# Objects of the $\mathcal{Pno}$ category?

The objects of $$\mathcal{Pno}$$ catergory are structures $$(S, \lambda, s)$$ where $$S$$ is a set, $$\lambda: S \mapsto S$$ is a functions and $$s \in S$$ is a nominated element. Given two such structures, a morphism $$f: (S, \lambda, s) \mapsto (S', \lambda', s')$$ is a function $$f: S \mapsto S'$$ which pereserves the structure in the sense that: $$f(s) = s'$$, $$f \circ \lambda = \lambda' \circ f$$, where $$(\circ)$$ represents function composition.

I need to show that for each $$\mathcal{Pno}$$-object $$(S, \lambda, s)$$ there exists a unique morphism: $$(\mathbb{N}, succ, 0) \rightarrow (S, \lambda, s)$$, where $$succ$$ is the successor function.

Given $$O = (S = \{ a, b \}, \lambda = \{ (a, b); (b, a) \}, a)$$. According to the abovementioned requirements, I see no reason why object $$O$$ is not a memeber of the $$\mathcal{Pno}$$. However there is no such structure-preserving morphism. Seems reasonable to define $$f$$ as follows: $$\{ (a, 0); (b, 1) \}$$. Than: $$(f \circ \lambda) b = 0 \not = (succ \circ f) b = 2$$. Redefining $$f$$ as the $$\{ (a, 0); (b, 0); \}$$ neither works: $$(f \circ \lambda) b = 0 \not = (succ \circ f) b = 1$$.

Hence, my question is rather simple: am I misunderstanding something or the abovementioned requirements for structure to be a $$\mathcal{Pno}$$-object are incomplete and thus my custom $$O$$ structure does not belong to this category?

You have the morphisms going the wrong way. You want to show that there is a unique morphism from the naturals to your $$\mathcal{Pno}$$.
In your example, this morphism sends all even naturals to $$a$$ and all odd naturals to $$b$$. You can check that this satisfies the preservation requirements. Meanwhile, what you've shown is that there is no morphism the other way.
Incidentally, this should suggest how to define $$f$$ in general ...
It might help to first consider a close analogue of this fact in a more classical context: that there is a unique group homomorphism from $$\mathbb{Z}$$ to $$\mathbb{Z}/2\mathbb{Z}$$, but no homomorphism at all the other way.
• Does it make $(\mathbb{N}, succ, 0)$ initial object of that category? – Sereja Bogolubov Sep 25 '18 at 18:13