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π1(S1)\pi_1(S^1)计算过程的理解.

题面

A group GG which is also a topology space is called a topological group if μ:G×GG:(x,y)xy\mu:G\times G\to G:(x,y)\mapsto xy and i:GG:xx1i:G\to G:x\mapsto x^{-1} are continuous. It's then easy to demonstrate that μx:GG:yxy\mu_x:G\to G:y\mapsto xy is homeomorphism and if HGH\triangleleft G then G/HG/H with quotient map ν:xxH\nu:x\mapsto xH is topological group. Now if in addition HH is discrete and closed, and GG is simply connected, show that π1(G/H,1)H\pi_1(G/H,1)\cong H by imitating the computation of π1(S1)=π1(R/Z)\pi_1(S^1)=\pi_1(\mathbb R/\mathbb Z)




先想再看提示哦

提示

Recall how one computes π1(S1)\pi_1(S^1). One firstly shows that for any compact convex CRmC\subset\mathbb R^m and f:(C,t0)(S1,1)f:(C,t_0)\to (S^1,1) there exists a unique f~:(C,t0)(R,n)\tilde f:(C,t_0)\to\mathbb (R,n) for some integer nn s.t. e2πif~=fe^{2\pi i\tilde f}=f by utilizing the fact that ff is uniformly continuous and thus dividing [t0,x][t_0,x] into nn intervals [nknt0+knx,nk1nt0+k+1nx][{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)1f(y)S1{1}f(x)^{-1}f(y)\in S^1-\lbrace-1\rbrace, and notice that (12,12)S1{1}:xe2πix(-{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,I˙)(S1,1)f:(I,\dot I)\to (S^1,1), setting f~:(I,0)(R,0)\tilde f:(I,0)\to (\mathbb R,0) and degf=f~(1)\deg f=\tilde f(1) we can routinely prove d:π1(S1)Z:[f]degfd:\pi_1(S^1)\to\mathbb Z:[f]\mapsto\deg f is a well-defined isomorphism. For this problem we basically replace the e2πi:RS1=R/Ze^{2\pi i\cdot}:\mathbb R\to S^1=\mathbb R/\mathbb Z above with ν:GG/H\nu:G\to G/H: we firstly prove the unique existence of f~:CG\tilde f:C\to G s.t. νf~=f\nu\circ\tilde f=f by seeking a neighborhood UU of 1G1\in G s.t. restriction of ν\nu on which is homeomorphism. Then by same trick we divide interval into nn parts s.t. on each part we have f(x)1f(y)ν(U)f(x)^{-1}f(y)\in\nu(U); one familiar with analysis should see that this is equivalent to prove ff is uniformly continuous when viewing GG as a uniform space. Remaining parts are trivial.




愿你无需看标答

标答

Note that (x,y)x1y(x,y)\mapsto x^{-1}y is continuous, and H{1}H-\lbrace1\rbrace is closed (Why it is without GG being T2T_2?), so we can construct UU s.t. x,yU,x1yH{1}\forall x,y\in U, x^{-1}y\notin H-\lbrace1\rbrace. For uniformly continuous part just notice that f(x)1f(y)f(x)^{-1}f(y) is also continuous.

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作者:HDD
链接:https://blog.hellholestudios.top/archives/1577
来源:Hell Hole Studios Blog
文章版权归作者所有,未经允许请勿转载。

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