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The quotient topology:
\u
Recommended problems:  22: 2, 4

Defn: $p: X \rightarrow Y$ is a quotient map if $p$ is surjective and $U$ open in $Y$ if 
and only if $p^{-1}(U)$ open in $X$.


Note $p$ is continuous.


Defn:  $f: X \rightarrow Y$ is an open map if $U$ open in $X$ implies $f(U)$ open in $Y$.
$g:X \rightarrow Y$ is a closed map if $A$ closed in $X$ implies $f(A)$ closed in $Y$.

Lemma.  A continuous surjective open map is a quotient map. A continuous surjective 
closed map is a quotient map.
\w
\vfil
Let $A = \{(x, y) \in {\cal R} \times {\cal R}~|~ x \geq 0 ~or ~ y = 0\}$.   Then 
$\pi_1|_A: A \rightarrow 
{\cal R}$ is a quotient map which is neither open nor closed. 
\vfil
\w
\eject
Defn:  The fiber of $p$ over $y$ is $p^{-1}(y)$.
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$C \subset X$ is saturated (w.r.t $p$) if there exists $D \subset Y$ such that $C = 
p^{-1}(D).

\u
Thus a quotient map takes saturated open sets to open sets.
\u
Defn:  Given a surjective map $p: X \rightarrow Y$, the quotient topology on $Y$ = $\{ U 
~|~ p^{-1}(U)$ open$\}$


Thm 22.2:  Let $p: X \rightarrow Y$ be a quotient map.  Suppose $g: X \rightarrow Z$ is 
constant on each fiber (i.e. $g|_p^{-1}(y)$ is a constant function). 
Then $g$ induces a unique map $f: Y \rightarrow Z$ such that $f \circ p = g$.  $f$ is 
continuous if and only if $g$ is continuous.  $f$ is a quotient map if and only if $g$ 
is a quotient map.


\w
Cor 22.3:  Let $g: X \rightarrow Z$ be a continuous surjective map.  Let $X^* = 
\{g^{-1}(\{z\} ~|~ z \in Z\}$.  Give $X^*$ the quotient topology.
$g$ induces a bijective continuous map $f: X^* \rightarrow Z$ which is a homeomorphism if 
and only if $g$ is a quotient map.

\w
The quotient map does not behave well w.r.t subspaces, products, preserving topological 
properties.



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