Math4202 Topology II (Lecture 5)

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, 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 , $F:X\to F(X)$ 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.

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

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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.