Initial algebra

Remarks

Example: on $\mathbf{Set}$, let $F X = \set{\ast} \sqcup X$, and let $\iota \colon F \N \to \N$ send $\ast$ to $0$ and $n$ to $n + 1$. Then $(\N, \iota)$ is an initial $F$-algebra. Given $(B, \beta)$, the condition on $h$ says $h(0) = \beta(\ast)$ and $h(n + 1) = \beta(h(n))$, and these equations define exactly one function $h$ by recursion.

Non-example: $(\Z, \iota')$ with $\iota'(\ast) = 0$ and $\iota'(n) = n + 1$ is an $F$-algebra but not an initial one. There is no homomorphism $h$ from it to $(\N, \iota)$: it would satisfy $h(0) = 0$ and $h(-1) + 1 = h(0) = 0$, which has no solution in $\N$.

For an algebraic signature $\sigma$ and the functor $F_\sigma X = \bigsqcup_n \sigma_n \times X^n$, the initial $F_\sigma$-algebra is the algebra of closed terms of $\sigma$.