NoteNextra-origin/content/Math4201/Math4201_L39.md

Math4201 Topology I (Lecture 39)

Separation Axioms

Embedding manifolds

A d dimensional manifold is the topological space satisfying the following three properties:

1. Haudorff property (\forall x,y\in X, \exists U,V\in \mathcal{T}_X such that x\in U\cap V and y\notin U\cap V)
2. Second countable property (\exists \mathcal{B}\subseteq \mathcal{T}_X such that \mathcal{B} is a basis for X and \mathcal{B} is countable)
3. Local homeomorphism to \mathbb{R}^d (\forall x\in M, there is a neighborhood U of x such that U is homeomorphic to \mathbb{R}^d. \varphi:U\to \mathbb{R}^d is bijective, continuous, and open)

Example of manifold

\mathbb{R}^d is a $d$-dimensional manifold. And any open subspace of \mathbb{R}^d is also a manifold.

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S^1 is a $1$-dimensional manifold.

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T=\mathbb{R}^2/\mathbb{Z}^2 is a $2$-dimensional manifold.

Recall the Urysohn metirzation theorem. Any normal and second countable space is metrizable.

In the proof we saw that any such space can be embedded into \mathbb{R}^\omega with the product topology.

Question: What topological space can be embedded into \mathbb{R}^n with the product topology?

Theorem for embedding compact manifolds into \mathbb{R}^n

Any $d$-dimensional (compact, this assumption makes the proof easier) manifold can be embedded into \mathbb{R}^n with the product topology.

Definition for support of function

\operatorname{supp}(f)=f^{-1}(\mathbb{R}-\{0\})

Definition for partition of unity

Let \{U_i\}_{i=1}^n be an open covering of X. A partition of unity for X dominated by \{U_i\}_{i=1}^n is a set of functions \phi_i:X\to\mathbb{R} such that:

1. \operatorname{supp}(\phi_i)\subseteq U_i
2. \sum_{i=1}^n \phi_i(x)=1 for all x\in X

Theorem for existence of partition of unity

Let X be a normal space and \{U_i\}_{i=1}^n is an open covering of X. Then there is a partition of unity dominated by \{U_i\}_{i=1}^n.

Proof uses Urysohn's lemma.

Proof for embedding compact manifolds

Let M be a compact manifold.

For any point x\in M, there is an open neighborhood U_x of x such that U_x is homeomorphic to \mathbb{R}^d.

Let \{U_x\}_{x\in M} be an open cover of M.

Since M is compact, \{U_x\}_{x\in M} has a finite subcover.

then \{U_{x_i}\}_{i=1}^n is an open cover of M.

Therefore F_i:U_{x_i}\to \mathbb{R}^d is a homeomorphism.

Since M is compact and second countable, M is normal.

Then there sis a partition of unity \{\phi_i:X\to \mathbb{R}\}_{i=1}^n for M with support by \{U_{x_i}\}_{i=1}^n dominated by \{U_{x_i}\}_{i=1}^n. Where

• \sum_{i=1}^n \phi_i(x)=1
• \operatorname{supp}(\phi_i)\subseteq U_{x_i}

Define \Psi:X\to \mathbb{R}^d as

\Psi_i(x)=\begin{cases}
\phi_i(x)F_i(x) & \text{if } x\in U_{x_i} \\
0 & x\in X-\operatorname{supp}(\phi_i)
\end{cases}

Note that \operatorname{supp}(\phi_i)\subseteq U_{x_i}, this implies that (X-\operatorname{supp}(\phi_i))\cup U_{x_i}=X.

U_{x_i}\cap (X-\operatorname{supp}(\phi_i))= U_i-\operatorname{supp}(\phi_i)

In particualr, for any x in the intersection, \phi_i(x)=0\implies \phi_i(x)F_i(x)=0.

So on the overlap, \phi_i(x)F_i(x)=0 and hence \Psi_i is well defined.

Define \Phi:X\to \mathbb{R}\times \dots \times \mathbb{R}\times \mathbb{R}^d\times \dots \times \mathbb{R}^d\cong \mathbb{R}^{(1+d)n} as

\Phi(x)=(\phi_1(x),\dots,\phi_n(x),\Psi_1(x),\dots,\Psi_n(x))

This is continuous because \phi_i(x) and \Psi_i(x) are continuous.

Since M is compact, we just need to show that \Phi is one-to-one to verify that it is an embedding.

Let \Phi(x)=\Phi(x'), then \forall i,\phi_i(x)=\phi_i(x'), and \forall i,\Psi_i(x)=\Psi_i(x').

Since \sum_{i=1}^n \phi_i(x)=1, \exists i such that \phi_i(x)\neq 0, therefore x\in U_{x_i}.

Since \phi_i(x)=\phi_i(x'), then x'\in U_{x_i}.

This implies that \Psi_i(x)=\Psi_i(x'), \phi_i(x)F_i(x)=\phi_i(x')F_i(x').

So F_i(x)=F_i(x') since F_i is a homeomorphism.

This implies that x=x'.

So \Phi is one-to-one, it is injective.

Therefore \Phi is an embedding.