From Willard's "General Topology" (section 13.3), [edit: I thought that] it seems that the T1 separation axiom is preserved by the quotient topology, meaning that it is preserved by continuous maps.
(Edit: Willard is actually claiming exactly the opposite, and this is clarified in exercise 13B that I failed to look at before coming here...)
How is it so? The author is leaving it as an exercise (and I am failing).
I have tried to prove it by showing that, if $X$ is $T_1$, $\forall f: X \rightarrow Y$ a continuous map, $\forall y \in Y$ we would have $f^{-1}(y)$ closed in $X$.
But, while it would have been easy if $f^{-1}(y)$ was finite, this seems tough in the general case...
Could anybody give me a hint to the general proof?
(Edit: of course I should have looked at the book more carefully... Now, thank you anyway for the great help! I could have overlooked that several more days without this precious feedback...)
The exercise he refers to is 13.B and that says: show that the quotient space of a $T_1$ space need not be $T_0$ but that the closed image of a $T_1$ space is $T_1$.
The former can be shown by example (decompose $\Bbb R$ into two classes $\Bbb Q$ and the irrationals and we get a two point indiscrete quotient space) while the latter is a trivial consequence of $\{y\}= f[\{x\}]$ for some preimage point $x$ of $y$, where we use $Y$ (or $X$) is $T_1$ iff all singletons are closed.