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.

  • $\begingroup$ Does it make $(\mathbb{N}, succ, 0)$ initial object of that category? $\endgroup$ – Sereja Bogolubov Sep 25 '18 at 18:13
  • 1
    $\begingroup$ @SerejaBogolubov Yes, since "there is a unique morphism from --- to any arbitrary object" is exactly the definition of initial object. $\endgroup$ – Noah Schweber Sep 25 '18 at 18:13

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