In my General Topology course we were recently shown the theorem that says that Second Countability Preserved under Open Continuous Surjection and the natural question is if we can switch the "Open" condition for "Closed":


Let $T_A=(S_A,\tau_A)$ and $T_B = (S_B,\tau_B)$ be topological spaces. Let $p:T_A \rightarrow T_B$ be a surjective closed mapping that is also continuous. If $T_A$ is second countable then $T_B$ is second countable

I tried proving it and failed, because being closed is very limiting when I'm trying to apply the function to a basis so the proof for the open case can't be easily adapted :(

then I tried looking it up and nothing showed up and then I found this other thread which has a similar question but it adds the extra hypothesis that for every $y \in Y$, $p^{-1}(y)$ is compact. So now my current guess is that it's probably false without that extra assumption.

Also, I haven't been able to give a counterexample and I couldn't find on in Lynn Steen's Counterexamples in Topology

Can anyone hint at me in the direction of a counterexample or proof? Thanks.


It is indeed false in general. See this question: if we identify $\mathbb{Z} \subset \mathbb{R}$ to a point (so have the equivalence classes $\{\{x\}, x \notin \mathbb{Z}\}$ and $\mathbb{Z}$ and give the set of classes the quotient topology under the standard quotient function $q$ sending each point to its class), this is a closed map as $\mathbb{Z}$ is closed ($A$ closed, then $q^{-1}[q[A]]$ equals $A$ or $A \cup \mathbb{Z}$, so always closed, making $q[A]$ closed by definition), but the image is not even first countable at the class of $q(0)$, so certainly not second countable.

With continuous $f$ with compact fibres we sometimes get closedness of $f$ "for free", between "nice enough" spaces, not in general though.


Consider $\mathbb R$ with the usual topology, and consider the quotient space $Y = \mathbb R / \mathbb Z$ (that is, all integers identified to a point). The natural quotient mapping $q : \mathbb R \to Y$ is clearly continuous and surjective. It is also fairly straightforward to show that it is closed.

Note that the space $Y$ is not second countable. In fact, the collapsed point, let's call it $\star$, has no countable local basis, so $Y$ is not even first countable. Note that for each neighborhood $U$ of $\star$ there are positive numbers $( \delta_n : n \in \mathbb Z )$ such that $( n - \delta_n , n ) \cup ( n , n + \delta_n ) \subseteq U$ for each $n \in \mathbb Z$. If we had a countable family $\{ U_i : i \in \mathbb Z \}$ of open neighborhoods of $\star$, then for each $i$ we can take these positive numbers $( \delta_{i,n} : n \in \mathbb Z )$ as above. Now consider the set $$V = \{ \star \} \cup \bigcup_{n \in \mathbb Z} \left( ( n - \delta_{n,n}/2 , n ) \cup ( n , n + \delta_{n,n}/2 ) \right).$$ $V$ is clearly an open neighborhood of $\star$, but no $U_i$ can be a subset of $V$, since $i + \delta_{i,i}/2$ is in $U_i$, but not in $V$.


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