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:

- $P_x \to x$ for all $x\in X$ (we write ${\cal F}\to x$ instead of $({\cal F},x)\in \to$), and
- 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 $x=y$.