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120 lines
4.2 KiB
Markdown
120 lines
4.2 KiB
Markdown
# Math4201 Topology I (Lecture 39)
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## Separation Axioms
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### Embedding manifolds
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A $d$ dimensional manifold is the topological space satisfying the following three properties:
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1. Haudorff property ($\forall x,y\in X, \exists U,V\in \mathcal{T}_X$ such that $x\in U\cap V$ and $y\notin U\cap V$)
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2. Second countable property ($\exists \mathcal{B}\subseteq \mathcal{T}_X$ such that $\mathcal{B}$ is a basis for $X$ and $\mathcal{B}$ is countable)
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3. Local homeomorphism to $\mathbb{R}^d$ ($\forall x\in M$, there is a neighborhood $U$ of $x$ such that $U$ is homeomorphic to $\mathbb{R}^d$. $\varphi:U\to \mathbb{R}^d$ is bijective, continuous, and open)
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<details>
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<summary> Example of manifold</summary>
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$\mathbb{R}^d$ is a $d$-dimensional manifold. And any open subspace of $\mathbb{R}^d$ is also a manifold.
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---
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$S^1$ is a $1$-dimensional manifold.
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---
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$T=\mathbb{R}^2/\mathbb{Z}^2$ is a $2$-dimensional manifold.
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</details>
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Recall the [Urysohn metirzation theorem](./Math4201_L38.md/#urysohn-metirzation-theorem). Any normal and second countable space is metrizable.
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In the proof we saw that any such space can be embedded into $\mathbb{R}^\omega$ with the product topology.
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Question: What topological space can be embedded into $\mathbb{R}^n$ with the product topology?
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#### Theorem for embedding compact manifolds into $\mathbb{R}^n$
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Any $d$-dimensional (compact, this assumption makes the proof easier) manifold can be embedded into $\mathbb{R}^n$ with the product topology.
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#### Definition for support of function
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$\operatorname{supp}(f)=f^{-1}(\mathbb{R}-\{0\})$
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#### Definition for partition of unity
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Let $\{U_i\}_{i=1}^n$ be an open covering of $X$. A partition of unity for $X$ dominated by $\{U_i\}_{i=1}^n$ is a set of functions $\phi_i:X\to\mathbb{R}$ such that:
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1. $\operatorname{supp}(\phi_i)\subseteq U_i$
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2. $\sum_{i=1}^n \phi_i(x)=1$ for all $x\in X$
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#### Theorem for existence of partition of unity
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Let $X$ be a normal space and $\{U_i\}_{i=1}^n$ is an open covering of $X$. Then there is a partition of unity dominated by $\{U_i\}_{i=1}^n$.
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Proof uses Urysohn's lemma.
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<details>
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<summary>Proof for embedding compact manifolds</summary>
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Let $M$ be a compact manifold.
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For any point $x\in M$, there is an open neighborhood $U_x$ of $x$ such that $U_x$ is homeomorphic to $\mathbb{R}^d$.
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Let $\{U_x\}_{x\in M}$ be an open cover of $M$.
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Since $M$ is compact, $\{U_x\}_{x\in M}$ has a finite subcover.
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then $\{U_{x_i}\}_{i=1}^n$ is an open cover of $M$.
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Therefore $F_i:U_{x_i}\to \mathbb{R}^d$ is a homeomorphism.
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Since $M$ is compact and second countable, $M$ is normal.
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Then there sis a partition of unity $\{\phi_i:X\to \mathbb{R}\}_{i=1}^n$ for $M$ with support by $\{U_{x_i}\}_{i=1}^n$ dominated by $\{U_{x_i}\}_{i=1}^n$. Where
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- $\sum_{i=1}^n \phi_i(x)=1$
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- $\operatorname{supp}(\phi_i)\subseteq U_{x_i}$
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Define $\Psi:X\to \mathbb{R}^d$ as
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$$
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\Psi_i(x)=\begin{cases}
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\phi_i(x)F_i(x) & \text{if } x\in U_{x_i} \\
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0 & x\in X-\operatorname{supp}(\phi_i)
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\end{cases}
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$$
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Note that $\operatorname{supp}(\phi_i)\subseteq U_{x_i}$, this implies that $(X-\operatorname{supp}(\phi_i))\cup U_{x_i}=X$.
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$U_{x_i}\cap (X-\operatorname{supp}(\phi_i))= U_i-\operatorname{supp}(\phi_i)$
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In particualr, for any $x$ in the intersection, $\phi_i(x)=0\implies \phi_i(x)F_i(x)=0$.
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So on the overlap, $\phi_i(x)F_i(x)=0$ and hence $\Psi_i$ is well defined.
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Define $\Phi:X\to \mathbb{R}\times \dots \times \mathbb{R}\times \mathbb{R}^d\times \dots \times \mathbb{R}^d\cong \mathbb{R}^{(1+d)n}$ as
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$$
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\Phi(x)=(\phi_1(x),\dots,\phi_n(x),\Psi_1(x),\dots,\Psi_n(x))
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$$
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This is continuous because $\phi_i(x)$ and $\Psi_i(x)$ are continuous.
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Since $M$ is compact, we just need to show that $\Phi$ is one-to-one to verify that it is an embedding.
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Let $\Phi(x)=\Phi(x')$, then $\forall i,\phi_i(x)=\phi_i(x')$, and $\forall i,\Psi_i(x)=\Psi_i(x')$.
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Since $\sum_{i=1}^n \phi_i(x)=1$, $\exists i$ such that $\phi_i(x)\neq 0$, therefore $x\in U_{x_i}$.
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Since $\phi_i(x)=\phi_i(x')$, then $x'\in U_{x_i}$.
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This implies that $\Psi_i(x)=\Psi_i(x')$, $\phi_i(x)F_i(x)=\phi_i(x')F_i(x')$.
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So $F_i(x)=F_i(x')$ since $F_i$ is a homeomorphism.
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This implies that $x=x'$.
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So $\Phi$ is one-to-one, it is injective.
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Therefore $\Phi$ is an embedding.
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</details> |