J. Munkres. Topology (2013) Chapter 2: Topological Spaces and Continuous Functions. 22: The Quotient Topology
Definition
Let \(X\) and \(Y\) be topological spaces. Let \(p : X \rightarrow Y\). Then \(p\) is strongly continuous iff, for all \(U \subseteq Y\), we have \(U\) is open in \(Y\) if and only if \(p^{-1}(U)\) is open in \(X\).
Definition
Let \(X\) and \(Y\) be topological spaces. Let \(p : X \rightarrow Y\). Then \(p\) is a quotient map iff \(p\) is surjective and strongly continuous.
Definition
Let \(X\) and \(Y\) be sets. Let \(p : X \rightarrow Y\). Let \(C \subseteq X\). Then \(C\) is saturated iff, for all \(x,y \in X\), if \(x \in C\) and \(p(x) = p(y)\) then \(y \in C\).
Proposition
Let \(X\) and \(Y\) be topological spaces. Let \(p : X \rightarrow Y\) be surjective. Then the following are equivalent.
- \(p\) is a quotient map.
- \(p\) is continuous and maps saturated open sets to open sets.
- \(p\) is continuous and maps saturated closed sets to closed sets.
Definition
Let \(X\) and \(Y\) be topological spaces. Let \(f : X \rightarrow Y\). Then \(f\) is a closed map iff it maps closed sets to closed sets.
Proposition
Every continuous open map is a quotient map. Every continuous closed map is a quotient map.
Example
\(\pi_1 : \{ (x,y) \in \mathbb{R}^2 \mid x \geq 0 \text{ or } y = 0 \} \rightarrow \mathbb{R}\) is a quotient map that is neither an open map nor a closed map.
Example 1
Define \(p : [0,1] \cup [2,3] \rightarrow [0,2]\) by \(p(x) = x\) if \(x \in [0,1]\) and \(p(x) = x-1\) if \(x \in [2,3]\). Then \(p\) is a quotient map and a closed map, but not an open map.
Example 2
The projection \(\pi_1 : \mathbb{R}^2 \rightarrow \mathbb{R}\) is a quotient map and an open map, but not a closed map.
Definition
Let \(X\) be a topological space, \(A\) a set, and \(p : X \rightarrow A\) a surjection. Then the quotient topology on \(A\) is the unique topology such that \(p\) is a quotient map, namely \(\{U \in \mathcal{P} A \mid p^{-1}(U) \text{ is open in } X\}\).
Definition
Let \(X\) be a topological space. Let \(X^*\) be a partition of \(X\). Let \(\pi : X \rightarrow X^*\) be the canonical projection. Then \(X^*\) under the quotient topology is called a quotient space of \(X\).
The following is the first difficult theorem in Munkres. At the moment (paraphrasing Thorsten Altenkirch) I understand each step of the proof, but I do not understand the proof.
Theorem 22.1
Let \(p : X \rightarrow Y\) be a quotient map. Let \(A\) be a subspace of \(X\) that is saturated with respect to \(p\). Let \(q : A \rightarrow p(A)\) be the function obtained by restricting \(p\).
- If \(A\) is either open or closed in \(X\), then \(q\) is a quotient map.
- If \(p\) is either an open map or a closed map, then \(q\) is a quotient map.
Proof
- \(\langle 1 \rangle 1\) For all \(V \subseteq p(A)\), we have \(q^{-1}(V) = p^{-1}(V)\).
- Let \(V \subseteq p(A)\).
- \(q^{-1}(V) \subseteq p^{-1}(V)\)
- Set theory.
- \(p^{-1}(V) \subseteq q^{-1}(V)\)
- Let \(x \in p^{-1}(V)\)
- \(p(x) \in V\)
- \(p(x) \in p(A)\)
- \(x \in A\)
- \(A\) is saturated.
- \(q(x) = p(x) \in V\)
- \(x \in q^{-1}(V)\)
- \( \langle 1 \rangle 2 \) For all \(U \subseteq X\), we have \(p(U \cap A) = p(U) \cap p(A)\).
- Let \(U \subseteq X\).
- \(p(U \cap A) \subseteq p(U) \cap p(A)\)
- Set theory.
- \(p(U) \cap p(A) \subseteq p(U \cap A)\)
- Let \(y \in p(U) \cap p(A)\)
- Pick \(u \in U\) and \(a \in A\) such that \(p(u) = p(a) = y\)
- \(u \in A\)
- \(A\) is saturated.
- \(u \in U \cap A\)
- \(y \in p(U \cap A)\)
- If \(A\) is open then \(q\) is a quotient map.
- Assume \(A\) is open.
- For any \(V \subseteq p(A)\), if \(q^{-1}(V)\) is open in \(A\) then \(V\) is open in \(p(A)\).
- Let \(V \subseteq p(A)\).
- Assume \(q^{-1}(V)\) is open in \(A\).
- \(q^{-1}(V)\) is open in \(X\).
- Lemma 16.2
- \(p^{-1}(V)\) is open in \(X\).
- By \(\langle 1 \rangle 1 \).
- \(V\) is open in \(Y\).
- \(p\) is a quotient map.
- \(V\) is open in \(p(A)\).
- \(V = V \cap p(A)\)
- \(q\) is a quotient map.
- If \(p\) is an open map then \(q\) is a quotient map.
- \( \langle 2 \rangle 1 \) Assume \(p\) is an open map.
- For any \(V \subseteq p(A)\), if \(q^{-1}(V)\) is open in \(A\) then \(V\) is open in \(p(A)\).
- Let \(V \subseteq p(A)\).
- Assume \(q^{-1}(V)\) is open in \(A\).
- \(p^{-1}(V)\) is open in \(A\).
- \(\langle 1 \rangle 1 \)
- \( \langle 3 \rangle 1 \) Pick \(U\) open in \(X\) such that \(p^{-1}(V) = U \cap A\).
- \( \langle 3 \rangle 2 \) \(p(p^{-1}(V)) = V\)
- Because \(p\) is surjective.
- \(V = p(U) \cap p(A)\) \[ \begin{align} V & = p(p^{-1}(V)) & (\langle 3 \rangle 2) \\ & = p(U \cap A) & (\langle 3 \rangle 1) \\ & = p(U) \cap p(A) & (\langle 1 \rangle 2) \end{align} \]
- \(p(U)\) is open in \(Y\).
- \( \langle 2 \rangle 1 \)
- \(V\) is open in \(p(A)\).
- \(q\) is a quotient map.
- If \(A\) is closed then \(q\) is a quotient map.
- Assume \(A\) is closed.
- For any \(V \subseteq p(A)\), if \(q^{-1}(V)\) is closed in \(A\) then \(V\) is closed in \(p(A)\).
- Let \(V \subseteq p(A)\).
- Assume \(q^{-1}(V)\) is closed in \(A\).
- \(q^{-1}(V)\) is closed in \(X\).
- Theorem 17.3
- \(p^{-1}(V)\) is closed in \(X\).
- By \(\langle 1 \rangle 1 \).
- \(V\) is closed in \(Y\).
- \(p\) is a quotient map.
- \(V\) is closed in \(p(A)\).
- \(V = V \cap p(A)\)
- \(q\) is a quotient map.
- If \(p\) is a closed map then \(q\) is a quotient map.
- \( \langle 2 \rangle 1 \) Assume \(p\) is a closed map.
- For any \(V \subseteq p(A)\), if \(q^{-1}(V)\) is closed in \(A\) then \(V\) is closed in \(p(A)\).
- Let \(V \subseteq p(A)\).
- Assume \(q^{-1}(V)\) is closed in \(A\).
- \(p^{-1}(V)\) is closed in \(A\).
- \(\langle 1 \rangle 1 \)
- \( \langle 3 \rangle 1 \) Pick \(U\) closed in \(X\) such that \(p^{-1}(V) = U \cap A\).
- Theorem 17.2
- \( \langle 3 \rangle 2 \) \(p(p^{-1}(V)) = V\)
- Because \(p\) is surjective.
- \(V = p(U) \cap p(A)\) \[ \begin{align} V & = p(p^{-1}(V)) & (\langle 3 \rangle 2) \\ & = p(U \cap A) & (\langle 3 \rangle 1) \\ & = p(U) \cap p(A) & (\langle 1 \rangle 2) \end{align} \]
- \(p(U)\) is closed in \(Y\).
- \( \langle 2 \rangle 1 \)
- \(V\) is open in \(p(A)\).
- \(q\) is a quotient map.
- \(\Box\)
Proposition
The composite of two quotient maps is a quotient map.
Example 7
The product of two quotient maps is not necessarily a quotient map. Let \(X = \mathbb{R}\) and let \(X^*\) be the quotient space formed by identifying all the points in \(\mathbb{Z}_+\). Let \(p : X \rightarrow X^*\) be the quotient map. We show that \(p \times \mathrm{id}_\mathbb{Q} : X \times \mathbb{Q} \rightarrow X^* \times \mathbb{Q}\) is not a quotient map.
- For \(n \in \mathbb{Z}_+\), let \(c_n = \sqrt{2} / n\).
- For \(n \in \mathbb{Z}_+\), let \[ U_n = \{(x,y) \in \mathbb{R} \times \mathbb{Q} \mid n-1/4 < x < n + 1/4, (y < c_n + n - x \wedge y < c_n - n + x) \vee (y > c_n + n - x \wedge y > c_n - n + x)\} \]
- For all \(n \in \mathbb{Z}_+\), \(U_n\) is open in \(\mathbb{R} \times \mathbb{Q}\)
- For all \(n \in \mathbb{Z}_+\), \(\{n\} \times \mathbb{Q} \subseteq U_n\)
- Let \(U = \bigcup_{n \in \mathbb{Z}_+} U_n\)
- \(U\) is saturated with respect to \(p \times \mathrm{id}_{\mathbb{Q}}\)
- Since \((n,q) \in U\) for all \(n \in \mathbb{Z}_+\) and \(q \in \mathbb{Q}\).
- \((p \times \mathrm{id}_\mathbb{Q})(U)\) is not open in \(X^* \times \mathbb{Q}\)
- Let \(U' = (p \times \mathrm{id}_\mathbb{Q})(U)\)
- Let \(b = p(n)\) for all \(n \in \mathbb{Z}_+\).
- \((b,0) \in U'\)
- Pick neighbourhoods \(W\) of \(b\) and \(\delta > 0\) such that \(W \times ((-\delta,\delta) \cap \mathbb{Q}) \subseteq U'\)
- \(p^{-1}(W) \times ((-\delta, \delta) \cap \mathbb{Q}) \subseteq U\)
- Pick \(n \in \mathbb{Z}_+\) such that \(c_n < \delta\)
- Pick \(0 < \epsilon < 1/4\) such that \((n-\epsilon, n+\epsilon) \subseteq p^{-1}(W)\)
- Pick a rational number \(q\) such that \(|q - c_n| < \epsilon / 2\)
- \((n + \epsilon / 2, q) \in U\)
- This is a contradiction.
- \(\Box\)
Theorem 22.2
Let \(p : X \rightarrow Y\) be strongly cont. Let \(Z\) be a topological space. Let \(f : Y \rightarrow Z\).
- If \(f \circ p\) is continuous then \(f\) is continuous.
- If \(f \circ p\) is a quotient map then \(f\) is a quotient map.
From the Exercises
Definition
Let \(X\) be a topological space and \(A \subseteq X\). A retraction of \(X\) onto \(A\) is a continuous map \(r : X \rightarrow A\) such that \(r(a) = a\) for all \(a \in A\).
Proposition
Every retraction is a quotient map.
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