3.7 KiB
Math4201 Topology I (Lecture 18)
Quotient topology
Let (X,\mathcal{T}) be a topological space and X^* be a set, q:X\to X^* is a surjective map. The quotient topology on X^*:
U\subseteq X^* is open \iff q^{-1}(U) is open in X.
Equivalently,
Z\subseteq X^* is closed \iff q^{-1}(Z) is closed in X.
Open maps
Let (X,\mathcal{T}) and (Y,\mathcal{T}') be two topological spaces
Let f:X\to Y is a quotient map if and only if f is surjective and
U\subseteq Y is open \iff f^{-1}(U) is open
or equivalently
Z\subseteq Y is closed \iff f^{-1}(Z) is closed.
Definition of open map
Let X\to Y be continuous. We say f is open if for any V\subseteq X be open, f(V) is open in Y.
Let X\to Y be continuous. We say f is closed if for any V\subseteq X be closed, f(V) is closed in Y.
ff^{-1}(U)=U\text{ if }f \text{ is surjective}=U\cap f(X)
Examples of open maps
Let X,Y be topological spaces. Define the projection map \pi_X:X\times Y\to X, \pi_X(x,y)=x.
This is a surjective continuous map (Y\neq \phi)
This map is open. If U\subseteq X is open and V\subseteq Y is open, then U\times V is open in X\times Y and such open sets form a basis.
$\pi_X(U\times V)=\begin{cases} U&\text{ if }V\neq \emptyset\ \emptyset &\text{ if }V=\emptyset \end{cases}$
In particular, image of any such open set is open. Since any open W\subseteq X\times Y is a union of such open sets.
W=\bigcup_{\alpha\in I}U_\alpha\times V\alpha
\pi_X(W)=\pi_X(\bigcup_{\alpha\in I}U_\alpha\times V_\alpha)=\bigcup_{\alpha\in I}\pi_X(U_\alpha\times V_\alpha)=\bigcup_{\alpha\in I}U_\alpha
is open in X.
However, \pi_X is not necessarily a closed map.
Let X=Y=\mathbb{R} and X\times Y=\mathbb{R}^2
Z\subseteq \mathbb{R}^2=\{(x,y)\in\mathbb{R}^2|x\neq 0, y=\frac{1}{x}\} is a closed set in \mathbb{R}^2
\pi_X(Z)=\mathbb{R}\setminus \{0\} is not closed.
Let X=[0,1]\cup [2,3], Y=[0,2] with subspace topology on \mathbb{R}
Let f:X\to Y be defined as:
f(x)=\begin{cases}
x& \text{ if } x\in [0,1]\\
x-1& \text{ if }x\in [2,3]
\end{cases}
f is continuous and surjective, f is closed Z\subseteq [0,1]\cup [2,3]=Z_1\cup Z_2, Z_1\subseteq [0,1],Z_2\subseteq [2,3] is closed, f(Z)=f(Z_1)\cup f(Z_2) is closed in X.
But f is not open. Take U=[0,1]\subseteq X, f=[0,1]\subseteq [0,2] is not open because of the point 1.
In general, and closed surjective map is a quotient map. In particular, this is an example of a closed surjective quotient map which is not open.
Let f be a surjective open map. Then f is a quotient map:
U\subseteq Y is open and f is continuous, \implies f^{-1}(U)\subseteq X is open
f^{-1}(U)\subseteq X is open and f is surjective and open, \implies f(f^{-1}(U))=U is open.
Proposition of continuous and open maps
If f is a continuous bijection, then f is open. if and only if f^{-1} is continuous.
Proof
To show f^{-1} is continuous, we have to show for U\subseteq X open. (f^{-1})^{-1}(U)=f(U)\subseteq Y is open.
This is the same thing as saying that f is open.
Let f be a quotient map f: X \to Y, and g be a continuous map g:X\to Z.
We want to find \hat{g} such that g=\hat{g}\circ f.
If x_1,x_2\in X, such that f(x_1)=f(x_2) and g(x_1)\neq g(x_2), then we cannot find \hat{g}.
Proposition for continuous and quotient maps
Let f and g be as above. Moreover, for any y\in Y, all the points in f^{-1}(y) are mapped to a single point by g. Then there is a unique continuous map \hat{g} such that g=\hat{g}\circ f.
Continue next week.