Why quotient map maps open sets to open sets?

Question

My book says that the quotient map of a topological space $X$ into its equivalence classes under some equivalence relation maps open sets to open sets.However my intuition tells me that this is wrong for the following reason.Let the situation be as above and suppose the set of points in $X$ forming one equivalence class is not open, name this set $C$.Now suppose you can find an open subset $U$ of $C$. Then $f(U)$ is just the equivalence class we started with. To decide if $f(U)$ is open by definition we must have $f^{-1}f(U)$ is open but it is not since it is the set $C$ which by construction is not open.

What am I saying wrong here?

Thanks in advance

Your argument is fine. Does your book make special assumptions about the equivalence classes? — Hagen von Eitzen, Dec 28 '14 at 15:48
No it does not, actually it's just some printed notes that a student gave me. I guess the notes are just wrong then — TheGeometer, Dec 28 '14 at 15:52
@user142775 The open sets in $X/{\sim}$ are precisely those which are open viewed in $X$. — user2345215, Dec 28 '14 at 16:12

score 1 · Answer 1 · answered Jan 31 '25 at 15:11

It is false. Each open map (i.e. one which maps open sets to open sets) is a quotient map, as it is shown here, but the converse is not true.

As a counterexample consider the map $f : [0,1] \to S^1 = \{ p \in \mathbb R^2 \mid \lVert p \rVert = 1 \}, f(t) = (\cos 2 \pi t, \sin 2 \pi t)$. This is a continuous surjection. Since $[0,1]$ is compact and $S^1$ is Hausdorff, it is a closed map (i.e. one which maps closed sets to closed sets). As a closed map it is a quotient map (see again here). However, $f$ is not an open map. For example $U= [0,1/2)$ is open in $[0,1]$, but $f(U)$ is not open in $S^1$.

Why quotient map maps open sets to open sets?

1 Answers1