# Math4202 Topology II (Lecture 5)

## Manifolds

### Imbedding of Manifolds

> [!NOTE]
>
> Suppose $f: X \to Y$ is an injective continuous map, where $X$ and $Y$ are topological spaces. Let $Z$ be the image set $f(X)$, considered as a subspace of $Y$, then the function $f’: X \to Z$ obtained by restricting the range of f is bijective. If f happens to be a homeomorphism of X with Z, we say that the map $f: X \to Y$ is a topological imbedding, or simply imbedding, of X in Y.

Recall from last lecture

#### Whitney's Embedding Theorem

If $X$ is a compact $m$-manifold, then $X$ can be imbedded in $\mathbb{R}^N$ for some positive integer $N$.

_In general, $X$ is not required to be compact. And $N$ is not too big. For non compact $X$, $N\leq 2m+1$ and for compact $X$, $N\leq 2m$._

#### Definition for partition of unity

Let $\{U_i\}_{i=1}^n$ be a finite open cover of topological space $X$. An indexed family of **continuous** function $\phi_i:X\to[0,1]$ for $i=1,...,n$ is said to be a **partition of unity** dominated by $\{U_i\}_{i=1}^n$ if

1. $\operatorname{supp}(\phi_i)=\overline{\{x\in X: \phi_i(x)\neq 0\}}\subseteq U_i$ (the closure of points where $\phi_i(x)\neq 0$ is in $U_i$) for all $i=1,...,n$
2. $\sum_{i=1}^n \phi_i(x)=1$ for all $x\in X$ (partition of function to $1$)

#### Existence of finite partition of unity

Let $\{U_i\}_{i=1}^n$ be a finite open cover of a normal space $X$ (Every pair of closed sets in $X$ can be separated by two open sets in $X$).

Then there exists a partition of unity dominated by $\{U_i\}_{i=1}^n$.

_A more generalized version, If the space is paracompact, then there exists a partition of unity dominated by $\{U_i\}_{i\in I}$ with locally finite. (Theorem 41.7)_

<details>
<summary>Proof for Whithney's Embedding Theorem</summary>

Since $X$ is a $m$ compact manifold, $\forall x\in X$, there is an open neighborhood $U_x$ of $x$ such that $U_x$ is homeomorphic to $\mathbb{R}^m$. That means there exists $\varphi_i:U_x\to \varphi(U_x)\subseteq \mathbb{R}^m$.

Where $\{U_x\}_{x\in X}$ is an open cover of $X$. Since $X$ is compact, there is a finite subcover $\bigcup_{i=1}^k U_{x_i}=X$.

Apply the [existence of partition of unity](#existence-of-finite-partition-of-unity), we can find a partition of unity dominated by $\{U_{x_i}\}_{i=1}^k$. With family of functions $\phi_i:\mathbb{R}^d\to[0,1]$.

Define $h_i:X\to \mathbb{R}^m$ by

$$
h_i(x)=\begin{cases}
\phi_i(x)\varphi_i(x) & \text{if }x=x_i\\
0 & \text{otherwise}
\end{cases}
$$

We claim that $h_i$ is continuous using pasting lemma.

On $U_i$, $h_i=\phi_i\varphi_i$ is product of two continuous functions therefore continuous.

On $X-\operatorname{supp}(\phi_i)$, $h_i=0$ is continuous.

By pasting lemma, $h_i$ is continuous.

Define

$$
F: X\to (\mathbb{R}^m\times \mathbb{R})^n
$$

where $x\mapsto (h_1(x),\varphi_1(x),h_2(x),\varphi_2(x),\dots,h_n(x),\varphi_n(x))$

We want to show that $F$ is imbedding map.

**(a). $F$ is continuous**

since it is a product of continuous functions.

**(b). $F$ is injective**

that is, if $F(x_1)=F(x_2)$, then $x_1=x_2$.

By partition of unity, we have, 

$h_1(x_1)=h_1(x_2), h_2(x_1)=h_2(x_2), \dots, h_n(x_1)=h_n(x_2)$.

And $\varphi_1(x_1)=\varphi_1(x_2), \varphi_2(x_1)=\varphi_2(x_2), \dots, \varphi_n(x_1)=\varphi_n(x_2)$.

Because $\sum_{i=1}^n \varphi_i(x_1)=1$, therefore the exists $\varphi_i(x_1)=\varphi_i(x_2)>0$.

Therefore $x1,x_2\in \operatorname{supp}(\phi_i)\subseteq U_i$.

By definition of $h$, $h_i(x_1)=h_i(x_2)$, $\varphi_i(x_1)\phi_i(x_1)=\varphi_i(x_2)\phi_i(x_2)$.

Using cancellation, $\phi_i(x_1)=\phi_i(x_2)$.

Therefore $x_1=x_2$ since $\phi_i(x_1)=\phi_i(x_2)$ is a homeomorphism.

_In this proof, $\varphi$ ensures the imbedding is properly defined on the open sets_

**(c). $F$ is a homeomorphism.**

Note that by [Theorem 26.6 on Munkres](https://notenextra.trance-0.com/Math4201/Math4201_L25/#theorem-of-closed-maps-from-compact-and-hausdorff-spaces), $F:X\to F(X)$ is a bijective map from a compact space to a Hausdorff space, therefore $F$ is a closed map.

Since $F$ is continuous, then $F^{-1}(C)$ where $C$ is a closed set in $F(X)$, $F^{-1}(C)$ is closed in $X$.

Therefore $F$ is a homeomorphism.

</details>

Then we will prove for the finite partition of unity.

<details>
<summary>Proof for finite partition of unity</summary>

Some intuitions:

By definition for partition of unity, consider the sets $W_i,V_i$ defined as

$$
W_i=f^{-1}_i((\frac{1}{2n},1])\subseteq f^{-1}_i([\frac{1}{2n},1])\subseteq V_i=f^{-1}_i((0,1])\subseteq \operatorname{supp}(f_i)\subseteq U_i
$$

Note that $V_i$ is open and $\overline {V_i}\subseteq U_i$.

And $\bigcup_{i=1}^n V_i=X$.

and $W_i$ is open and $\overline{W_i}\subseteq V_i$.

And $\bigcup_{i=1}^n W_i=X$.

---

Step 1: $\exists$ V_i$ ope subsets $i=1,\dots,n$ such that $\overline{V_i}\subseteq U_i$, and $\bigcup_{i=1}^n V_i=X$.

For $i=1$, consider $A_1=X-(U_2\cup U_3\cup \dots \cup U_n)$. Therefore $A_1$ is closed, and $A_1\cup U_1=X$.

So $A_1\subseteq U_1$.

Note that $A_1$ and $X-U_1$ are disjoint closed subsets of $X$.

Since $X$ is normal, we can separate disjoint closed subsets $A_1$ and $X-U_1$.

So we have $A_1\subset V_1\subseteq \overline{V_1}\subseteq U_1$.

For $i=2$, note that $V_1\cup\left( \bigcup_{i=2}^n U_i\right)=X$,

Take $A_2=X-\left(V_1\cup\left( \bigcup_{i=3}^n U_i\right)\right)$ (skipping $U_2$).

Then we have $V_2\subseteq \overline{V_2}\subseteq U_2$.

For $i=j$, we have

$$
A_j=X-\left(\left(\bigcup_{i=1}^{j-1}V_i\right)\cup \left(\bigcup_{i=j+1}^n U_i\right)\right)
$$

Continue next lecture.
</details>
