Can the quotient topology be defined in terms of neighbourhoods rather than open sets?

Question

If $f:X\rightarrow Y$ is surjective, define that a set $U$ in $Y$ is a neighbourhood of $y \in Y$ if and only if its preimage is a saturated neighbourhood containing the preimage of y. Does this work to define the usual quotient topology? Because in treating a cylinder as quotient of a square, it is much easier to define neighbourhoods of the identified edges as union of two half discs of equal radii.

Answer 1

Unfortunately, this does not work in general. For instance, let $X=\mathbb{R}$ and let $f:X\to Y$ be the quotient map for the equivalence relation which identifies $1/n$ with $n$ for each $n\in\mathbb{Z}_+$. Let $V=(-1,1)\cup\mathbb{N}$. Then $V$ is a saturated neighborhood of the equivalence class of $0$ (which is just $\{0\}$), but $f(V)$ is not a neighborhood of $f(0)$ in the quotient topology since any saturated open set containing $0$ must contain an open interval around $n$ for all sufficiently large $n\in\mathbb{N}$.

It is true if $X$ is compact Hausdorff and the equivalence relation $\sim$ corresponding to $f$ is closed. In fact, more strongly, if $y\in Y$ and $V\subseteq X$ is any neighborhood (not necessarily saturated!) of $f^{-1}(\{y\})$, then $f(V)$ is a neighborhood of $y$ in the quotient topology. To prove this, we must show that $V$ contains a saturated open neighborhood of $A=f^{-1}(\{y\})$. Let $W$ be open such that $A\subseteq W\subseteq V$. Note that the saturation $C$ of $X\setminus W$ is closed, since it is the image of the compact set ${\sim}\cap (X\setminus W)\times X\subseteq X\times X$ under the second projection $X\times X\to X$. Then $X\setminus C$ is a saturated open neighborhood of $A$ contained in $V$, as desired.