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 $f:X\to Y$ 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 $f:X\to Y$) 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 $f:X\to Y$ 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 $f:X\to Y$ 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 $f:X\to Y$ 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 $f:X\to A$ 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 $f:X\to X^{*}$ 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 $f:X\to Y$ be a quotient map; let A be a subspace of X that is saturated with respect to f ; let $g:A\to f(A)$ 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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