Are there any $\mathbf{Z}[x]$-algebra maps of the following form? An exercise I am computing reduces to the following question:

Is there a map $f$ of $\mathbf{Z}[x]$-algebras such that the composition
  $$
\mathbf{Z}[x,x^{-1}]\xrightarrow{f}\mathbf{Z}[x]\to\mathbf{Z}[x,x^{-1}]
$$
  is the identity on $\mathbf{Z}[x,x^{-1}]$, where the map $\mathbf{Z}[x]\to\mathbf{Z}[x,x^{-1}]$ is the unique map since $\mathbf{Z}[x]$ is initial in the category $\mathbf{Alg}_{\mathbf{Z}[x]}$?

I believe that the answer is no, because the unique map $\mathbf{Z}[x]\to\mathbf{Z}[x,x^{-1}]$ sends a polynomial
$$
\sum_{i\geq0}c_{i}x^{i}\in\mathbf{Z}[x]
$$
to a Laurent polynomial with no negative powers. Said another way, an $R$-algebra map
$$
R\to R[y]
$$
must send $r\mapsto r$. Thus, there is no morphism $f:\mathbf{Z}[x,x^{-1}]\to\mathbf{Z}[x]$ which will work, because it will necessarily send negative powers of $x$ to elements in $\mathbf{Z}[x]$ which cannot be recovered.

Another way of expressing it is like this: a homomorphism $f$ of algebras from $\mathbb{Z}[x, x^{-1}]$ to $\mathbb{Z}[x]$ must map $x$ to a unit (invertible element) in $\mathbb{Z}[x]$ and the only units in $\mathbb{Z}[x]$ are $\pm 1$. So the image of such a homomorphism would have $x - 1$ or $x + 1 $ in its kernel and the homomorphism could not be one-to-one and so could not have a right inverse. 
