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There are several similar questions on math.SE, but I couldn't find a resolution of this specific case. I apologize if it turns out this is duplicate.

I'm trying to prove the following property concerning quotient maps: the restriction of a quotient map to a saturated open or closed subset is a quotient map.

Here is my attempt for the case where the restricted domain is open.

Let $X,Y$ be topological spaces, $A$ be a saturated open subset of $X$, and $p:X \to Y$ be a quotient map. Define $q:A \to p(A)$ by $q = p|_A$. We prove that $q$ maps saturated open subsets to open subsets. Let $U \subset A$ be saturated w.r.t. $q$ and open in $A$. We prove it's saturated w.r.t. $p$ and open in $X$. If it weren't saturated, there would be some point in $p^{-1}(p(U))$ which does not lie in $U$. It must be contained in either $A$ or the complement of $A$ in $X$. It cannot lie in $A$, because $U$ is saturated w.r.t. $q$, which agrees with $p$ on $A$. And it cannot lie outside of $A$, because $A$ itself is saturated w.r.t. $p$, and our point is in $U \subset A$. So it must be in $U$. And $U$ is open by definition of the subspace topology, since $A$ is open. So $p$ takes $U$ to an open set in $Y$, and by definition of the subspace topology, $p(U) \cap p(A) = p(U)$ is open in $p(A)$. So $q$ is a quotient map.

But this finagling with the definition of the subspace topology doesn't work as nicely when $A$ is closed instead. Can anyone give me some pointers?

And my other question is: can anyone give a good counterexample for when $A$ is saturated but neither open nor closed?

lanf
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  • Can you think of a characterisation of quotient maps in terms of closed sets? – Daniel Fischer Jan 11 '21 at 18:45
  • Ah! Yes, a quotient map sends saturated closed subsets to closed subsets. But then to use the same strategy, we want to show that a saturated (w.r.t. $q$) closed subset $U \subset A$ is saturated (w.r.t. $p$) and closed in $X$, where $A$ is a saturated (w.r.t. $p$) closed subset of $X$. We can get that $U$ is saturated w.r.t. $p$, but seeing that it is closed in $X$ is escaping me. – lanf Jan 11 '21 at 21:31

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I figured it out with Daniel Fisher's hint in the comments. A quotient map sends saturated closed subsets to closed subsets. But then to use the same strategy, we want to show that a saturated (w.r.t. ) closed subset $U\subset A$ is saturated (w.r.t. ) and closed in , where is a saturated (w.r.t. ) closed subset of . We can get that is saturated w.r.t. $p$ by the same argument as above. We can see that it is closed in X via: Is it true that a subset that is closed in a closed subspace of a topological space is closed in the whole space?

Then $p$ sends $U$ to a closed subset $p(U) \subset p(A)$, since $p$ is quotient. We know $p(A)$ is closed in $Y$, so $p(U)$ is the intersection of a closed subset in $X$ (itself) with $p(A)$, hence closed in $p(A)$.

lanf
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