crimboi[21] 模仿大赛

Rotman algebraic topology上的题, 个人认为出的非常好, 极大增进了读者对$\pi_1(S^1)$计算过程的理解.

题面

A group $G$ which is also a topology space is called a topological group if $\mu:G\times G\to G:(x,y)\mapsto xy$ and $i:G\to G:x\mapsto x^{-1}$ are continuous. It's then easy to demonstrate that $\mu_x:G\to G:y\mapsto xy$ is homeomorphism and if $H\triangleleft G$ then $G/H$ with quotient map $\nu:x\mapsto xH$ is topological group. Now if in addition $H$ is discrete and closed, and $G$ is simply connected, show that $\pi_1(G/H,1)\cong H$ by imitating the computation of $\pi_1(S^1)=\pi_1(\mathbb R/\mathbb Z)$

先想再看提示哦

提示

Recall how one computes $\pi_1(S^1)$. One firstly shows that for any compact convex $C\subset\mathbb R^m$ and $f:(C,t_0)\to (S^1,1)$ there exists a unique $\tilde f:(C,t_0)\to\mathbb (R,n)$ for some integer $n$ s.t. $e^{2\pi i\tilde f}=f$ by utilizing the fact that $f$ is uniformly continuous and thus dividing $[t_0,x]$ into $n$ intervals $[{n-k\over n}t_0+{k\over n}x,{n-k-1\over n}t_0+{k+1\over n}x]$ s.t. on any interval $f(x)^{-1}f(y)\in S^1-\lbrace-1\rbrace$, and notice that $(-{1\over 2},{1\over 2})\to S^1-\lbrace-1\rbrace:x\mapsto e^{2\pi ix}$ is a homeomorphism. Then for any $f:(I,\dot I)\to (S^1,1)$, setting $\tilde f:(I,0)\to (\mathbb R,0)$ and $\deg f=\tilde f(1)$ we can routinely prove $d:\pi_1(S^1)\to\mathbb Z:[f]\mapsto\deg f$ is a well-defined isomorphism. For this problem we basically replace the $e^{2\pi i\cdot}:\mathbb R\to S^1=\mathbb R/\mathbb Z$ above with $\nu:G\to G/H$: we firstly prove the unique existence of $\tilde f:C\to G$ s.t. $\nu\circ\tilde f=f$ by seeking a neighborhood $U$ of $1\in G$ s.t. restriction of $\nu$ on which is homeomorphism. Then by same trick we divide interval into $n$ parts s.t. on each part we have $f(x)^{-1}f(y)\in\nu(U)$; one familiar with analysis should see that this is equivalent to prove $f$ is uniformly continuous when viewing $G$ as a uniform space. Remaining parts are trivial.

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标答

Note that $(x,y)\mapsto x^{-1}y$ is continuous, and $H-\lbrace1\rbrace$ is closed (Why it is without $G$ being $T_2$?), so we can construct $U$ s.t. $\forall x,y\in U, x^{-1}y\notin H-\lbrace1\rbrace$. For uniformly continuous part just notice that $f(x)^{-1}f(y)$ is also continuous.