Free Boolean algebra

Remarks

It exists for every $S$ and is unique up to a unique isomorphism compatible with $\iota$. For finite $S = \set{1, \ldots, n}$, one model is the algebra of all Boolean functions $\set{0, 1}^n \to \set{0, 1}$, with $\iota(i)$ the $i$-th coordinate function; it has $2^{2^n}$ elements and $2^n$ atoms. For any $S$, one model is the algebra of clopen subsets of the product space $\set{0, 1}^S$. The free Boolean algebra on a countably infinite set is atomless.

Example: for $S = \set{x}$, the algebra $\set{0, x, \neg x, 1}$ with $x$ and $\neg x$ incomparable is free. A map $g$ extends by $\neg x \mapsto \neg g(x)$, $0 \mapsto 0$, $1 \mapsto 1$, and this extension is forced.

Non-example: the power set of $\set{1, 2}$ with $\iota(x) = \set{1}$ and $\iota(y) = \set{2}$ is generated by the image of $\iota$ but is not free on $\set{x, y}$. For the map $g(x) = g(y) = 1$ into the two-element algebra, a homomorphism $\bar g$ would send $\set{1} \cap \set{2} = \emptyset$ to $1 \wedge 1 = 1$, but a homomorphism sends $\emptyset$ to $0$.