# Topology/Quotient Spaces

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The quotient topology is not a natural generalization of anything studied in analysis, however it is easy enough to motivate. One motivation comes from geometry. For example, the torus can be constructed by taking a rectangle and pasting the edges together.

## Definition: Quotient Map

Let ${\displaystyle X}$ and ${\displaystyle Y}$ be topological spaces; let ${\displaystyle f:X\rightarrow Y}$ be a surjective map. The map f is said to be a quotient map provided a ${\displaystyle U\subseteq Y}$ is open in Y if and only if ${\displaystyle f^{-1}(U)}$ is open in X .

## Definition: Quotient Map Alternative

There is another way of describing a quotient map. A subset ${\displaystyle C\subset X}$ is saturated (with respect to the surjective map ${\displaystyle f:X\rightarrow Y}$) if C contains every set ${\displaystyle f^{-1}(\{y\})}$ that it intersects. To say that f is a quotient map is equivalent to saying that f is continuous and f maps saturated open sets of X to open sets of Y . Likewise with closed sets.

There are two special types of quotient maps: open maps and closed maps .

A map ${\displaystyle f:X\rightarrow Y}$ is said to be an open map if for each open set ${\displaystyle U\subseteq X}$, the set ${\displaystyle f(U)}$ is open in Y . A map ${\displaystyle f:X\rightarrow Y}$ is said to be a closed map if for each closed ${\displaystyle A\subseteq X}$, the set ${\displaystyle f(A)}$ is closed in Y . It follows from the definition that if ${\displaystyle f:X\rightarrow Y}$ is a surjective continous map that is either open or closed, then f is a quotient map.

## Definition: Quotient Topology

If X is a topological space and A is a set and if ${\displaystyle f:X\rightarrow A}$ is a surjective map, then there exist exactly one topology ${\displaystyle \tau }$ on A relative to which f is a quotient map; it is called the quotient topology induced by f .

## Definition: Quotient Space

Let X be a topological space and let ,${\displaystyle X^{*}}$ be a partiton of X into disjoint subsets whose union is X . Let ${\displaystyle f:X\rightarrow X^{*}}$ be the surjective map that carries each ${\displaystyle x\in X}$ to the element of ${\displaystyle X^{*}}$ containing it. In the quotient topology induced by f the space ${\displaystyle X^{*}}$ is called a quotient space of X .

## Theorem

Let ${\displaystyle f:X\rightarrow Y}$ be a quotient map; let A be a subspace of X that is saturated with respect to f ; let ${\displaystyle g:A\rightarrow f(A)}$ be the map obtained by restricting f , then g is a quotient map.

1.) If A is either opened or closed in X .

2.) If f is either an open map or closed map.

Proof: We need to show:
${\displaystyle f^{-1}(V)=g^{-1}(V)}$ when V ${\displaystyle \subset f(A)}$

and

${\displaystyle f(U\cap A)=f(U)\cap f(A)}$ when ${\displaystyle U\subset X}$.

Since ${\displaystyle V\subset f(A)}$ and A is saturated, ${\displaystyle f^{-1}(V)\subset A}$. It follows that both ${\displaystyle f^{-1}(V)}$ and ${\displaystyle g^{-1}(V)}$ equal all points in A that are mapped by f into V . For the second equation, for any two subsets U and ${\displaystyle A\subset X}$

${\displaystyle f(U\cap A)\subset f(U)\cap f(A).}$

In the opposite direction, suppose ${\displaystyle y=f(u)=f(a)}$ when ${\displaystyle u\in U}$ and ${\displaystyle a\in A}$. Since A is saturated, ${\displaystyle A\subset f^{-1}(f(a))}$, so that in particular ${\displaystyle A\subset u}$. Then ${\displaystyle y=f(u)}$ where ${\displaystyle u\in U\cap A}$.

Suppose A or f is open. Since ${\displaystyle V\subset f(A)}$, assume ${\displaystyle g^{-1}(V)}$ is open in ${\displaystyle A}$ and show V is open in ${\displaystyle f(A)}$.

First, suppose A is open. Since ${\displaystyle g^{-1}(V)}$ is open in A and A is open in X , ${\displaystyle g^{-1}(V)}$ is open in X . Since ${\displaystyle f^{-1}(V)=g^{-1}(V)}$, ${\displaystyle f^{-1}(V)}$ is open in X . V is open in Y because f is a quotient map.

Now suppose f is open. Since ${\displaystyle g^{-1}(V)=f^{-1}(V)}$ and ${\displaystyle g^{-1}(V)}$ is open in A, ${\displaystyle f^{-1}(V)=U\cap A}$ for a set U open in X . Now ${\displaystyle f(f^{-1}(V))=V}$ because f is surjective; then

${\displaystyle V=f(f^{-1}(V))=f(U\cap A)=f(U)\cap f(A).}$

The set ${\displaystyle f(U)}$ is open in Y because f is an open map; hence V is open in ${\displaystyle f(A)}$. The proof for closed A or f is left to the reader.

 Topology ← Product Spaces Quotient Spaces Continuity and Homeomorphisms →