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I read that a quotient mapping is not necessarily open. I wonder why that is.

Say we have a quotient mapping $f$ between $(X,\mathfrak{A})$ and $(Y,\mathfrak{B})$. Let us take an open set $O\subset X$. If $f(O)\neq\emptyset_Y$, then $f(O)$ is open in $Y$. If $f(O)=\emptyset_Y$, then, as $\emptyset_Y$ is open in $Y$, we still have an open mapping.

Thanks in advance!

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    $\begingroup$ math.stackexchange.com/questions/655797/… (BTW this question was shown as the first one among the related questions on the right.) $\endgroup$
    – Martin Sleziak
    Feb 22, 2014 at 9:02
  • $\begingroup$ Sorry about that. $\endgroup$
    – algebraically_speaking
    Feb 22, 2014 at 9:11
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    $\begingroup$ In the link that you've posted, I don't understand what $x=y\wedge\{x,y\}\subset \Bbb{Z}$ means. Could you kindly explain? $\endgroup$
    – algebraically_speaking
    Feb 22, 2014 at 9:20
  • $\begingroup$ Also, could you kindly point out the flaw in my argument? The definition of quotient mapping that I have read is "if we have a quotient mapping $f:X\to Y$, then for $U\subset Y$, $U$ is open only if $f^{-1}(U)$ is open." $\endgroup$
    – algebraically_speaking
    Feb 22, 2014 at 9:29
  • $\begingroup$ The wedge $\wedge$ stands for $\mathrm{AND}$. The vee $\vee$ (as it appears there) stands for $\mathrm{OR}$. Also, if you'd like a user to be notified of your comments, you need to address them, like so: @algebraically_speaking $\endgroup$
    – Jonathan Y.
    Feb 22, 2014 at 10:08

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Your argument fails because $f^{-1}(f(U))$ is different from $U$. The easiest example I can come up with is the following: Let $X=\mathbb R$, $Y=\mathbb R\big/{\sim}$ where $\sim$ identifies $0$ and $2$. Let $f\colon X\to Y$ be the quotient map $x\mapsto [x]$. Then $U=(-1,1)$ is open in $X$, but $$ f^{-1}(f(U)) = (-1,1) \cup \{2\} $$ is not open in $X$, so $f(U)$ is not open in $Y$ by the definition of the quotient topology.

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