Math4202 Topology II (Lecture 2)

Reviewing quotient map

Recall from last lecture example (Example 4 form Munkers):

A map of wrapping closed unit circle to S^2, where f:\mathbb{R}^2\to S^2 maps everything outside of circle to south pole s.

To show it is a quotient space, we need to show that f:

is continuous (every open set in S^2 has reverse image open in \mathbb{R}^2)
surjective (trivial)
with the property that U\subset S^2 is open if and only if f^{-1}(U) is open in \mathbb{R}^2.

If A\subseteq S^2 is open, then f^{-1}(A) is open in \mathbb{R}^2. (consider the basis, the set of circle in \mathbb{R}^2, they are mapped to closed sets in S^2)
If f^{-1}(A) is open in \mathbb{R}^2, then A is open in S^2.
- If s\notin A, then f is a bijection, and A is open in S^2.
- If s\in A, then f^{-1}(A) is open and contains the complement of set S=\{(x,y)|x^2+y^2\geq 1\}=f^{-1}(\{s\}), therefore there exists U=\bigcup_{x\in S} B_{\epsilon _x}(x) is open in \mathbb{R}^2, U\subseteq f^{-1}(A), f^{-1}(\{s\})\subseteq U.
- Since \partial f^{-1}(\{s\}) is compact, we can even choose U to be the set of the following form
- \{(x,y)|x^2+y^2>1-\epsilon\} for some 1>\epsilon>0.
- So f(U) is an open set in A and contains s.
- s is an interior point of A.
- Other oint y in A follows the arguments in the first case.

Quotient space

Definition of quotient topology induced by quotient map

If X is a topological space and A is a set and if p:X\to A is surjective, there exists exactly one topology \mathcal{T} on A relative to which p is a quotient map.


\mathcal{T} \coloneqq \{U|f^{-1}(U)\text{ is open in }X\}

and \mathcal{T} is called the quotient topology on A induced by p.

Definition of quotient topology induced by equivalence relation

Let X be a topological space, and let X^* be a partition of X into disjoint subsets whose union is X. Let p:X\to X^* be the surjective map that sends each x\in X to the unique A\in X^* such that each point of X to the subset containing the point. In the quotient topology induced by p, the space X^* is called the associated quotient space.

Example of quotient topology induced by equivalence relation

Consider S^n and x\sim -x, then the induced quotient topology is \mathbb{R}P^n (the set of lines in \mathbb{R}^n passing through the origin).

Theorem about a quotient map and quotient topology

Let p:X\to Y be a quotient map; and A be a subspace of X, that is saturated with respect to p: Let q:A\to p(A) be the restriction of p to A.

If A is either open or closed in X, then q is a quotient map.
If p is either open or closed, then q is a quotient map.

Note

Recall the definition of saturated set:

\forall y\in Y, consider the set f^{-1}(\{y\})\subset X, if f^{-1}(\{y\})\cap A\neq \emptyset, then f^{-1}(\{y\})\subseteq A. sounds like connectedness

That is equivalent to say that A is a union of f^{-1}(\{y\}) for some y\in Y.

3.2 KiB Raw Blame History