《微分拓扑》复习笔记(4)-Brouwer不动点定理 "Differential Topology" review notes (4)-Brouwer's fixed point theorem
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Brouwer不动点定理

引理 1. 设$M$为紧致光滑带边流形, 则不存在保持边界的光滑映射$f:M\rightarrow \partial M.$

不然, 由Sard定理, 存在$f^{-1}(q)$为$1$维紧致带边子流形, 有偶数个端点. 而

$$ \partial (f^{-1}(q))=f^{-1}(q)\cap \partial M=\{q\}, $$

矛盾.

引理 2. 设$f:B^n\rightarrow B^n$为光滑映射, 则存在不动点$x\in B^n.$

不然, 取$g(x)=x+t\frac{x-f(x)}{|x-f(x)|},$ $t>0$使得$g(x)\in S^{n-1}$即可. 这是与前一引理矛盾的映射.

定理 3 (Brouwer不动点定理). 任何闭圆盘到自身的连续映射有不动点.

不然, 由光滑函数逼近连续映射, 构造矛盾.

一维光滑流形分类

引理 4. 设$f:I\rightarrow M,$ $g:J\rightarrow M$为$M$的两个弧长参数表示, 则$f(I)\cap g(J)$至多有两个连通分支. 若有一个连通分支, 弧长参数可延拓到$f(I)\cup g(J)$上. 若有两个连通分支, 则$M$微分同胚于$S^1.$

证明考虑$g^{-1}\circ f$的图像. 由此可证明分类定理.

定理 5. 任何连通的一维光滑流形微分同胚于圆周$S^1$或区间$\{0,1\}.$

Brouwer区域不变性定理

定理 6 (区域不变性). 若$U\subset \mathbb{R}^n$为开集, $f:U\rightarrow \mathbb{R}^n$为连续单射, 则$f(U)$为$\mathbb{R}^n$的开集.

定理 7 (维数不变性). 设$U\subset \mathbb{R}^n,$ $V\subset \mathbb{R}^m$为开集. 若$m\neq n,$ 则$U,V$不同胚.

设$n>m,$ 考虑映射$U\approx V\hookrightarrow \mathbb{R}^n$为连续单射, 与Brouwer区域不变性定理矛盾.

区域不变性定理由如下定理得证.

定理 8. 设$f:B^n\rightarrow \mathbb{R}^n$为连续单射, 则$f(0)$为$f(B^n)$内点.

引理 9 (零点稳定性). 设$f:B^n\rightarrow \mathbb{R}^n$为任意连续映射. 若连续映射$h:f(\mathbb{R}^n)\rightarrow \mathbb{R}^n$满足$|h(f(x))-x|\le 1,$ $\,\forall\,x\in B^n,$ 则$\,\exists\,x_0\in B^n,$ 使得$h(f(x_0))=0.$

取$g:=\mathrm{id}-h\circ f$即可, 由Brouwer不动点定理得证.

对于定理证明, 用反证法, 若$f(0)$不是内点, 则总存在$c\in \mathbb{R}^n\setminus f(B^n)$靠近$f(0).$ 由此构造无零点的映射逼近逆映射即可.

文章最后更新于 2023-06-12 15:19:22

Brouwer's fixed point theorem

Lemma 1. Assume $M$ is a compact smooth manifold with edges, then there is no boundary-preserving smooth map $f:M\rightarrow \partial M.$

Otherwise, according to Sard's theorem, there exists $f^{-1}(q)$ as a $1$-dimensional compact edged submanifold with an even number of endpoints. And

$$ \partial (f^{-1}(q))=f^{-1}(q)\cap \partial M=\{q\}, $$

Contradiction.

Lemma 2. Assume $f:B^n\rightarrow B^n$ is a smooth mapping, then there is a fixed point $x\in B^n.$

Otherwise, take $g(x)=x+t\frac{x-f(x)}{|x-f(x)|},$ $t>0$ makes $g(x)\in S^{n-1}$ suffice. This is a mapping that contradicts the previous lemma.

Theorem 3 (Brouwer’s fixed point theorem). Any continuous mapping of a closed disk onto itself has fixed points.

Otherwise, the smooth function approximates the continuous mapping and constructs a contradiction.

One-dimensional smooth manifold classification

Lemma 4. Let $f:I\rightarrow M,$ $g:J\rightarrow M$ be the two arc length parameter representations of $M$, then $f(I)\cap g(J)$ has at most two connected components. If there is one connected component, the arc length parameter can be extended to $f(I)\cup g(J)$. If there are two connected components, then $M$ is diffeomorphic to $S^1.$

Proof Consider the image of $g^{-1}\circ f$. From this we can prove the classification theorem.

Theorem 5. Any connected one-dimensional smooth manifold is diffeomorphic to the circumference $S^1$ or the interval $\{0,1\}.$

Brouwer's zone invariance theorem

Theorem 6 (regional invariance). If $U\subset \mathbb{R}^n$ is an open set and $f:U\rightarrow \mathbb{R}^n$ is a continuous injective, then $f(U)$ is the open set of $\mathbb{R}^n$.

Theorem 7 (Dimensionality invariance). Let $U\subset \mathbb{R}^n,$ $V\subset \mathbb{R}^m$ be open sets. If $m\neq n,$ then $U,V$ is not an embryo.

Let $n>m,$ consider the mapping $U\approx V\hookrightarrow \mathbb{R}^n$ to be a continuous injective, It contradicts Brouwer's zone invariance theorem.

The regional invariance theorem is proved by the following theorem.

Theorem 8. Assume $f:B^n\rightarrow \mathbb{R}^n$ is a continuous injective, then $f(0)$ is the interior point of $f(B^n)$.

Lemma 9 (Zero point stability). Let $f:B^n\rightarrow \mathbb{R}^n$ be any continuous mapping. If the continuous mapping $h:f(\mathbb{R}^n)\rightarrow \mathbb{R}^n$ satisfies $|h(f(x))-x|\le 1,$ $\,\forall\,x\in B^n,$, then $\,\exists\,x_0\in B^n,$ makes $h(f(x_0))=0.$

Just take $g:=\mathrm{id}-h\circ f$, which is proved by Brouwer’s fixed point theorem.

For theorem proof, use proof by contradiction. If $f(0)$ is not an interior point, Then there is always $c\in \mathbb{R}^n\setminus f(B^n)$ close to $f(0).$ From this, a zero-free mapping can be constructed to approximate the inverse mapping.

The article was last updated on 2023-06-12 15:19:22

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