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Let $X$ and $Y$ be topological spaces; let $p \colon X \to Y$ be a surjective map.

Then $p$ is said to be a quotient map provided a subset $U$ of $Y$ is open in $Y$ if and only if $p^{-1}(U)$ is open in $X$ (or equivalently a subset $B$ of $Y$ is closed in $Y$ if and only if the set $p^{-1}(B)$ is closed in $X$).

And, a subset $C$ of $X$ is said to be saturated with respect to the surjective map $p \colon X \to Y$ if $C$ contains every set $p^{-1}(\{ y \})$ that it intersects.

Then how to establish the following?

The map $p$ is a quotient map if and only if $p$ is continuous and $p$ maps saturated open sets of $X$ to open sets of $Y$ (or saturated closed sets of $X$ to closed sets of $Y$).

An afterthought:

If $C$ is a saturated subset of $X$, then, for every $x \in C$, we have $$ p^{-1}(\{ p(x) \} ) \subset C.$$ So $$C = \bigcup_{x \in C} \{ x \} \subset \bigcup_{x \in C} p^{-1}( \{ p(x) \} ) \subset C;$$ that is, $$C = \bigcup_{x \in C} p^{-1}( \{ p(x) \} ) = p^{-1} \left( \bigcup_{x \in C} \{ p(x) \} \right) = p^{-1} \left( p(C) \right).$$

Am I right?

Hoot
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Saaqib Mahmood
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1 Answers1

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Assume $p$ is a quotient map:

  • As any open set $U \subset Y$ is open in $Y \iff p^{-1}(U)$ is open in $X$, we have that $p$ is continuous.
  • For any saturated open subset, $V \subset X$, consider $p(V)$. As $V$ is saturated, we can write $$V = \cup_{y \in p(V)}p^{-1}(y)$$ That is: $$V = p^{-1}(p(V))$$ Thus as $p^{-1}(p(V))$ is open, $p(V)$ must be too, and so $p$ maps saturated open sets to open sets.

Lets go the other way, assume $p$ is continuous and takes saturated open sets of X to open sets of Y:

  • If $U \subset Y$ is open in $Y$, $p^{-1}(U)$ is open in $X$ by the continuity of $p$. $$U \text{ open } \Rightarrow p^{-1}(U) \text{ open}$$ Now to show that if $p^{-1}(U)$ is open in $X$, then $U$ is open in $Y$

  • $p^{-1}(U)$ is open in $X$ by assumption, and further is saturated - it is the union of the $p^{-1}(y)$ s.t. $y \in U$. I claim that $$U = p(p^{-1}(U))$$ This is true because $p$ is surjective. Thus as $p^{-1}(U)$ is a saturated open set, $U$ is open by the assumption that saturated open sets are mapped to open sets. That is $$p^{-1}(U) \text{ open } \Rightarrow U \text{ open}$$

Thus we are done.

Krishan Bhalla
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  • To claim that $U = p(p^{-1}(U))$ is true, we need to have $p$ being surjective. But you haven't proved it yet. To prove $p$ is a quotient map, you still need to prove $p$ is surjective, then we can say $U = p(p^{-1}(U))$ is true. – user398843 Mar 24 '18 at 03:59
  • "And, a subset C of X is said to be saturated with respect to the surjective map p:X→Y if C contains every set p−1({y}) that it intersects."

    I believe at the time I read this to mean that p is assumed to be surjective.

     – Krishan Bhalla Apr 11 '18 at 09:37