On Wikipedia it says

A map $f: X\to Y$ is a quotient map if it is surjective, and a subset $U$ of $Y$ is open if and only if $f^{-1}(U)$ is open.

So by definition, if $f^{-1}(U) \subset X$ is open, then $U \subset Y$ is open. But I also read that a quotient map is not necessarily an open map. I saw some counter-example maps, but they were too difficult for me to understand intuitively. But if a map is not open, then there exists an open set $f^{-1}(U) \subset X$ such that $U \subset Y$ is not open. But by definition of quotient map, this is impossible, so I don't understand how this can be.

  • $\begingroup$ If the map is not open, then there exists an open set $V\subset X$ such that $f(V)$ is not open. Why do you think $V=f^{-1}(U)$? What is $U$ anyway? $\endgroup$ – bof Sep 8 '15 at 21:57

$f : X \to Y$ be your quotient map, $U \subset X$ be an open set and $f(U)$ be the image of $U$ in $Y$.

By definition, $U$ is open iff $f^{-1}(f(U))$ is open. Note here that $f^{-1}(f(U))$ needn't be the same as $U$, which is probably the source of your confusion.

For example, consider the map $\Bbb R \to S^1$ given by $x \mapsto e^{2\pi i x}$, where $S^1$ is realized as a subspace of $\Bbb C$. $[a, b]$ be an arbitrary small ($|a - b| < 1$, say) interval in $\Bbb R$. Image of $[a, b]$ via $f$ is an arc in $S^1$, while preimage of that arc is a union of translates of $[a, b]$, that is, $$f^{-1}(f([a, b])) = \bigcup_{n \in \Bbb Z} [a + n, b + n] \neq [a, b]$$

That said, $f$ is open iff for every $U \subset X$, $f^{-1}(f(U))$ is open.

For an example of a quotient map which is not open, consider the quotient map $f : \Bbb R \to \Bbb R/\Bbb Q$, $\Bbb R$ having the standard topology. Consider the open set $(1, -1) \in \Bbb R$. $f^{-1}(f((1, -1))) = (1, -1) \cup \Bbb Q$. This is not open in $\Bbb R$, hence $f$ cannot possibly be an open map.

  • $\begingroup$ $x \mapsto e^{2\pi ix}: \mathbb{R} \to \mathbb{C}$ is an open map, so it's not a very relevant example here. $\endgroup$ – Rob Arthan Sep 8 '15 at 22:11
  • $\begingroup$ @RobArthan Not my point. I was trying to give an example of a quotient map $f : X \to Y$ such that $f^{-1}(f(U)) \neq U$. $\endgroup$ – Balarka Sen Sep 8 '15 at 22:12
  • 1
    $\begingroup$ If you have read OP's question, it is relevant. He wasn't asking for a quotient map that is not open -- he was trying to contradict that there exists quotient maps which aren't open by using the flawed argument which relied on the false fact that $f^{-1}(f(U)) = U$ $\endgroup$ – Balarka Sen Sep 8 '15 at 22:16
  • $\begingroup$ Fair enough, I agree there's ambiguity about what the question asks. I will perhaps just add an example of a non-open quotient map in the answer. $\endgroup$ – Balarka Sen Sep 8 '15 at 22:21
  • $\begingroup$ @RobArthan Not true. Image of $\pi$ isn't in the image of $(-1, 1)$. $\endgroup$ – Balarka Sen Sep 8 '15 at 22:50

Consider a function $f : \mathbb{R} \to \mathbb{R}$ that preserves order as best it can but scrunches the interval $[-1, 1]$ to a point:

$$ f(x) = \left\{ \begin{array}{ll} x + 1 & \mbox{if $x < -1$} \\ 0 & \mbox{if $-1 \le x < 1$} \\ x - 1 & \mbox{if $x \ge 1$} \end{array}\right. $$ Then $f$ is a quotient mapping, but if $A$ is any non-empty open subset of $[-1, 1]$, $f(A) = \{0\}$ which is not open in $\mathbb{R}$. Such an $A$ is not equal to $f^{-1}(U)$ for any $U \subseteq \mathbb{R}$.


Definition of open maps says that:

$f$ $:$ $X$ $\to$ $Y$ is said to be open map if for each open set $U$ of $X$, the set $f(U)$ is open in $Y$.

So to prove that a map $f$ is a quotient map but not an open map take the set $X$ $=${ ${1,2,3}$} and the set

$Y$ $=$ {$a,b$}.

define $$ f(x) = \left\{ \begin{array}{ll} a & \mbox{if $x =1$ OR $2 $} \\ b & \mbox{if $ x = 3$} \ \end{array}\right. $$ Now let open sets in $X$ be {$2$}, {$3$}, and open set in $Y$ be {$b$}.

For this topology $f$ is a quotient map but not an open map, because {$2$} maps to a set not open in $Y$.

But {$2$} is not of the type $f^{-1}(U)$ for any $U$ in $Y$ so it does not violate the definition of quotient maps.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.