Let $z\in R$ be fixed then the map $\phi:R[x]\rightarrow R $ defined as $\phi(f(x))=f(z)$ is surjective? Could someone please explain me why it is.
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Let $r \in R$; where does $\phi$ send the polynomial $f$ given by $f(x)=x+r-z$?
[Edit] Or, far more simply, consider $f(x)=r$, as egreg points out in the comments.
Hint $\,\ \phi\ $ restricts to the identity map on $\rm\,R,\:$ i.e. constants evaluate to themselves.