I’m working on the following question. Describe all of the ring homeomorphisms from $\mathbb{Z}[x]$ to an arbitrary unitary ring $R$. As I understand this question could be answered with the following proposition
Proposition 1: Let $R$ be an arbitrary unitary ring, then $f\in \mathrm{Hom}\left(\mathbb{Z}[x],R\right)$ ($f$ is a ring homomorphism) $\Longleftrightarrow$ $\exists! r\in R \ \forall P(x)\in\mathbb{Z}[x] (f(P(x)) = P(r))$.
This means that any homomorphism $f:\mathbb{Z}[x]\to R$ could be described by unique $r\in R$ such that $f$ maps any polynomial from $\mathbb{Z}[x]$ to its value at $x = r$.
Proof: The sketch of proof looks like this:
($\Leftarrow$) For the fixed $r$, the substitution of $r$ in $P(x)$ is obviously a ring homomorphism, as manipulations with polynomials preserve the unitary ring structure.
($\Rightarrow$) For a fixed $f:\mathbb{Z}[x] \to R$ let’s denote $f(x) = r$ (here $x$ is a polynomial $P(x) = x$). Now the proof is a straightforward calculation. $\square$
Somehow I still have some doubts about this reasoning. And also I wonder if the proof is the same for the generalized proposition
Proposition 2: Let $R$ be an arbitrary unitary ring, then $f\in \mathrm{Hom}\left(\mathbb{Z}[x_{1},x_{2},\dots,x_{n}],R\right)$ $\Longleftrightarrow$ $\exists! \vec{r}=(r_{1},r_{2},\dots,r_{n})\in R^{n} \ \forall P(x_{1},x_{2},\dots,x_{n})\in\mathbb{Z}[x_{1},x_{2},\dots, x_{n}] (f(P(x_{1},x_{2},\dots, x_{n})) = P(r_{1},r_{2},\dots, r_{n}))$.