Proving a polynomial ring isomorphism From Leinster - Basic  Category Theory 2016 p8
My question is how to solve (b) below. In particular it was my thought that, using the homomorphism property proved in (a), the supplied definition of $\psi$ in (b) has no relevance to the solution. Would someone please tell me how am I mistaken?
Denote by $\mathbb{Z}[x]$ the polynomial ring over $\mathbb{Z}$ in one variable.
(a) Prove that for all rings $R$ and all $r ∈ R$, there exists a unique ring homomorphism $φ: \mathbb{Z}[x] \rightarrow R$ such that $φ(x) = r.$
(b) Let $A$ be a ring and $a ∈ A$. Suppose that for all rings $R$ and all $r ∈ R$, there
exists a unique ring homomorphism $φ: A → R$ such that $φ(a) = r$. Prove
that there is a unique isomorphism $ι: \mathbb{Z}[x] → A$ such that $ι(x) = a.$
Thank you
 A: Hint: This is an example of a universal property. Using part (a), you have a unique morphism from $\mathbb{Z}[x]$ to $A$ sending $x$ to $a;$ call this $\iota.$ To prove $\iota$ is an isomorphism, you just need to construct its inverse. By what property of $A$ can you construct a candidate map for $\iota^{-1} : A \rightarrow\mathbb{Z}[x],$ and how can you show they are inverses?
A: $\iota \colon \mathbb{Z}[x] \rightarrow A$ maps
$\sum_{i=1}^n p_ix^i$ to $\sum_{i=1}^n p_ia^i$, using $\iota(x) = a$,
the multiplicative property of a homomorphism to get $\iota(x^i)=\iota(x)^i$,
and the additive property to get $\iota(p_i)\iota(x)^i = p_i\iota(x)^i$ remembering $p_i$ is in $\mathbb{Z}$.
Going in the direction $A \rightarrow \mathbb{Z}[x]$ we know as provided in (b) above that,
taking $R=\mathbb{Z}[x]$, and $\psi=\iota^\prime$, there exists a unique ring homomorphism such that $\iota^\prime(a)=x$.
So $\iota^\prime $ maps $\sum_{i=1}^np_ia^i$ to $\sum_{i=1}^{n}p_ix^i$ and $\iota^\prime \circ \iota = 1_{\mathbb{Z}[x]}$.
Also using definitions of $\iota$ and $\iota^\prime$ easily yields $\iota \circ \iota^\prime = 1_A$
