F-algebra

Remarks

Example: on $\mathbf{Set}$, let $F X = \set{\ast} \sqcup X$. The natural numbers with $\alpha(\ast) = 0$ and $\alpha(n) = n + 1$ form an $F$-algebra $(\N, \alpha)$. More generally, for an algebraic signature $\sigma$ with $\sigma_n$ the set of $n$-ary symbols, the $F_\sigma$-algebras for $F_\sigma X = \bigsqcup_n \sigma_n \times X^n$ are exactly the $\sigma$-algebras, and their homomorphisms are the usual ones.

Non-example: a morphism $\gamma \colon A \to F A$ in the other direction does not make $A$ an $F$-algebra; such a pair $(A, \gamma)$ is an F-coalgebra. For $F X = \set{\ast} \sqcup X$ on $\mathbf{Set}$, a coalgebra describes a system in which each state either stops or has one next state.

If $(T, \eta, \mu)$ is a monad, every algebra over the monad is a $T$-algebra in this sense, but not conversely, since no unit or associativity law is required here.