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 together the edges together.

Definition: Quotient Map[edit]

Let  \emph{X} and  \emph{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[edit]

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[edit]

If X is a topological space and A is a set and if  f : X \rightarrow Y 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[edit]

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[edit]

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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