NoteNextra-origin/content/Math4201/Math4201_L18.md

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.

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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 X,Y,Z be topological spaces. p is a quotient map from X to Y and g is a continuous map from X to Z.

Moreover, if for any y\in Y, the map g is constant on p^{-1}(y), then there is a continuous map f: Y\to Z satisfying f\circ p=g.

Proof

For any y\in Y, take x\in X such that p(x)=y (since p is surjective).

Define f(y)\coloneqq g(x).

Note that this is well-defined and it doesn't depend on the specific choice of x that p(x)=y because g is constant on p^{-1}(y).

Then we check that f is continuous.

Let U\subseteq Z be open. Then we want to show that f^{-1}(U)\subseteq Y is open.

Since p is a quotient map, this is equivalent to showing that p^{-1}(f^{-1}(U))\subseteq X is open. Note that p^{-1}(f^{-1}(U))=g^{-1}(U).

Since g is continuous, g^{-1}(U) is open in X.

Since g^{-1}(U) is open in X, p^{-1}(g^{-1}(U)) is open in Y.

In general, p^{-1}(y) is called the fiber of p over y. The g must be constant on the fiber.

We may define p^{-1}(y) as the equivalence class of y if p is defined using the equivalence relation. By definition p^{-1}([x]) is the element of x that are \sim x.

Additional to the proposition

Note that f is unique.

It is not hard to see that f is a quotient map if and only if g is a quotient map. (check book for detailed proofs)

Definition of saturated map

Let p:X\to Y be a quotient map. We say A\subseteq X is saturated by p if A=p^{-1}(B) for some B\subseteq Y.

Equivalently, if x\in A, then p^{-1}(p(x))\subseteq A.

Proposition for quotient maps from saturated sets

Let p:X\to Y be a quotient map and q be given by restriction of p to A\subseteq X. q:A\to p(A), q(x)=p(x),x\in A.

Assume that A is saturated by p.

1. If A is closed or open, then q is a quotient map.
2. If p is closed or open, then q is a quotient map.

Proof

We prove 1 and assume that A is open, (the closed case is similar).

clearly, q:A\to p(A) is surjective.

In general, restricting the domain and the range of a continuous map is continuous.

Since A is saturated by p, then p^{-1}(p(A))=A is open, so p(A) is open because p is a quotient map. Let V\subseteq p(A) and q^{-1}(V)\subseteq A is open. Then q^{-1}(V)=p^{-1}(V).

(i) q^{-1}(V)\subseteq p^{-1}(V): x\in q^{-1}(V)\implies q(x)\in V. Then p(x)=q(x)\in V

(ii) p^{-1}(V)\subseteq q^{-1}(V): x\in p^{-1}(V)\implies p(x)\in V\subseteq p(A). This implies that x\in p^{-1}(p(A))=A since A is saturated by p. Therefore x\in q^{-1}(V).

Since A is open in X, any open subspace of A is open in X. In particular, q^{-1}(V)=p^{-1}(V) is open in X.

Since p is a quotient map, and p^{-1}(V) is open in X, V is open in Y. So V\subseteq p(A) is open in Y.

This shows q is a quotient map.

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We prove 2 next time...