# Adjoints for the functor ${\bf Top}\to {\bf Conv}$

Let $X$ be a set and let $\Phi(X)$ denote the collection of filters on $X$. For $x\in X$ we denote by $P_x$ the filter $P_x=\{A\subseteq X:x\in A\}$. A convergence space is a pair $(X,\to)$, where $X$ is a set, and $\to$ is a subset of $\Phi(X)\times X$ with the following properties:

1. $P_x \to x$ for all $x\in X$ (we write ${\cal F}\to x$ instead of $({\cal F},x)\in \to$), and
2. If ${\cal F}\subseteq {\cal G}\in \Phi(X)$ and ${\cal F}\to x$, then ${\cal G}\to x$.

If $(X,\to_X)$ and $(Y,\to_Y)$ are convergence spaces, then $f: X\to Y$ is said to be continuous if ${\cal F}\in \Phi(X), x\in X$ and ${\cal F}\to_X x$ imply $f({\cal F}) \to_Y f(x)$, where $f({\cal F})$ is the image filter of ${\cal F}$.

The category ${\bf Conv}$ consists of convergence spaces with continous maps between them. There is a functor ${\bf Top}\to {\bf Conv}$ constructed in the following way:

• Objects: $(X,\tau)$ maps to $(X,\to_\tau)$ where ${\cal F} \to_\tau x$ if and only if ${\cal F} \supseteq {\cal N}_x$, where ${\cal N}_x$ is the neighborhood filter of $x$ in $(X,\tau)$.

• Maps: It is easy to see that a continous map in the topological sense is continuous in the convergence space sense, so the functor is the identity on the morphisms.

Question. Does this functor have an adjoint going back from ${\bf Conv}$ to ${\bf Top}$?

Note. Many topological properties can be transferred to convergence spaces. For instance, a convergence space $(X,\to)$ is compact if for every ultrafilter ${\cal U}\in \Phi(X)$ there is $x\in X$ such that ${\cal U}\to x$. Also, Tychonoff's theorem can be proved for compact convergence spaces. Moreover, a convergence space is said to be $T_2$ if ${\cal F}\to x$ and ${\cal F} \to y$ imply $x=y$.

In a comment to a related question Todd Trimble gave a link to the nLab entry on relational $$\beta$$-modules which I think answers much of this question.
It [the construction of the functor $$\mathbf{Conv} \to \mathbf{Top}$$] is at that nLab page, just before Proposition 2.1: $$U$$ is open iff for any $$\mathcal F \to x$$ with $$\mathcal F$$ an ultrafilter, $$x \in U$$ implies $$U \in \mathcal F$$.