Math4201 Topology I (Lecture 19)

Quotient topology

More propositions

Proposition for continuous and quotient maps

Let $X,Y,Z$ be topological spaces. $p$ is a quotient map from $X$ to $Y$ and $g$ is a continuous map from $X$ to $Z$ .

Moreover, if for any $y\in Y$ , the map $g$ is constant on $p^{-1}(y)$ , then there is a continuous map $f: Y\to Z$ satisfying $f\circ p=g$ .

Proof

For any $y\in Y$ , take $x\in X$ such that $p(x)=y$ (since $p$ is surjective).

Define $f(y)\coloneqq g(x)$ .

Note that this is well-defined and it doesn’t depend on the specific choice of $x$ that $p(x)=y$ because $g$ is constant on $p^{-1}(y)$ .

Then we check that $f$ is continuous.

Let $U\subseteq Z$ be open. Then we want to show that $f^{-1}(U)\subseteq Y$ is open.

Since $p$ is a quotient map, this is equivalent to showing that $p^{-1}(f^{-1}(U))\subseteq X$ is open. Note that $p^{-1}(f^{-1}(U))=g^{-1}(U)$ .

Since $g$ is continuous, $g^{-1}(U)$ is open in $X$ .

Since $g^{-1}(U)$ is open in $X$ , $p^{-1}(g^{-1}(U))$ is open in $Y$ .

In general, $p^{-1}(y)$ is called the fiber of $p$ over $y$ . The $g$ must be constant on the fiber.

We may define $p^{-1}(y)$ as the equivalence class of $y$ if $p$ is defined using the equivalence relation. By definition $p^{-1}([x])$ is the element of $x$ that are $\sim x$ .

Additional to the proposition

Note that $f$ is unique.

It is not hard to see that $f$ is a quotient map if and only if $g$ is a quotient map. (check book for detailed proofs)

Definition of saturated map

Let $p:X\to Y$ be a quotient map. We say $A\subseteq X$ is saturated by $p$ if $A=p^{-1}(B)$ for some $B\subseteq Y$ .

Equivalently, if $x\in A$ , then $p^{-1}(p(x))\subseteq A$ .

Proposition for quotient maps from saturated sets

Let $p:X\to Y$ be a quotient map and $q$ be given by restriction of $p$ to $A\subseteq X$ . $q:A\to p(A)$ , $q(x)=p(x),x\in A$ .

Assume that $A$ is saturated by $p$ .

If $A$ is closed or open, then $q$ is a quotient map.
If $p$ is closed or open, then $q$ is a quotient map.

Proof

We prove 1 and assume that $A$ is open, (the closed case is similar).

clearly, $q:A\to p(A)$ is surjective.

In general, restricting the domain and the range of a continuous map is continuous.

Since $A$ is saturated by $p$ , then $p^{-1}(p(A))=A$ is open, so $p(A)$ is open because $p$ is a quotient map. Let $V\subseteq p(A)$ and $q^{-1}(V)\subseteq A$ is open. Then $q^{-1}(V)=p^{-1}(V)$ .

(i) $q^{-1}(V)\subseteq p^{-1}(V)$ : $x\in q^{-1}(V)\implies q(x)\in V$ . Then $p(x)=q(x)\in V$

(ii) $p^{-1}(V)\subseteq q^{-1}(V)$ : $x\in p^{-1}(V)\implies p(x)\in V\subseteq p(A)$ . This implies that $x\in p^{-1}(p(A))=A$ since $A$ is saturated by $p$ . Therefore $x\in q^{-1}(V)$ .

Since $A$ is open in $X$ , any open subspace of $A$ is open in $X$ . In particular, $q^{-1}(V)=p^{-1}(V)$ is open in $X$ .

Since $p$ is a quotient map, and $p^{-1}(V)$ is open in $X$ , $V$ is open in $Y$ . So $V\subseteq p(A)$ is open in $Y$ .

This shows $q$ is a quotient map.

We prove 2 next time…