This is a follow-up question to [Existence of injective neighborhood selection function as separation axiom](https://mathoverflow.net/questions/199555/existence-of-injective-neighborhood-selection-function-as-separation-axiom).


Let $(X, \tau)$ be a topological space. If there is an injective map $f:X\to\tau$ such that $x\in f(x)$ for all $x\in X$, we call $f$ an *injective neighborhood selection map*. 


We say a space is *critical* if it has an injective neighborhood selection map and every such map $f:X\to\tau$ is "almost surjective", i.e. $\text{im}(f) = \tau\setminus\{\emptyset\}$.


For example, if $\alpha$ is an ordinal, consider $\tau\_\alpha = \{\emptyset\} \cup \{\alpha\setminus \beta: \beta \in \alpha\}$, which is the topology of upper sets. Then $(\alpha,\tau\_\alpha)$ is critical.


Is every critical space isomorphic to $(\alpha, \tau\_\alpha)$ for some ordinal $\alpha$? If not, is there an example of a critical space that is $T\_1$ or even Hausdorff? 


[**EDIT**: The second question has become obsolete, see Eric Wofsey's comment below.]



Critical topological spaces
The answer is no. Consider the space $X=\{0,1,2\}$ with the topology
$\tau=\{\emptyset,\{0\},\{1,2\},X\}$. There are precisely two
injective neighborhood selectors, both of which are almost surjective in your sense. Namely, we must 
map $0\mapsto\{0\}$ and then $1$ and $2$ get mapped to $\{1,2\}$
and $X$, in either way.


So this space is critical in your sense, but the topology is not
of your form $\tau\_\alpha$ for any ordinal $\alpha$. For example, it
is not $T\_0$, since we cannot separate $1$ and $2$.


