Soberification of a topological space In Johnstone´s  Stone Spaces, he introduces the concept of soberification of a topological space:

Let $X$ be a topological space and $\Omega(X)$ the lattice of open subsets of $X$, the soberification of $X$ is $pt(\Omega(X))$, where $pt(A)$ is the set of prime elements of $A$, for any set $A$.

$pt(\Omega(X))$ is the of all principal prime open sets, i.e., open sets whose complements are irreducible closed sets.

The function $\psi: X \to pt(\Omega(X))$ sends a point $x\in X$ to  $\overline{ \{ x\} }^c$.
$pt(\Omega(X))$ has the topology of subspace of $Spec(\Omega(X))$.

I would like to prove that this function is always continuous, and it is injective if and only if $X$ is $T_0$.

To prove that $\psi$ is injective if and only if $X$ is $T_0$ is really simple, because $\psi(x)=\psi(y)$ iff $\overline{ \{ x\} }^c$=$\overline{ \{ y\} }^c$ iff $x, y$ are contained in exactly the same open sets.
So, $\psi(x)=\psi(y)$ implies $x=y$ iff $X$ is $T_0$.

So, the proof of continuity is left.
I was also wondering when this function is onto.
Any help or hint is appreciated.
Edit: So far I've managed to come up with a characterization of $X$ so that $\psi$ is onto:

$\psi$ is onto if and only if for any $x,y\in X$ such that $y\notin\overline{\{x\}}$ we have $\overline{\{x,y\}}$ is not connected.

Is there a name for this kind of spaces?
 A: The most important thing to note in order to prove that $\psi$ is continuous is the fact that opens of $pt(\Omega(X))$ are defined by the opens of $X$, i.e. the opens of $pt(\Omega(X))$ are precisely sets of the form $\phi(U) = \{V \in pt(\Omega(X)) \mid U$ is not contained in $V\}$, where $U \in \Omega(X)$ (see the wikipedia page on Stone duality). Take such an open $\phi(U)$ - we are going to consider $\psi^{-1}(\phi(U))$.
It holds that $x \in \psi^{-1}(\phi(U))$ if and only if $\overline{\{x\}}^c \in \phi(U)$, so if and only if $U$ is not contained in $\overline{\{x\}}^c$. It follows that $U \subset \psi^{-1}(\phi(U))$, since $x \notin \overline{\{x\}}^c$ for all $x \in U$. Suppose that $x \notin U$. Then $x \in U^c$, so $\overline{\{x\}} \subset U^c$ and $U \subset \overline{\{x\}}^c$, so $x \notin \psi^{-1}(\phi(U))$. Thus $U = \psi^{-1}(\phi(U))$, which shows that $\psi$ is continuous.
The function $\psi$ is onto if and only if every closed irreducible subset of $X$ is the closure of a point $x \in X$. This holds if $X$ is an affine scheme, for example. I was unable to further specify when $\psi$ is onto.
