Example of quotient mapping that is not open I have the following definition: 
Let ($X$,$\mathcal{T}$) and ($X'$, $\mathcal{T'}$) be topological spaces. A surjection $q: X \longrightarrow X'$ is a quotient mapping if $$U'\in \mathcal{T'} \Longleftrightarrow q^{-1}\left( U'\right) \in \mathcal{T} \quad \text{i.e. if } \mathcal{T'}=\{ U' \subset X' : q^{-1}\left( U' \right) \in \mathcal{T} \}$$
and the properties:


*

*$q$ is a bijective quotient mapping $\Leftrightarrow$ $q$ is a homeomorphism 

*In general, $q$ quotient $\not \Rightarrow q$ open. 
If $U \in \mathcal{T}, q(U)\subset X'$ is open if $q^{-1}\left( q\left( U \right) \right) \in \mathcal{T}$ but not in general.


I could not find an example of quotient mapping for which $q^{-1}\left( q\left( U \right) \right)$ is not open. I would understand the idea better if you could show me one. 
 A: Here is a "visual" example: 
Wrap the unit interval $[0,1]$ around the unit circle $S^1$ by identifying $0$ and $1$. The interval $[0,1/2) $ is open in $[0,1]$, but its image in $S^1$ is not open.
A: Let $X$ be a Hausdorff space, and let $a,b$ be a pair of distinct points. Stipulate that $\{b\}$ is not open. Define $f:X\to X\setminus\{b\}$ by $f(z)=z$ if $z\ne b$ and $f(b)=a$. There is a unique topology we may assign the codomain such that $f$ is a quotient map, let's endow that topology on $X\setminus\{b\}$. 
Then, pick disjoint neighborhoods $U\ni a,V\ni b$, and note that $f^{-1}(f(U))=U\cup\{b\}$. This set is not open; for, if it were open, then there would be a neighborhood of $b$--call it $W$--contained in $U\cup\{b\}$. So, $\{b\}=(W\cap V)\cap(U\cup\{b\})$ is open, a contradiction.
A: Consider a space with three points $a,b,c$ and open sets
$\varnothing, \{a\},\{a,b\},\{a,b,c\}$. The projection that identifies $a$ and $c$ is not open (or closed, for that matter.)
A: Consider $\mathbb{R}$ with the standard topology. On $\mathbb{R}$, consider the equivalence relation
$$x\sim y \iff (x = y \lor \{x,y\} \subset \mathbb{Z}),$$
and let $(X',\mathcal{T}')$ the quotient space $\mathbb{R}/{\sim}$. By definition, $\pi \colon \mathbb{R}\to X';\; x \mapsto [x]_\sim$ is a quotient map, but that map is not open:
If $U \subset\mathbb{R}$ is an open set containing an integer, then $\pi^{-1}(\pi(U)) = U\cup \mathbb{Z}$ is in general not open.
A: Partition $X=ℝ$ (under the usual topology) into two equivalence classes $(-\infty, 0]$ and $(0,\infty)$ to get a two-point quotient space $X'$ (called the Sierpiński space) and let $q:X\to X'$ be the canonical projection mapping that takes each element of $ℝ$ to its equivalence class.  Then $q$ is a quotient mapping that satisfies $q^{-1}(q((-\infty,0)))=(-\infty,0]$.
A: An open map is:
for any open set in the domain, its image is open
A quotient map only requires:
for any open preimage in the domain, its image is open
A quotient map may not be an open map because there may exist a set such that it is open but is not a preimage, i.e., the preimage of the image of an open set may not be the open set itself.
