Show that for each $\mathbf{Pno}$ object $(A, \alpha, a)$ there is a unique arrow $(\mathbb{N}, \mathrm{succ}, 0) \to (A, \alpha, a)$ 
The objects of $\mathbf{Pno}$ are structures $(A, \alpha, a)$ where $A$ is a set, and where $\alpha \colon A \longrightarrow A$ is a function and $a \in A$ is a nominated element. Given two such structures a morphism
$$
  (A, \alpha, a) \longrightarrow (B, \beta, b)
$$
is a function $f \colon A \longrightarrow B$ which preserves the structure in the sense that
$$
  f \circ \alpha = \beta \circ f \,, \quad f(a) = b \,.
$$
Show that for each $\mathbf{Pno}$ object $(A, \alpha, a)$ there is a unique arrow
$$
  (\mathbb{N}, \mathrm{succ}, 0) \longrightarrow (A, \alpha, a)
$$
and describe the behaviour of the carrying function.

This exercise is in the Introduction to Category Theory by Harold Simmons. Context it is being a while since I have any mathematical classes, and I never had this course so I am just curious, and probably my attempt won't be formal:
Per the definition of a morphism in this category:
$$
  f(0) = a \,.
$$
We also know that
$$
  f \circ \mathrm{succ} = \alpha \circ f \,.
$$
Thus:
$$
  f(\mathrm{succ}(0)) = \alpha(f(0)) \,.
$$
Since $f(0) = a$, then
$$
  f(\mathrm{succ}(0)) = \alpha(a) \,.
$$
And here is where I am lost. I know that somehow I have to use the successor function, and the fact that the object we are mapping has $\mathbb{N}$ as its set. However, I am lost in how to use this information.
 A: The precise proof of this depends on a precise definition of $\mathbb{N}$. In set theory, we typically define $\mathbb{N}$ and $<$ first before getting to anything more complicated.

Lemma: for all $n \in \mathbb{N}$, there exists a unique $g_n : \{m \in \mathbb{N} \mid m < n\}$ such that (1) if $0 < n$, then $g_n(0) = a$, and (2) for all $j$, if $succ(j) < n$ then $g_n(succ(j)) = \alpha(g_n(j))$.

Proof: we proceed by induction on $n$. In the base case, the unique function $\emptyset \to A$ clearly satisfies (1) and (2).
For the inductive step, suppose we have the unique $g_k : \{0, \ldots, k - 1\} \to A\}$. Note that if we had $g_{k + 1} : \{0, \ldots, k\} \to A\}$ satisfying (1) and (2), we must have $g_{k + 1}|_{\{0, \ldots, k - 1\}} = g_k$. And we would necessarily have
$$g_{k + 1}(k) = \begin{cases}
  a & k = 0 \\
  \alpha(g_k(j)) & k = succ(j)
\end{cases}$$
This shows uniqueness. And we can simply define $g_{k + 1}$ in this way and verify it satisfies (1) and (2), showing existence. $\square$
With this Lemma in hand, we note that if we had such an $f$, then $f|_{\{0, \ldots, n - 1\}}(n) = g_n$. Thus, we would have $g_{n + 1}(n) = f(n)$ for all $n$. This shows uniqueness. Taking $f(n) = g_{n +1}(n)$ as a definition and checking properties shows existence.
