# 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 partition 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.

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