NoteNextra-origin/content/Math4202/Math4202_L6.md

Math4202 Topology II (Lecture 6)

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.

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

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).

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. 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.

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.