NoteNextra-origin/content/Math4202/Math4202_L6.md

# Math4202 Topology II (Lecture 5)

## Manifolds

### Imbedding of Manifolds

#### 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)_

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

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

$$
V_1\subseteq \overline{V_1}\subseteq U_1
$$

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 $A_1\subset V_1\subseteq \overline{V_1}\subseteq U_1$ (by [normal space proposition](https://notenextra.trance-0.com/Math4201/Math4201_L37/#proposition-of-normal-spaces)).

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

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

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

Repeat the above construction for $\{V_i\}_{i=1}^n$.

Then we have $\{W_i\}_{i=1}^n$ open and $W_i\subseteq \overline{W_i}\subseteq V_i\subseteq \overline{V_i}\subseteq U_i$.

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

**Step 2**:

Using [Urysohn's lemma](https://notenextra.trance-0.com/Math4201/Math4201_L37/#urysohn-lemma). To construct the partition of unity $\phi_i$.

> [!NOTE]
>
> Suppose
> 
> - $X$ be a normal space
> - $Z_1,Z_2\subseteq X$ are closed
> - $Z_1$ and $Z_2$ are disjoint
>
> Then:
>
> There exists $f:X\to[0,1]$ such that
>
> - $f(Z_1)=\{0\}$ and $f(Z_2)=\{1\}$
> - $f$ is continuous.

Since $W_1\subseteq \overline{W_1}\subseteq V_1\subseteq \overline{V_1}\subseteq U_1$,

Note that $\overline{W_1}$ and $X-V_1$ are two disjoint closed subsets of normal space $X$

Then we can have $f_1:X\to[0,1]$ such that $f_1(\overline{W_1})=\{0\}$ and $f_1(X-V_1)=\{1\}$.

Then we have the remaining list of function $f_2,\dots,f_n$.

Recall the definition for support of functions $\operatorname{supp}(f_i)=\overline{\{x\in X: f_i(x)>0\}}$. Since $f_i(x)=0$ for $x\in X-V_i$, we have $\operatorname{supp}(f_i)\subseteq \overline{V_i}$

Next we need to check $\sum_{i=1}^n f_i(x)=1$ for all $x\in X$.

Note that $\forall x\in X$, since $\bigcup_{i=1}^n W_i=X$, then there exists $i$ such that $x\in W_i$, thus $f_i(x)=1$.

And $\sum _{i=1}^n f_i(x)\geq 1$.

Then we do normalization for our value. Set $F(x)=\sum_{i=1}^n f_i(x)$.

Since $F(x)$ is sum of continuous functions, $F$ is continuous.

Then we define $\phi_i=f_i/F(x)$, since $F(x)\geq 1$, we are safe to divide by $F(x)$ and $\phi_i(x)$ is continuous.

And $\operatorname{supp}(\phi_i)=\operatorname{supp}(f_i)\subseteq \overline{V_i}\subseteq U_i$.

And $\sum_{i=1}^n \phi_i(x)=\frac{\sum_{i=1}^n f_i(x)}{F(x)}=\frac{F(x)}{F(x)}=1$ for all $x\in X$.

</details>

### Some Extension

#### Definition of paracompact space

Locally finite: $\forall x\in X$, $\exists$ open $x\in U$ such that $U$ only intersects finitely many open sets in $\mathcal{B}$.

A space $X$ is paracompact if every open cover $A$ of $X$ has a **locally finite** refinement $\mathcal{B}$ of $A$ that covers $X$.

## Algebraic Topology

Homeomorphism: A topological space $X$ is homeomorphic to a topological space $Y$ if there exists a homeomorphism $f:X\to Y$

- $f$ is continuous
- $f^{-1}$ is continuous
- $f$ is bijective

Equivalence relation: If $\sim$ satisfies the following:

- $\sim$ is reflexive $\forall x\in X, x\sim x$
- $\sim$ is symmetric $\forall x,y\in X, x\sim y\implies y\sim x$
- $\sim$ is transitive $\forall x,y,z\in X, x\sim y, y\sim z\implies x\sim z$

Homeomorphism is an equivalence relation.

- Reflexive: identity map
- Symmetric: inverse map is also homeomorphism
- Transitive: composition of homeomorphism is also homeomorphism

Main Question: classify topological space up to homeomorphism.

### Invariant in Mathematics

Quantities associated with topological spaces that don't change under homeomorphism.

We want to use some algebraic tools to classify topological spaces.