Example of quotient mapping that is not open

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 1

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.

Answer 2

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.

Answer 3

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.