An example of a closed quotient map that is not open

Question

We consider the map $$f\colon X:=[0,2\pi]\times [0,1]\to Y:=S^1\times [0,1],\quad\text{defined as}\quad f(t,s)=((\cos t, \sin t), s).$$

$f$ is a closed quotient map and corresponds to the equivalence relation it identifies the point $(0,s)$ with the point $(2\pi, s)$.

$f$ is an open quotient map if an open subset $A\subseteq X$ is such that $f(A)$ is open in $Y$, that is $f^{-1}(f(A))$ is open in $X$.

I cannot find an open $A$ ($A$ must not be saturated with respect to $f$) such that $f (A)$ is not open in $Y$

Could anyone help me find such an open?

$f$ is an open quotient map iff it's open and a quotient map, so it's enough to check that $f$ is not open as in the answer below. The reason for it is basically that the exponent $[0, 2\pi]\to S^1$ defined by $t\mapsto (\cos(t), \sin(t))$ is not open. — Jakobian, Feb 02 '22 at 16:48
@Jakobian Thanks for the answer, but I would like to see concretely why it is not open, giving an example with a particular open sets. — NatMath, Feb 02 '22 at 16:54
@Jakobian That's right, but I don't understand why the image isn't open. Could you help me please? How do you explicitly write that image? — NatMath, Feb 02 '22 at 17:20
You could have a look at the examples given here - and check whether some of them are closed: Example of quotient mapping that is not open — Martin Sleziak, Feb 03 '22 at 12:09

score 3 · Accepted Answer · answered Feb 02 '22 at 16:08

3

Take $A = [0,\pi) \times [0,1]$. Its image is not open in $S^1 \times [0,1]$.

answered Feb 02 '22 at 16:08

Thanks for the answer. But could you be more detailed? – NatMath Feb 02 '22 at 16:12
@NatMath What precisely do you want to know? An explanation why $A$ is open or why its image is not open? – Kritiker der Elche Feb 03 '22 at 09:58

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