Universal coalgebra

Remarks

Example: let $A$ be a set and $F(X) = A \times X$, with $F(h)(a, x) = (a, h(x))$. An $F$-coalgebra is a set $X$ with a map $\alpha(x) = (o(x), n(x))$, an output $o(x) \in A$ and a next state $n(x) \in X$. The set $A^{\N}$ of streams with $\beta(s) = (s_0, (s_1, s_2, \dots))$ is such a coalgebra. For every coalgebra $(X, \alpha)$, the map $h(x) = (o(x), o(n(x)), o(n(n(x))), \dots)$ satisfies $\beta \circ h = F(h) \circ \alpha$, and it is the only such map, so the streams form a final $F$-coalgebra.

Non-example of a homomorphism: let $A = \set{a, b}$, let $X = \set{x}$ with $\alpha(x) = (a, x)$, and let $Y = \set{y_1, y_2}$ with $\beta(y_1) = (a, y_2)$ and $\beta(y_2) = (b, y_2)$. The map $h(x) = y_1$ keeps the output of $x$, but $\beta(h(x)) = (a, y_2)$ while $F(h)(\alpha(x)) = (a, y_1)$, so the square does not commute.

The same definition, with any category in place of sets, gives the F-coalgebra.