Does the term "free" in "free ultrafilter" have a meaning related to category theory?

Question

I know that free ultrafilters are defined in contrast to principal/fixed ultrafilters. Nonetheless, is there some categorical way to view the use of the word "free" here (e.g. some pair of adjoint functors), or some universal property that free ultrafilters satisfy?

Answer 1

That's unlikely since free ultrafilters are not unique, but universal properties always imply a kind of uniqueness. MO/410462 has more information how different free ultrafilters can be (check also the comments there).

I think the term "free" is used in a different meaning here, it is more related to injectivity. In the context of vector spaces, or more general $R$-modules, linearly independent subsets $S \subseteq M$ are sometimes called free (in the sense of: free from any unexpected relations), and this means that the canonical map from the free module $R^{\oplus S} \to M$ injective, i.e. has trivial kernel.

The non-additive analogue are group actions. A group action $G \curvearrowright X$ is called free if $gx=x$ can only happen when $g=1$ (for every $x$), so again we are "free from unexpected relations". It is equivalent to the map $G \times X \to X \times X$, $(g,x) \mapsto (gx,x)$ being injective.

If we have an ultrafilter $\mathcal{U} \subseteq P(X)$, a "relation" could be thought of as a set $T \in P(X)$ that is a subset of all the sets of $U$. It turns out that an ultrafilter is free if and only if $T = \emptyset$ is the only such set, i.e. there is only the "trivial relation". More generally, if $\mathcal{U} \subseteq P(X)$ is any family, one defines its kernel $\ker(\mathcal{U}) := \bigcap_{A \in \mathcal{U}} A$, and $\mathcal{U}$ is called free if $\ker(\mathcal{U})=\emptyset$.