$X$ is regular iff $Y$ is regular

Question

Let $q:X\to Y$ be a quotient function (i,e $q$ is continuous, onto and $U\subseteq Y$ is open iff $q^{-1}(U)$ is open). I need to show that $X$ is a regular topological space iff $Y$ is regular too.

I don't know how attack my problem. I tried to use the quotient space over $X$ generated by the partition $\{q^{-1}(y):y\in Y\}$ which is homeomorphic to $Y$ but I didn't get nothing.

Can someone help me?

Answer 1

A quotient space of a regular space need not be regular. Let $X=\Bbb R$ and $\sim$ be the equivalence on it with $x\sim y$ if $x=y=0$ or if $xy\ne0$. Then $Y=X/\sim$ has two points (one closed, the other not), but is not regular.