I am asked prove or refute that if $X$ satisfies certain separation axiom and $f: X \longrightarrow Y$ is continuous and surjective then $Y$ also satisfies that certain axiom.
For now I have found counterexamples (if I am not mistaken) for $X$ being $T_0$, $T_1$, $T_2$, regular and $T_3$ but I am struggling to find one (if there's any) for $X$ being completely regular.
At first I thought that given $y \in Y$ and $K$ closed in $Y$ then by $f$ being surjective and continuous $f^{-1}(K)$ would be closed in $X$ and also there would exist $x \in X$ such that $f(x)=y$ (with $x \notin f^{-1}(K)$) and then using that by $X$ being completely regular there exists $g: X \longrightarrow [0,1]$ such that $g(x)=0$ and $g(f^{-1}(K))= \{1\}$ and using composition I would find a function $h: Y \longrightarrow [0,1]$ needed, but I cannot assure that it would be a function because $f$ not necessarily being injective.
So I think it's more likely that $f$ does not preserve being completely regular but I am struggling to find a counterexample. Any help? Thanks in advance
Let $Y$ be any space and let $X$ be $Y$ equipped with the discrete topology. Then $f$ is continuous and surjective, $X$ is completely regular, but of course $Y$ might not be.
This example might work for the other cases too.