Is Hausdorffness preserved under continuous surjective open mappings? Is is true that Hausdorffness is preserved under continuous surjective open mappings, I tried to prove it, but I couldn't since even though the images of open sets are open but they need not to be disjoint.
Also, I tried to find a counter example that contradicts the above statement but I couldn't find such an example.
Can someone help me.
Thanks in advance.
 A: Consider $X = \{a,b\}$ with the indiscrete topology, and $f \colon \mathbb{R}\to X$, 
$$f(x) = \begin{cases} a &, x \in \mathbb{Q}\\ b &, x \notin \mathbb{Q}. \end{cases}$$
$f$ is continuous, open and surjective.
A: Another example: Let $X$ be a compact space and $A$ an open subset which is not closed but contains a closed subset $B$. If $q:X\to X/A$ denotes the quotient map, then $q$ is an open surjection. $X/A$ cannot be Hausdorff, as this would make $q$ a closed map, however the saturation of the closed set $B$ is $A$ which is not closed.
As an example, take $X=[0,2],\ A=(0,2)$ and $B=\{1\}$
On the other hand, Hausdorffness is preserved by so-called perfect mappings, that are continuous closed surjections with compact fibers. They preserve many more properties, e.g. local compactness or regularity.
A: to determine continuity of a map $f$ we look at a condition imposed by the codomain (range) topology which must be satisfied by the domain topology. basically the domain topology must be a refinement of the topology defined by the pullback (via the map) of the codomain topology. if the codomain has the trivial topology then every function is continuous, but this would only have much chance of being Hausdorff if its cardinality were less than 2. 
