Question: 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:

1. $q$ is a bijective quotient mapping $\Leftrightarrow$ $q$ is a homeomorphism
2. 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.

Answer: 
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.
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](https://en.wikipedia.org/wiki/Sierpi%C5%84ski_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 $q^{-1}(q((-\infty,0)))=(-\infty,0]$.