Topology/Quotient Spaces

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 partition 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.