# Concrete category of topological spaces over preordered sets: what are the initial morphisms?

I am reading The Joy of Cats to become more familiar with category theory and I came upon the following question on concrete categories.

Let Top be the category of topological spaces and let Prost be the category of preordered sets, with monotone maps as morphisms. Since every topological space is equipped with the specialization preorder and since a continuous function is monotone with respect to this preorder, there is a natural forgetful functor from Top to Prost.

I now consider Top as a concrete category over Prost. According to The Joy of Cats (p. 134), a morphism $f: X \to Y$ in Top is called initial if for each topological space $Z$ and for each monotone map $g: Z \to X$, the condition $f \circ g$ continuous implies $g$ continuous.

It is not difficult to see that any continuous map $f: X \to Y$, where $X$ is equipped with the initial topology defined by $f$, is an initial morphism.

Question. Is the converse true? If not, what are the initial morphisms in this case?

N.B. The converse is true if Top is considered as a concrete category over Set, but this case is different.

My colleague Mai Gehrke found a positive answer to my question: if $f : X \to Y$ is an initial morphism of Top over Prost, then $X$ is equipped with the initial topology defined by $f$.
Proof. Let $f: X \to Y$ be an initial map.
Step 1. Let $x,y \in X$. If $f(x) \leqslant f(y)$, then $x \leqslant y$.
Consider the cofinite topology on $\mathbb{N}$ and let $g: \mathbb{N} \to X$ be the map defined by $g(0) = x$ and $g(n) = y$ if $n > 0$. Since the specialization order defined by the cofinite topology is the equality relation, $g$ is a monotone map. Let $h = f \circ g$. Then $h(0) = f(x)$ and $h(n) = f(y)$ if $n > 0$. I claim that $h$ is continuous. Indeed, let $U$ be an open subset of $Y$. Since $f(x) \leqslant f(y)$, the condition $f(x) \in U$ implies $f(y) \in U$. It follows that $h^{-1}(U)$ is either equal to $\emptyset$, $\mathbb{N}$ or $\mathbb{N} - \{0\}$ and hence is open in all cases, which proves the claim. Since $f$ is an initial map and $g$ is monotone, $g$ is continuous. Thus if $V$ is an open subset of $X$ containing $x$, $g^{-1}(V)$ is an open subset of $\mathbb{N}$ containing $0$. Therefore $g^{-1}(V)$ is cofinite and contains some $n > 0$. It follows that $y = g(n)$ belongs to $V$. Thus any open subset containing $x$ also contains $y$, that is, $x \leqslant y$.
Step 2. Let $\mathcal{T}$ be the topology on $X$ and let $\mathcal{I}$ be the initial topology on $X$ defined by $f$. Since $f$ is continuous, $\mathcal{I} \subseteq \mathcal{S}$. Let $i: (X, \mathcal{I}) \to (X, \mathcal{S})$ be the identity map. Then $f \circ i:(X, \mathcal{I}) \to Y$ is continuous by definition of the initial topology. Moreover, suppose that $x \leqslant_\mathcal{I} y$. Since $f \circ i$ is continuous, it is monotone with respect to the specialization orders and hence $f(x) \leqslant f(y)$. It follows by Step 1 that $x \leqslant_\mathcal{T} y$. Thus $i: (X, \mathcal{I}) \to (X, \mathcal{S})$ is monotone and since $f$ is initial and $f \circ i$ is continuous, $i$ is a continuous map from $(X, \mathcal{I})$ to $(X, \mathcal{S})$. It follows that $\mathcal{S} \subset \mathcal{I}$ and finally $\mathcal{S} = \mathcal{I}$. Thus the topology on $X$ is the initial topology defined by $f$.