# 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 $f\circ p=g$.

<details>
<summary>Proof</summary>

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 $Y$.

</details>

> 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 $p$.

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

<details>
<summary>Proof</summary>

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.

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We prove 2 next time...

</details>