Show that the restriction of a quotient map onto a saturated open subset is still a quotient map.

Question

Def: Let $X,Y$ be sets and $p:X \rightarrow Y$ be a surjective map, then $A\subset X$ is called saturated with respect to p if it is the preimage of its image under $p$, i.e. $A=p^{-1}(p(A)).$

Def et $X, Y$ be topological spaces, then a map $p:X \rightarrow Y$ is called a quotient map if $p$ is surjective, continuous and mapping saturated open sets of $X$ to open sets of $Y$.

Problem: Let $X, Y$ be topological spaces and $p:X \rightarrow Y$ be a quotient map (i.e. p is a continuous surjection s.t. $\forall$ open subset $A\subset X$ that is saturated with respect to $p$, $p(A)$ is an open subset of $Y$). Then for a saturated open subset $A\subset X$, $p|_A=q:A \rightarrow p(A)$ is a quotient map, where $A,p(A)$ are endowed with subspace topologies.

Attempt: It suffices to prove that for any saturated open subset $B\subset A$, i.e. $B=q^{-1}(q(B))$, we have $q(B)$ open in $p(A)$. Because $q$ is obviously surjective, and $q=p|_A\Rightarrow q$ is continuous.

Since for any open subset $B\subset A$, $B=A\cap U$ where $U$ is an open set in $X$, we may assume that we are given a saturated open set $A\cap U$. Observa that $A\cap U=q^{-1}\circ q(A\cap U)=p^{-1}\circ p(A\cap U)$, since $A$ is saturated with respect to $p$. So we have $A\cap U$ is saturated with respect to $p$. Also note that $A\cap U$ is open in $X$ because $A,U$ are open in $X$. By the assumption that $p$ is a quotient map, we know that $q(A\cap U)=p(A\cap U)$ is open in Y. Then $q(A\cap U)=p(A\cap U) \cap p(A)$ is open in $p(A)$.

Since the choice of $A\cap U$ is arbitrary, we conclude that $q$ is a quotient map.

Answer 1

I guess your question is whether your proof is correct although you didn't say this. Yes, it is.

Note that a function $p : X \to Y$ is a quotient map in the sense of your definition if and only if the following are satisfied:

(1) $p$ is surjective.

(2) A subset $U \subset Y$ is open in $Y$ if and only if $p^{-1}(U)$ is open in $X$.

This is the usual definition of a quotient map. The continuity of $p$ is nothing else than the "only if" part in (2). To verify the equivalence of both definitions observe that the saturated [open] subsets of $X$ are precisely the sets having the form $p^{-1}(B)$ [with $B \subset Y$ open]. This is true because $p(p^{-1}(B)) = B$ by surjectivity of $p$.

If you start with the above definition, then your problem is equivalent to showing that if $V \subset Y$ is open , then $q : p^{-1}(V) \to V$ satisfies the "if" part of (2). So let $U \subset V$ be a set such that $W = q^{-1}(U)$ is open in $p^{-1}(V)$. Since $p^{-1}(V)$ is open in $X$, also $W$ is open in $X$. But $W = p^{-1}(U)$, hence $U$ is open in $Y$. This implies that $U$ is open in $V$.

Answer 2

I was looking at this problem in Appendix A of Smooth Manifolds by John M. Lee (Exercise A.28). Lee uses the definition given by Paul Frost above. I think the proof I arrived at is essentially the same, just worded differently.

Problem: Suppose $\pi: X \rightarrow Y$ is a quotient map. Choose an arbitrary open, saturated subset $U \subset X$. Denote $\left. \pi\right|_U = q: U \rightarrow \pi(U)$. Show that when $U$ and $\pi(U)$ are endowed with subspace topologies, $q$ is a quotient map.

Proof: Clearly $q$ is surjective. So, we need only show that $q^{-1}(v) \subset U$ is open if and only if $v \subset \pi(U)$ is open.

"$\Leftarrow$"

Choose an arbitrary open set $v \subset \pi(U)$. Because $U$ is open and saturated and $\pi$ is a quotient map, we know that $\pi(U)$ is open. Since $v$ is open in the subspace topology, $v = \pi(U) \cap \tilde{v}$ for some open set $\tilde{v} \subset Y$. Because $v$ is the intersection of two open sets, we conclude that $v$ is open in $Y$ and, because $\pi$ is a quotient map, $\pi^{-1}(v) = \tilde{u} \subset U$ is open in $X$. Moreover, $\tilde{u} = \tilde{u} \cap U$, and we conclude that $\tilde{u}$ is open in $U$ per the definition of the subspace topology. It just remains to note that, since $U$ is saturated, $\pi^{-1}(v) = q^{-1}(v) = \tilde{u}$ to demonstrate that $v$ open $\Rightarrow q^{-1}(v)$ open.

"$\Rightarrow$"

Now choose an arbitrary set $v \subset \pi(U)$ such that $q^{-1}(v) \subset U$ is open. Since $q^{-1}(v)$ open in $U$, we know that there exists $\tilde{u} \subset X$ such that $q^{-1}(v) = \tilde{u} \cap U$. Because $U$ is open, $q^{-1}(v) = \pi^{-1}(v)$ is the intersection of two open sets and therefore open in $X$. Thus, $v$ is open in $Y$. Finally, $v \cap \pi(U) = v$ implies that $v$ satisfies the subspace topology definition of an open set and $v$ is open in $\pi(u)$. Hence, $q^{-1}(v)$ open $\Rightarrow v$ open.

Answer 3

It seems to me that the proof is more fluid if we use this lemma :
Lemma. Let $p:X \to Y$ a map, $A$ a subset of $X$, and $q:A \to p\left( A \right)\;;\;\;x \mapsto p\left( x \right)$.
a) If $A$ is saturated with $p$, then the subsets of $A$ saturated with $q$ are the subsets of $X$ saturated with $p$ and included in $A$.
b) If $A$ is an open subset of $X$ saturated with $p$, then the open subsets of $A$ saturated with $q$ are the open subsets of $X$ saturated with $p$ and included in $A$.
c) If $A$ is an closed subset of $X$ saturated with $p$, then the closed subset of $A$ saturated with $q$ are the closed subset of $X$ saturated with $p$ and included in $A$.
Proof. a) Let $S$ a subset of $A$. $A$ is saturated with $p$ thus $\left\{ {x \in X,\;\exists a \in S,\;\;p\left( x \right) = p\left( a \right)} \right\} \subset \left\{ {x \in X,\;\exists a \in A,\;\;p\left( x \right) = p\left( a \right)} \right\} = A$.
We deduce that $\left\{ {x \in X,\;\exists a \in S,\;\;p\left( x \right) = p\left( a \right)} \right\} = \left\{ {x \in A,\;\exists a \in S,\;\;p\left( x \right) = p\left( a \right)} \right\} = \left\{ {x \in A,\;\exists s \in S,\;\;q\left( x \right) = q\left( a \right)} \right\}$.
Thus $S = \left\{ {x \in X,\;\exists a \in S,\;\;p\left( x \right) = p\left( a \right)} \right\}\quad \Leftrightarrow \quad S = \left\{ {x \in A,\;\exists s \in S,\;\;q\left( x \right) = q\left( a \right)} \right\}$.
Thus $S$ is saturated with $p$ ssi $S$ is saturated with $q$.
b) If $A$ is an open subset of $X$, then the open subsets of $A$ are the open subsets of $X$ included in $A$. We apply a).
c) If $A$ is a closed subset of $X$ then the closed subsets of $A$ are the closed subsets of $X$ included in $A$.
Proof of the theorem. Let $p:\left( {X\;,\;\mathcal{O}} \right) \to \left( {Y\;,\;{\mathcal{U}_p}} \right)$ a quotient map.
We notice ${\mathcal{U}_p}$ the quotient topology on $Y$ and ${\mathcal{O}_s}$ the set of the open subsets of $X$ saturated with $p$.
We have $p\left( {{\mathcal{O}_s}} \right) = {\mathcal{U}_p}$. Let $A$ an open subset of $X$ saturated with $p$.
$q\left( A \right) = p\left( A \right)$ thus $q$ is surjective. $p$ is continuous thus $q$ est continuous.
To prove that $q$ is a a quotient map, it suffices to show that, if $S$ is an open subset of $A$ saturated with $q$, then $q\left( S \right)$ is an open subset of $p\left( A \right)$.
Since $p\left( {{\mathcal{O}_s}} \right) = {\mathcal{U}_p}$ and $A \in {\mathcal{O}_s}$, we have $p\left( A \right) \in {\mathcal{U}_p}$. $p\left( A \right)$ is an open subset of $Y$ thus the open subsets of $p\left( A \right)$ are the open subsets of $Y$ included in $p\left( A \right)$.
Let $S$ an open subset of $A$ saturated with $q$ ; by the lemma, $S$ is an open subset of $X$ saturated with $p$, and since $p\left( {{\mathcal{O}_s}} \right) = {\mathcal{U}_p}$, thus $q\left( S \right) = p\left( S \right) \in {\mathcal{U}_p}$. $q\left( S \right)$ is an open subset of $Y$ included in $p\left( A \right)$, thus $q\left( S \right)$ is an open subset of $p\left( A \right)$. Thus $q$ is a quotient map.

Answer 4

Let ${ X \overset{q}{\to} Y }$ be a quotient map (i.e. a surjection such that ${ V \subseteq _{\text{open}} Y }$ if and only if ${ q ^{-1} (V) \subseteq _{\text{open}} X }$).

Let ${ A \subseteq X }.$ This gives a map ${ A \overset{f}{\to} q(A) , }$ ${ a \mapsto q(a) .}$

We can ask ourselves:
Q) Can we impose a "general enough" constraint on ${ A }$ such that ${ f }$ is also a quotient map ?

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The statement "${ f }$ is a quotient map" is equivalent to the one below.
"For ${ V \subseteq q(A) }$ we have ${ V \subseteq _{\text{open}} q(A) }$ if and only if ${ f ^{-1} (V) \subseteq _{\text{open}} A }$".

But ${ \big( V \subseteq _{\text{open}} q(A) }$ implies ${ f ^{-1}(V) \subseteq _{\text{open}} A \big) }$ anyways holds because ${ f }$ is continuous.

The map ${ A \overset{q \big\vert _{A}}{\to} Y }$ which is ${ q \circ i _A }$ is continuous. Restricting its codomain to ${ q(A) }$ we get ${ f }.$

So "${ f }$ is a quotient map" is equivalent to "${ V \subseteq q(A) }$ and ${ f ^{-1} (V) \subseteq _{\text{open}} A }$ implies ${ V \subseteq _{\text{open}} q(A) }$".
Rewriting, its equivalent to "${ V \subseteq q(A) }$ and ${ A \cap q ^{-1} (V) \subseteq _{\text{open}} A }$ implies ${ V \subseteq _{\text{open}} q(A) }$".

• Say we impose that ${ \color{green}{ A = q ^{-1} (\tilde{A}) } }$ for some ${ \tilde{A} \subseteq Y }.$

In this situation, above statement simplifies as "${ V \subseteq \tilde{A} }$ and ${ q ^{-1} (\tilde{A} \cap V) \subseteq _{\text{open}} q ^{-1} (\tilde{A}) }$ implies ${ V \subseteq _{\text{open}} \tilde{A} }$".
That is, the statement becomes "${ V \subseteq \tilde{A} }$ and ${ q ^{-1} (V) \subseteq _{\text{open}} q ^{-1} (\tilde{A}) }$ implies ${ V \subseteq _{\text{open}} \tilde{A} }$".

• Say we further impose that ${ \color{purple}{q ^{-1} (\tilde{A}) \subseteq _{\text{open}} X } }.$

Now the statement is true.

Let ${ V \subseteq \tilde{A} }$ and ${ q ^{-1} (V) \subseteq _{\text{open}} q ^{-1} (\tilde{A}) }.$
But ${ \color{purple}{ q ^{-1} (\tilde{A}) \subseteq _{\text{open}} X } }$ so ${ q ^{-1} (V) \subseteq _{\text{open}} X }.$ Now because ${ q }$ is a quotient map, ${ V \subseteq _{\text{open}} Y }.$
So ${ V }$ ${ = V \cap \tilde{A} }$ ${ \subseteq _{\text{open}} \tilde{A} , }$ as needed.

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To summarise,

Let ${ X \overset{q}{\to} Y }$ be a quotient map, and ${ A \subseteq X }.$
If ${ A }$ is saturated (wrt ${ q }$) and open (in ${ X }$), then ${ A \overset{f}{\to} q(A) ,}$ ${ a \mapsto q (a) }$ is also a quotient map.