NoteNextra-origin/content/Math4201/Math4201_L20.md

Math4201 Topology I (Lecture 20)

Quotient topology

More propositions

Proposition for quotient maps in restrictions

Let X,Y be topological spaces and p:X\to Y is surjective and open/closed. Let A\subseteq X be saturated by p, (p^{-1}(p(A))=A).

Then q: A\to p(A) given by the restriction of p is open/closed surjective map (In particular, it's a quotient map).

Proof

q is surjective and continuous. Now assume p is open and we will show that q is also open. Any open subspace of A is given as U\cap A where U is open in X. By definition, q(U\cap A)=p(U\cap A)=p(U)\cap p(A)

To see the second identity:

1. p(U\cap A)\subseteq p(U)\cap p(A)

\forall y\in p(U\cap A), y=p(x) with x\in U\cap A, since x\in A and x\in U, y=p(x)\in p(U)\cap p(A)

2. p(U)\cap p(A)\subseteq p(U\cap A)

\forall y\in p(U)\cap p(A), y=p(x_1) with x_1\in U and y=p(x_2) with x_2\in A, since x_1\in U and x_2\in A, y=p(x_1)=p(x_2)\in p(U\cap A)

So x_1=x_2\in U\cap A, y=p(x_1)=p(x_2)\in p(U)\cap p(A), y\in p(U\cap A).

Note that p(U)\subseteq X is open by p is an open map.

So p(U)\cap p(A) is open in p(A).

q(U\cap A)=p(U\cap A)=p(U)\cap p(A) is open.

So q is open in p(A).

Simplicial complexes (extra chapter)

Definition for simplicial complexes

Simplicial complexes are topological space with simplices (n dimensional triangles) as their building blocks.

Definition for n dimensional simplex

Let v_0,\dots,v_n be points in \mathbb{R}^m such that v_n-v_0, v_{n-1}-v_0, \cdots, and v_1-v_0 are linearly independent in \mathbb{R}^m. (in particular n\leq m).

The $n$-dimensional simplex determined by \{v_0,\dots,v_n\} is given as:

\Delta^n\coloneqq [v_0,\dots,v_n]=\{t_0v_0+t_1v_1+\cdots+t_nv_n\vert t_i\geq 0, \sum_{i=0}^n t_i=1\}

The coefficients t_0,\dots,t_n are called barycentric coordinates.

Example of simplicial complex

n=0,

\Delta^0=\{v_0\}

n=1,

\Delta^1=\{t_0v_0+t_1v_1\vert t_0+t_1=1\}, this is the line segment between v_0 and v_1.

n=2,

\Delta^2=\{t_0v_0+t_1v_1+t_2v_2\vert t_0+t_1+t_2=1\}, this is the triangle with vertices v_0,v_1,v_2.

Note

Every non-empty subset \{v_{i_0},\dots,v_{i_k}\} of \{v_0,\dots,v_n\} determines a k dimensional simplex [v_{i_0},\dots,v_{i_k}]\subseteq \Delta^n=[v_0,\dots,v_n]. Inside the n dimensional simplex t_{i_0}v_{i_0}+\cdots+t_{i_n}v_{i_k}\in \Delta^n. Where the coefficient t_j of v_j\notin \{v_{i_0},\dots,v_{i_n}\} is 0.

Any such k dimensional simplex is called a face of the simplex [v_{i_0},\dots,v_{i_n}].

Example of faces for simplicial complex

\

For a triangle [v_0,v_1,v_2], the faces are [v_0,v_1], [v_0,v_2], and [v_1,v_2] (the edges of the triangle).

Definition for abstract simplicial complex

Let V be a finite or countable set, an abstract simplicial complex on V is a collection of finite non-empty subset of V, denoted by K. And the two conditions are satisfied:

1. If \sigma\in K and \tau\subseteq \sigma, then \tau\in K.

2. For any v\in V, \{v\}\in K.

Example of abstract simplicial complex

Let V=\{a,b,c,d\}.

If we want to include \{a,b,c\}, then we need to include \{a,b\} and \{b,c\}, so we have K=\{\{a,b,c\},\{a,b\},\{b,c\},\{a\},\{b\},\{c\},\{d\}\} is an abstract simplicial complex.

Topological realization of abstract simplicial complex

Let \bigsqcup_{\sigma\in K}\Delta^{|\sigma|-1} be the disjoint union of all |\sigma|-1 dimensional simplices in K.

\tilde{X_k}=\bigsqcup_{\sigma\in K}\Delta^{|\sigma|-1}

We use subspace topology to define a topology on \Delta^n and the union of such topology for each \Delta^{|\sigma|-1} defines a topology on \tilde{X_k}.

We define the equivalence relation x\in \Delta_{\sigma}^{|\sigma|-1}\sim x'\in \Delta_{\sigma'}^{|\sigma'|-1} if x\in \Delta_{\sigma'\cap \sigma}^{|\sigma'\cap \sigma|-1}\subseteq \Delta_{\sigma}^{|\sigma|-1}. and x'\in \Delta_{\sigma'\cap \sigma}^{|\sigma'\cap \sigma|-1}\subseteq \Delta_{\sigma'}^{|\sigma'|-1}.

are the sample points of \Delta_{\sigma\cap \sigma'}^{|\sigma\cap \sigma'|-1}.

X_K is the quotient space of \tilde{X_k} by the equivalence relation.

Continue next time.