Confused about definition of quotient map 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.
 A: $f : X \to Y$ 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.
A: 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 $U \subseteq \mathbb{R}$.
A: 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.
