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Definition. If $p:X\rightarrow Y$ is a function, a subset $U\subseteq X$ is said to be saturated (with respect to $\mathbf{p}$) if $U=p^{-1}(p(U))$.

Proposition. Let $X$ and $Y$ be topological spaces and let $F:X\rightarrow Y$ be a surjective continuous function. The following are equivalent:

  1. $F$ is a quotient map.

  2. $F$ maps saturated open subsets of X to open subsets of Y.

  3. $F$ maps saturated closed subsets of $X$ to closed subsets of $Y$.

I know that $1$ easily implies $2$ and $3$ and I'm sure I can prove that the reverse implication is true but I can't prove that $2$ implies $3$. I think it involves showing that the complement of a saturated set is saturated but I haven't been able to show that.

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    $\begingroup$ If you can prove the reverse implication doesn't that automatically prove equivalence? $\endgroup$ – user160738 Dec 22 '16 at 9:57
  • $\begingroup$ By reverse implication, I mean that 2 and 3 imply 1. $\endgroup$ – Eigenfield Dec 22 '16 at 11:36
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As you mentioned, we only need to prove that the complement of a saturated set is saturated. Let $U^c$ be a saturated set. We need to show $U=p^{-1}(p(U))$. For any set $U$ it's evident that $U\subset p^{-1}(p(U))$, so what remains is $p(x)\in p(U)\implies x\in U$. Suppose on the contrary that $x\not\in U$, then we would have $x\in U^c$, thus $p(x)\in p(U^c)$. Since $U^c$ is saturated, $p^{-1}(p(x))\subset p^{-1}(p(U^c))=U^c$. But $p(x)\in p(U)$, which indicates $p^{-1}(p(U))\cap U\not=\emptyset$, a contradiction.

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  • $\begingroup$ I am having trouble seeing that the complement of a saturated set is saturated will help prove that 2 implies 3. $\endgroup$ – Eigenfield Dec 22 '16 at 12:54
  • $\begingroup$ @Eigenfield The argument above shows that $p(U)\cap p(U^c)=\emptyset$. And note that $p$ is surjective, which indicates $p(U)\cup p(U^c)=Y$. $\endgroup$ – Cave Johnson Dec 22 '16 at 12:58
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Define an equivalence relation on $X$ by $x\sim y$ iff $p(x)=p(y)$. If you can prove that: $$U \textrm{ is saturated } \Leftrightarrow \forall u\in U \,\, \forall v\in X \big( u\sim v \Rightarrow v\in U\big)$$ then the fact that the complement of a saturated set is saturated will follow immediately.

Note that the right hand side just says that $U$ is the union of the equivalence classes of its members.

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