# Math4201 Topology I (Lecture 39)

## Separation Axioms

### Embedding manifolds

A $d$ dimensional manifold is the topological space satisfying the following three properties:

1. Haudorff property ($\forall x,y\in X, \exists U,V\in \mathcal{T}_X$ such that $x\in U\cap V$ and $y\notin U\cap V$)
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)
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)

<details>
<summary> Example of manifold</summary>

$\mathbb{R}^d$ is a $d$-dimensional manifold. And any open subspace of $\mathbb{R}^d$ is also a manifold.

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$S^1$ is a $1$-dimensional manifold.

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$T=\mathbb{R}^2/\mathbb{Z}^2$ is a $2$-dimensional manifold.

</details>

Recall the [Urysohn metirzation theorem](./Math4201_L38.md/#urysohn-metirzation-theorem). Any normal and second countable space is metrizable.

In the proof we saw that any such space can be embedded into $\mathbb{R}^\omega$ with the product topology.

Question: What topological space can be embedded into $\mathbb{R}^n$ with the product topology?

#### Theorem for embedding compact manifolds into $\mathbb{R}^n$

Any $d$-dimensional (compact, this assumption makes the proof easier) manifold can be embedded into $\mathbb{R}^n$ with the product topology.

#### Definition for support of function

$\operatorname{supp}(f)=f^{-1}(\mathbb{R}-\{0\})$

#### Definition for partition of unity

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:

1. $\operatorname{supp}(\phi_i)\subseteq U_i$
2. $\sum_{i=1}^n \phi_i(x)=1$ for all $x\in X$

#### Theorem for existence of partition of unity

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 $\{U_i\}_{i=1}^n$.

Proof uses Urysohn's lemma.

<details>
<summary>Proof for embedding compact manifolds</summary>

Let $M$ be a compact manifold.

For any point $x\in M$, there is an open neighborhood $U_x$ of $x$ such that $U_x$ is homeomorphic to $\mathbb{R}^d$.

Let $\{U_x\}_{x\in M}$ be an open cover of $M$.

Since $M$ is compact, $\{U_x\}_{x\in M}$ has a finite subcover.

then $\{U_{x_i}\}_{i=1}^n$ is an open cover of $M$.

Therefore $F_i:U_{x_i}\to \mathbb{R}^d$ is a homeomorphism.

Since $M$ is compact and second countable, $M$ is normal.

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

- $\sum_{i=1}^n \phi_i(x)=1$
- $\operatorname{supp}(\phi_i)\subseteq U_{x_i}$

Define $\Psi:X\to \mathbb{R}^d$ as

$$
\Psi_i(x)=\begin{cases}
\phi_i(x)F_i(x) & \text{if } x\in U_{x_i} \\
0 & x\in X-\operatorname{supp}(\phi_i)
\end{cases}
$$

Note that $\operatorname{supp}(\phi_i)\subseteq U_{x_i}$, this implies that $(X-\operatorname{supp}(\phi_i))\cup U_{x_i}=X$.

$U_{x_i}\cap (X-\operatorname{supp}(\phi_i))= U_i-\operatorname{supp}(\phi_i)$

In particualr, for any $x$ in the intersection, $\phi_i(x)=0\implies \phi_i(x)F_i(x)=0$.

So on the overlap, $\phi_i(x)F_i(x)=0$ and hence $\Psi_i$ is well defined.

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

$$
\Phi(x)=(\phi_1(x),\dots,\phi_n(x),\Psi_1(x),\dots,\Psi_n(x))
$$

This is continuous because $\phi_i(x)$ and $\Psi_i(x)$ are continuous.

Since $M$ is compact, we just need to show that $\Phi$ is one-to-one to verify that it is an embedding.

Let $\Phi(x)=\Phi(x')$, then $\forall i,\phi_i(x)=\phi_i(x')$, and $\forall i,\Psi_i(x)=\Psi_i(x')$.

Since $\sum_{i=1}^n \phi_i(x)=1$, $\exists i$ such that $\phi_i(x)\neq 0$, therefore $x\in U_{x_i}$.

Since $\phi_i(x)=\phi_i(x')$, then $x'\in U_{x_i}$.

This implies that $\Psi_i(x)=\Psi_i(x')$, $\phi_i(x)F_i(x)=\phi_i(x')F_i(x')$.

So $F_i(x)=F_i(x')$ since $F_i$ is a homeomorphism.

This implies that $x=x'$.

So $\Phi$ is one-to-one, it is injective.

Therefore $\Phi$ is an embedding.

</details>