Math4201 Topology I (Lecture 19)

Quotient topology

More propositions

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.

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

We prove 2 next time...

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