 있고, 이 때 다음 네 가지를 보이자.\\ $(1)\,\,\phi_{\sharp}$ is well defined : 첫번째 equivalence relation $\overset{s}{\sim}$에 대해 $\phi_{\sharp}(v_0\cdot\cdot v_iv_i\cdot\cdot v_0)=\phi_{\sharp}(v_0\cdot\cdot v_i\cdot\cdot v_0) $ 임을 알 수 있고, 나머지 두개에 대해서도 마찬가지로 확인할 수 있다.\\ $(2)\,\,\phi_{\sharp}$ is a homomorphism : $\phi(v_0v_{i_1}\cdot\cdot v_0)=\alp\,$ 와 $\,\phi(v_0v_{j_1}\cdot\cdot v_0)=\beta$ 에 대해 $\phi((v_0v_{i_1}\cdot\cdot v_0)\cdot(v_0v_{j_1}\cdot\cdot v_0))$ 은 juxtaposition에 의해 $\alp$와 $\beta$의 juxtaposition으로 가고 이는 $\phi((v_0v_{i_1}\cdot\cdot v_0))\cdot \phi((v_0v_{j_1}\cdot\cdot v_0))$ 와 같다.\\\\ $(3)\,\,\phi_{\sharp}$ is onto : 임의의 $\alp\in \pi_1(|K|,v_0)$에 대해 $\alp$의 simplicial approximation $\overline{\alp}$ 가 존재한다. $\phi_{\sharp}(\overline{\alp})=\alp$ 가 되어 onto이다.\\ $(4)\,\,\phi_{\sharp}$ is 1-1 : $\alp\,,\,\beta\,\in \Omega_{s}$에 대해 $\phi_{\sharp}(\alp)\sim\phi_{\sharp}(\beta)$이라고 가정하고 $\alp\overset{s}{\sim}\beta$ 임을 보이자. 먼저 $\alp,\beta$ 에 대해 각각 다음과 같이 $\alp',\beta'$을 잡자. $\phi_{\sharp}(\alp)$와 $\phi_{\sharp}(\beta)$ 사이의 homotopy $F\,,\,F(0,t)=\alp(t),F(1,t)=\beta(t)$에 대해 $F$의 simplicial approximation($G$라 두자)이 존재하도록 다음과 같이 subdivision한다. \begin{center} \psset{unit=1.5cm} \begin{pspicture}(0,-1.0)(6.0,2.5) %\psgrid[gridwidth=0.1pt,subgridwidth=0.1pt,gridcolor=red,subgridcolor=green] \psline{|-|}(0,0)(0.4,0)\psline{|-|}(0.4,0)(0.6,0)\psline{|-|}(0.6,0)(1.1,0) \psline{|-|}(1.1,0)(1.4,0)\psline{|-|}(1.4,0)(1.6,0)\psline{|-|}(1.6,0)(2.0,0) \psline{|-|}(0,2.0)(0.2,2.0)\psline{|-|}(0.2,2.0)(0.5,2.0)\psline{|-|}(0.5,2.0)(0.9,2.0) \psline{|-|}(0.9,2.0)(1.4,2.0)\psline{|-|}(1.4,2.0)(1.5,2.0)\psline{|-|}(1.5,2.0)(1.7,2.0) \psline{|-|}(1.7,2.0)(2.0,2.0)\psline(0,0)(0,2.0)\psline(2.0,0)(2.0,2.0) \rput(1.0,-0.3){$\alp$} \rput(1.0,2.3){$\bet$} \rput(2.5,0){\psline{|-|}(0,0)(0.4,0)\psline{|-|}(0.4,0)(0.6,0)\psline{|-|}(0.6,0)(1.1,0) \psline{|-|}(1.1,0)(1.4,0)\psline{|-|}(1.4,0)(1.6,0)\psline{|-|}(1.6,0)(2.0,0) \psline{|-|}(0,2.0)(0.2,2.0)\psline{|-|}(0.2,2.0)(0.5,2.0)\psline{|-|}(0.5,2.0)(0.9,2.0) \psline{|-|}(0.9,2.0)(1.4,2.0)\psline{|-|}(1.4,2.0)(1.5,2.0)\psline{|-|}(1.5,2.0)(1.7,2.0) \psline{|-|}(1.7,2.0)(2.0,2.0)\psline(0,0)(0,2.0)\psline(2.0,0)(2.0,2.0) \rput(1.0,-0.3){$\alp'$} \rput(1.0,2.3){$\bet'$} \psline(0.4,0)(0.4,2.0)\psline(0.6,0)(0.6,2.0)\psline(1.1,0)(1.1,2.0)\psline(1.4,0)(1.4,2.0) \psline(1.6,0)(1.6,2.0)\psline(0.2,0)(0.2,2.0)\psline(0.5,0)(0.5,2.0)\psline(0.9,0)(0.9,2.0) \psline(1.5,0)(1.5,2.0)\psline(1.7,0)(1.7,2.0) \psline(0.2,0)(0.2,2.0)\psline(0.8,0)(0.8,2.0)\psline(1.8,0)(1.8,2.0) \psline(0,0.2)(2.0,0.2)\psline(0,0.4)(2.0,0.4)\psline(0,0.6)(2.0,0.6) \psline(0,0.8)(2.0,0.8)\psline(0,1.0)(2.0,1.0)\psline(0,1.2)(2.0,1.2) \psline(0,1.4)(2.0,1.4)\psline(0,1.6)(2.0,1.6)\psline(0,1.8)(2.0,1.8)} \rput(4.7,1.0){$\overset{G}{\longrightarrow}$} \rput(5.0,0){\pspolygon(0,1.0)(0.3,0.4)(0.7,0.2)(1.1,0.5)(1.5,0.9)(1.7,1.4)(1.4,1.7)(1.0,1.7) (0.6,1.5)(0.2,1.3)(0,1.0)(0.4,0.8)(0.8,0.5)(1.2,1.0)(0.7,1.4)(0.4,1.2)(0,1.0)} \end{pspicture} \end{center} 여기서 $G(0,t)=\alp'(t),G(1,t)=\beta'(t)$ 으로 두었을 때, $\alp \overset{s}{\sim} \alp' , \bet \overset{s}{\sim} \bet'$라 가정하여도 무방하다. 왜냐하면, $\alp$의 한 소구간이 1-simplex $$로 갔다면, 이것의 subdivision은 예컨대 $,,\cdots $로 보내면 $\alp \overset{s}{\sim} \alp'$이고 $\phi(\alp) = \phi(\alp')$이 된다. $\phi(\alp')$와 $\phi(\bet')$사이의 homotopy $F$의 simplicial approximation $G$는 $\alp' , \bet'$가 이미 simplicial 이므로 $G_{0}=\alp'$이고 $G_{1}=\bet'$라고 가정해도 좋다. 따라서 $\alp'\overset{s}{\sim}\beta'$를 보이면 $\alp\overset{s}{\sim}\beta$임을 보일 수 있다. \begin{center} \psset{unit=2.5cm} \begin{pspicture}(-0.5,-0.5)(2.5,2.5) %\psgrid[gridwidth=0.1pt,subgridwidth=0.1pt,gridcolor=red,subgridcolor=green] \pspolygon(0,0)(2.0,0)(2.0,2.0)(0,2.0)(0,0) \psline(0.2,0)(0.2,2.0)\psline(0.4,0)(0.4,2.0)\psline(0.6,0)(0.6,2.0) \psline(0.8,0)(0.8,2.0)\psline(1.0,0)(1.0,2.0)\psline(1.2,0)(1.2,2.0) \psline(1.4,0)(1.4,2.0)\psline(1.6,0)(1.6,2.0)\psline(1.8,0)(1.8,2.0) \psline(0,0.2)(2.0,0.2)\psline(0,0.4)(2.0,0.4)\psline(0,0.6)(2.0,0.6) \psline(0,0.8)(2.0,0.8)\psline(0,1.0)(2.0,1.0)\psline(0,1.2)(2.0,1.2) \psline(0,1.4)(2.0,1.4)\psline(0,1.6)(2.0,1.6)\psline(0,1.8)(2.0,1.8) \psline(0,0)(2.0,2.0)\psline(0,0.2)(1.8,2.0)\psline(0,0.4)(1.6,2.0) \psline(0,0.6)(1.4,2.0)\psline(0,0.8)(1.2,2.0)\psline(0,1.0)(1.0,2.0) \psline(0,1.2)(0.8,2.0)\psline(0,1.4)(0.6,2.0)\psline(0,1.6)(0.4,2.0) \psline(0,1.8)(0.2,2.0)\psline(0.2,0)(2.0,1.8)\psline(0.4,0)(2.0,1.6) \psline(0.6,0)(2.0,1.4)\psline(0.8,0)(2.0,1.2)\psline(1.0,0)(2.0,1.0) \psline(1.2,0)(2.0,0.8)\psline(1.4,0)(2.0,0.6)\psline(1.6,0)(2.0,0.4) \psline(1.8,0)(2.0,0.2) \pspolygon[fillstyle=solid,fillcolor=lightgray](1.8,0)(2.0,0)(2.0,0.2)(1.8,0) \rput(1.0,-0.2){$\alp'$} \rput(1.0,2.2){$\bet'$} \rput(-0.3,1.0){$v_0$} \rput(2.2,1.0){$v_0$} \end{pspicture} \end{center} 위 그림에서 빗금친 삼각형부터 하나씩 $\overset{s}{\sim}$을 적용하면, $\alp'\overset{s}{\sim}\beta'$ 임을 알 수 있다. \end{proof}\\ \textit{{\bf 4. Graph}} A 1-dimensional simplicial complex is called a graph.\\ A simply connected simplicial complex is called a tree.\\ \begin{center} \psset{unit=2.5cm} \begin{pspicture}(-0.5,-0.5)(5.5,2.5) %\psgrid[gridwidth=0.1pt,subgridwidth=0.1pt,gridcolor=red,subgridcolor=green] \pspolygon(0.5,1.0)(0.75,0.5)(1.25,1.0)(1.0,1.5)(0.5,1.0) \psline(0.5,1.0)(1.25,1.0) \pspolygon(1.25,1.0)(1.5,0.5)(1.5,1.25)(1.25,1.0) \pspolygon(0.5,1.0)(1.0,1.5)(0.5,1.75)(0.5,1.0) \psline(0.75,0.5)(0.5,0)\psline(0.5,1.0)(0.25,0.5)\psline(0.5,1.0)(0,1.20) \psline(1.0,1.5)(1.3,2.0)\psline(1.0,1.5)(1.6,1.75)\psline(1.5,0.5)(1.5,0) \psline(1.5,0.5)(2.0,0.4)\psline(1.5,1.25)(2.0,1.5) \rput(3.0,0){ \pscircle(0.4,1){0.4} \pscircle(1.6,1){0.4} \psline(0.8,1.0)(1.2,1.0)\psdots(0.8,1.0)\psdots(1.2,1.0) \psdots(0.3,1.4)\psdots(0.3,0.6)\psdots(1.7,1.4)\psdots(1.7,0.6)} \end{pspicture} \end{center} \begin{lem} A graph $K$ is a tree if and only if it is contractible. \end{lem} \begin{proof} ($\Leftarrow$) clear.\\ ($\Rightarrow$) Fix $v_{0} \in V(K). \forall v \in V(K)$, choose a path $\alp_{v}$ from $v$ to $v_{0}$.\\ Define $F : V \times I \rightarrow |K|$ by $F(v,t)=\alp_{v}(t)$.Then\\ For each edge $\sigma$, define $F:|\sigma| \times I \rightarrow |K|$ as the following picture.\\ \begin{center} \psset{unit=2.0cm} \begin{pspicture}(-0.5,-0.5)(3.5,2.5) %\psgrid[gridwidth=0.1pt,subgridwidth=0.1pt,gridcolor=red,subgridcolor=green] \pspolygon[fillstyle=solid,fillcolor=lightgray](0,0)(2.0,0)(2.0,2.0)(0,2.0)(0,0) \rput(-0.2,1.0){$I$} \rput(-0.1,-0.1){$v_{i}$} \rput(2.1,-0.1){$v_{i+1}$} \rput(1.0,-0.2){$\sigma$} \rput(1.8,1.0){$\alp_{v_{i+1}}$} \rput(0.2,1.0){$\alp_{v_{i}}$} \rput(1.0,2.2){$v_{0}$} \rput(2.5,1.0){$\overset{F}\longrightarrow$} \rput(3.2,1.0){$|K|$} \end{pspicture} \end{center} 여기서 $|K|$가 simply connected이고 이미 boundary condition을 알고 있으므로 $|\sigma| \times I$의 내부로 extend시킬 수 있음에 주의하자.\\ 그러면 $F|_{|\sigma| \times I}$가 연속이므로 $F:|K| \times I \rightarrow |K|$도 연속이고 $F_{0} = id , F_{1} = v_{0}$이므로 contractible인 것을 확인하였다. \end{proof} \begin{lem} Let $K$ be a connected simplicial complex. Then\\ (1) $K$ contains a maximal tree.\\ (2) Every maximal tree contains all the vertices of $K$.\\ \end{lem} \begin{proof} (1) Use Zorn's lemma.\\ Let ${\bf \mathcal{T}}$ be the collection of all the trees in $K$. Then ${\bf \mathcal{T}}$ is a partially ordered set with respect to inclusion.\\ Let $\{T_{j}\}$ be a simply ordered set of trees in $K$. And $T = \bigcup T_{j}$. Then $T$ is simply connected.\\ $\because$ For given $\alp : S^{1} \rightarrow T, \alp(S^{1})$ is compact. Hence it is contained $T_{j}$ for some $j$. Consequently, $\alp \simeq const.$ in $T_{j}$, hence in $T$.\\ By Zorn's lemma, there exists a maximal element in ${\bf \mathcal{T}}$.\\ (2) Let $T$ be the maximal tree. Suppose that $T$ does not contain all the vertices of $K$. Then $\exists v_{1}, v_{2} \in V(K)$ s.t $v_{1} \in T$ and $v_{2} \notin T$ and $\{ v_{1},v_{2} \}$ is a 1-simplex in $K$. \\ $T_{1} := T \cup \{ \{ v_{1},v_{2} \} , \{ v_{2} \} \}$. Then $|T|$ is a strong deformation retract of $|T_{1}|$. Hence $T_{1}$ is simply connected and hence a tree containing $T$ properly. This contradicts to the normality of $T$. \end{proof} {\bf Note} (1) In fact, given a tree $T$, $\exists$ a maximal tree containing $T$.\\ (2) A tree containing all the vertices of $K$ is maximal.\\ (1)의 증명은 lemma의 증명과 같은 것이고, (2)는 edge-path group을 참조하면 쉽게 확인할 수 있다.(exercises)\\ Let $T$ be a maximal tree in $K$.\\ Let $G$ be the group generated by the oriented edges $(v,v')$ of $K$ with the following relations.\\ \hspace*{2.0em} (a) $(v,v') \in T \Rightarrow (v,v')=1$\\ \hspace*{2.0em} (b) $(v,v')(v',v) =1$\\ \hspace*{2.0em} (c) $v_1 , v_2 , v_3$ are vertices of a 2-simplex in $K$.Then\\ \hspace*{4.0em} $(v_{1},v_{2})(v_{2},v_{3}) = (v_{1},v_{3})$\\ \begin{thm} $E(K,v_{0}) \cong G = F/R$ \end{thm} \begin{proof} \hspace*{2.0em}$\phi : E(K,v_0) \longrightarrow G$라고 정의하면, \\ $[v_{0}v_{i_{1}}v_{i_{2}} \cdots v_{i_{k}} v_{0}] \mapsto (v_{0}v_{i_{1}})(v_{i_{1}}v_{i_{2}}) \cdots (v_{i_{k}}v_{0})$\\ 이는 junxtaposition과 equivalence relation을 보존하므로 잘 정의된다.또\\ \hspace*{2.0em}$\psi : F \longrightarrow E(K,v_{0})$를 정의하자.\\ \hspace*{2.0em}$(v,v') \mapsto \alp_{v}vv'\alp_{v'}^{-1}$\\ 이 때 각 vertex $v$마다 $v_{0}$에서 $v$로 가는 simplicial path $\alp_v$를 $T$에서 선택하고 고정시키자. $T$가 모든 vertex를 다 포함하므로 $\psi$는 $R$의 원소들을 모두 0으로 보내므로 \\ $F/R \overset{\bar{\psi}}\longrightarrow E(K,v_{0})$를 induce한다.\\ $\phi \circ \bar{\psi} =id. \hspace{1.0em} \bar{\psi} \circ \phi = id.$임은 쉽게 알 수 있으므로 정리가 증명되었다. \end{proof} \begin{cor} Let $K$ be a finite connected simplicial complex. Then $\pi_{1}(|K|,v_{0})$ is finitely presented. \end{cor} \begin{proof} $\pi_{1}(|K|,v_{0}) = E = G$인데 edge의 갯수가 유한이므로 generator의 갯수도 유한이고, $G$의 relation이 유한이므로 relation의 갯수도 유한이다. 따라서 finitely presented이다. \end{proof} \begin{cor} Let $K$ be a connected graph. If $T$ is a maximal tree in $K$, then $E(K,v_{0})$ is a free group generated by the 1-simplices in $K-T$. \end{cor} \begin{proof} 이는 $G$의 정의에 의해서 당연하다. \begin{center} \psset{unit=1.0cm} \begin{pspicture}(-0.5,-0.5)(14.5,3.5) %\psgrid[gridwidth=0.1pt,subgridwidth=0.1pt,gridcolor=red,subgridcolor=green] \psline(0,2.0)(0.6,1.0) \psline(0.6,1.0)(0.6,0.4) \psline(0.6,0.4)(1.2,0.8) \psline(1.2,0.8)(1.3,0) \psline(1.3,0)(1.6,0.2) \psline(1.6,0.2)(1.8,0) \psline(1.2,0.8)(1.5,1.0) \psline[linecolor=red](1.5,1.0)(1.3,0) \psline(0.6,1.0)(1.3,2.0) \psline(1.3,2.0)(1.8,1.5) \psline(1.8,1.5)(1.8,0.9) \psline(1.8,0.9)(2.5,0.4) \psline[linecolor=red](1.8,0.9)(2.3,1.7) \psline(2.3,1.7)(1.8,1.5) \psline(2.3,1.7)(2.7,1.0) \psline(1.3,2.0)(2.2,2.5) \psline(2.2,2.5)(2.4,3.0) \psline(2.2,2.5)(2.5,2.0) \psline(2.5,2.0)(2.7,1.7) \psline(2.7,1.7)(3.4,1.8) \psline(3.4,1.8)(3.6,1.2) \psline[linecolor=red](3.6,1.2)(3.8,1.9) \psline(3.8,1.9)(3.4,1.8) \psline[linecolor=red](2.7,1.7)(3.2,2.8) \psline(3.2,2.8)(2.5,2.0) \psline(3.2,2.8)(3.5,2.5) \psline(3.5,2.5)(3.6,2.4) \psline(3.5,2.5)(3.7,2.9) \rput(5.0,0){ \psline(0,2.0)(0.6,1.0) \psline(0.6,1.0)(0.6,0.4) \psline(0.6,0.4)(1.2,0.8) \psline(1.2,0.8)(1.3,0) \psline(1.3,0)(1.6,0.2) \psline(1.6,0.2)(1.8,0) \psline(1.2,0.8)(1.5,1.0) \psline(0.6,1.0)(1.3,2.0) \psline(1.3,2.0)(1.8,1.5) \psline(1.8,1.5)(1.8,0.9) \psline(1.8,0.9)(2.5,0.4) \psline(2.3,1.7)(1.8,1.5) \psline(2.3,1.7)(2.7,1.0) \psline(1.3,2.0)(2.2,2.5) \psline(2.2,2.5)(2.4,3.0) \psline(2.2,2.5)(2.5,2.0) \psline(2.5,2.0)(2.7,1.7) \psline(2.7,1.7)(3.4,1.8) \psline(3.4,1.8)(3.6,1.2) \psline(3.8,1.9)(3.4,1.8) \psline(3.2,2.8)(2.5,2.0) \psline(3.2,2.8)(3.5,2.5) \psline(3.5,2.5)(3.6,2.4) \psline(3.5,2.5)(3.7,2.9) \rput(2.0,-0.3){$T$}} \rput(10.0,0){\psline[linecolor=red](1.5,1.0)(1.3,0) \psline[linecolor=red](1.8,0.9)(2.3,1.7) \psline[linecolor=red](3.6,1.2)(3.8,1.9) \psline[linecolor=red](2.7,1.7)(3.2,2.8) \rput(2.0,-0.3){$K-T$}} \end{pspicture} \end{center} \end{proof} {\bf Note} Let $K$ be a finite connected graph, $n_{1}$ be the number of 1-simplices in $K$ and $n_{0}$ be the number of 0-simplices in $K$. Then \\ The number of 1-simplices in $K-T = n_{1} - (n_{0} - 1) = 1-(n_{0}-n_{1}) = 1-\chi(K)$. \begin{thm} Let $p : \widetilde{X} \rightarrow X$ be a covering. If $X=|K|$ for some simplicial complex $K$, then the simplicial complex structure of $X$ can be lifted to $\widetilde{X}$ in such a way that $p$ becomes a simplicial map. \end{thm} \begin{proof} simplex의 fundamental group은 trivial이다. 따라서 lifting할 수 있고 $\widetilde{X}$에서도 locally 똑같이 할 수 있다.\\ {\bf 숙제 18} (Prove in detail.) \end{proof} \begin{cor} 1. Any subgroup of a free group is free.\\ 2. Let $F$ be a free group on $n$ generators and $F'$ be a subgroup of $F$ of index $m$. Then $F'$ is a free group in $1-m+mn$ generators. \end{cor} \begin{proof} 1. $\exists K$, a connected graph such that $\pi_{1}(|K|) = F$.\\ $F'