# 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 MapEdit

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 AlternativeEdit

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 TopologyEdit

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 SpaceEdit

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 .

## TheoremEdit

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.

Topology
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