NoteNextra-origin/content/Math4202/Math4202_L5.md

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)

Proof for Whithney's Embedding Theorem

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.

Then we will prove for the finite partition of unity.

Proof for finite partition of unity

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.