Is it valid to make an admission of a topological space by a "partial quotient map"? It is well-known that the Sierpiński space, $\{F,T\}$ endowed with topology $\{\emptyset, \{F\},\{F,T\}\}$, is admissible. I tried to implement it in Haskell.
First I implement $\mathbb{N}$ (including zero; some topologists might prefer denoting it by $S_\omega$) by Peano definition:
data Peano = Zero | Succ Peano deriving (Eq, Ord)

This encodes $\mathbb{N}$, but there is one additional value lurking behind: fix Succ. I denote it by $\omega$, and I denote $\mathbb{N} \cup \{\omega\}$ by $\overline{S_\omega}$. We observe that $\overline{S_\omega}$ is in order topology.
I abuse this fact and implement the Sierpiński space by taking a quotient space. The quotient map $q$ is:
$$
q(o) =
\begin{cases}
F &amp; \text{if } o &lt; \omega \\
T &amp; \text{o.w.}
\end{cases}
$$
In Haskell, This can be realized by:
newtype Sierpinski = Sierpinski Peano

instance Eq Sierpinski where
    Sierpinski m == Sierpinski n = let
        q m = case m of
            Zero   -&gt; False
            Succ n -&gt; q n
        in q m == q n

But to think about it, q doesn't halt on $\omega$. In other words, q is partial, and doesn't match $q$. Is this really a valid implementation of the Sierpiński space?
When we implement a space we do not actually implement the space itself, but rather a representation of it. That is, an implementation of $X$ consists of a datatype $T$ and a partial surjection $\delta : T \to X$, as in Type Two Effectivity. When $\delta(p) = x$ we say that the datatype value $p \in T$ represents or realizes the point $x \in X$. Cruicially, there may be many values that represent the same point. Only $T$ exists in the computer, $X$ and $\delta$ are mathematical objects. A particularly well-behaved kind of representation is an admissible one, as you mention.
In your case, you correctly implemented an admissible representation of the Sierpinski space: $X$ is the Sierpinski space, $T$ is Sierpinski, and the surjection $\delta$ maps $\omega$ to $\bot$ and $\mathtt{Succ}^n \,\mathtt{Zero}$ to $\top$. There is no need for further quotienting.
The attempt to define == on Sierpinski is not about quotienting, but rather trying to define the characteristic map $\mathrm{eq} : X \to \{0,1\}$ of equality as a map into Booleans, which is only possible when the space is discrete. You discovered that the Sierpinski space is not discrete.
If we have a representation $\delta : T \to X$ then we can form the representation of a quotient $Y = X/{\sim}$ as $\delta' : T \to Y$ where $\delta'(p) = [\delta(p)]_{\sim}$. That is, quotienting affects the map but not the datatype.