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Let $\mathcal{K}$ be a class of algebraic systems (of the same type $\tau$). Consider an algebra $A\in\mathcal{K}$ generated by a set $X=\{x_{i}\}$ indexed by $i\in I$. $A$ is said to be a free algebra over $\mathcal{K}$, with free generating set $X$, if for any algebra $B\in\mathcal{K}$ with any subset $\{y_{i}\mid i\in I\}\subseteq B$, there is a homomorphism $\phi:A\to B$ such that $\phi(x_{i})=y_{i}$.
If we define $f:I\to A$ to be $f(i)=x_{i}$ and $g:I\to B$ to be $g(i)=y_{i}$, then freeness of $A$ means the existence of $\phi:A\to B$ such that $\phi\circ f=g$.
Note that $\phi$ above is necessarily unique, since $\{x_{i}\}$ generates $A$. For any $n$ary polynomial $p$ over $A$, any $z_{1},\ldots,z_{n}\in\{x_{i}\mid i\in I\}$, $\phi(p(z_{1},\ldots,z_{n}))=p(\phi(z_{1}),\ldots,\phi(z_{n}))$.
For example, any free group is a free algebra in the class of groups. In general, however, free algebras do not always exist in an arbitrary class of algebras.
Remarks.

$A$ is free over itself (meaning $\mathcal{K}$ consists of $A$ only) iff $A$ is free over some equational class.

If $\mathcal{K}$ is an equational class, then free algebras exist in $\mathcal{K}$.

Any term algebra of a given structure $\tau$ over some set $X$ of variables is a free algebra with free generating set $X$.
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