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

## Definition: Quotient Map Alternative

There is another way of describing a quotient map. A subset $C\subset X$  is saturated (with respect to the surjective map $f:X\rightarrow Y$ ) if C contains every set $f^{-1}(\{y\})$  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 $f:X\rightarrow Y$  is said to be an open map if for each open set $U\subseteq X$ , the set $f(U)$  is open in Y . A map $f:X\rightarrow Y$  is said to be a closed map if for each closed $A\subseteq X$ , the set $f(A)$  is closed in Y . It follows from the definition that if $f:X\rightarrow Y$  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 $f:X\rightarrow A$  is a surjective map, then there exist exactly one topology $\tau$  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 ,$X^{*}$  be a partiton of X into disjoint subsets whose union is X . Let $f:X\rightarrow X^{*}$  be the surjective map that carries each $x\in X$  to the element of $X^{*}$  containing it. In the quotient topology induced by f the space $X^{*}$  is called a quotient space of X .

## Theorem

Let $f:X\rightarrow Y$  be a quotient map; let A be a subspace of X that is saturated with respect to f ; let $g:A\rightarrow f(A)$  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:
$f^{-1}(V)=g^{-1}(V)$  when V $\subset f(A)$

and

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

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

$f(U\cap A)\subset f(U)\cap f(A).$

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

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

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

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

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

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

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