Free algebra in a variety

Remarks

Clause 1 makes the homomorphism $h$ in clause 2 unique, since two homomorphisms that agree on a generating set are equal. A free algebra on $X$ is unique up to a unique isomorphism compatible with $\eta$. If $\mathcal{V}$ contains an algebra with at least two elements, then $\eta$ is injective, and $X$ is usually regarded as a subset of $F$. Example: in the variety of abelian groups, $F(X) = \Z^{(X)}$, the group of finitely supported functions $X \to \Z$, with $\eta(x)$ the indicator function of $x$. A map $f \colon X \to A$ extends by $h(c) = \sum_{x} c(x) f(x)$, and this is a homomorphism. Non-example: $\Z / 2\Z$ with $\eta(x) = 1$ is not free on $\set{x}$ in abelian groups, since the map $x \mapsto 1 \in \Z$ has no homomorphic extension $\Z / 2\Z \to \Z$. When the signature has no constants and empty algebras are excluded, one takes $X$ nonempty.