Let $\mathfrak{g}$ be a 3-dimensional Lie algebra with basis $\{X,Y,Z\}$ such that $Z$ is central and $[X,Y] = Z$.
Let $U(\mathfrak{g}) = k \langle X,Y,Z \rangle / \langle \{YZ - ZY, XZ-ZX, XY-YX-Z\} \rangle$ denote its universal enveloping algebra.
Let $k$ be a field and $A_1(k) = k\langle x,y \rangle/\langle xy - yx -1 \rangle$ be the first Weyl algebra over $k$.
$A_1(k)$ has a filtering by $R_i := \sum_{j=1}^jk[x]y^j$.
Form the Rees ring $\tilde{R} := \oplus_{i \in \mathbb{Z}}R_it^i$ (as a subring of Laurent polynomials).
There is a $k$-algebra isomorphism $\tilde{R} \cong U(\mathfrak{g})$
One can note that the map $\phi:k\langle X,Y,Z \rangle \to \tilde{R}$ sending $X \mapsto x$, $Y \mapsto Yt$, $Z \mapsto t$, factors through $U(\mathfrak{g})$.
We get a homomorphism $\psi: U(\mathfrak{g}) \to \tilde{R}$. It seems to be clearly surjective. To show it is injective I will try to work by induction on the length of a word. Any word of length $1$ isn't sent to $0$ by $\psi$. Let $l \in U(\mathfrak{g})$ and write $l = l_0r$ where $r \in \{X,Y,Z\}$. If $\psi(l) = 0 = \psi(l_0)\psi(r)$.
Then, if $r = Z$ clearly $\psi(l_0) = 0$ and by induction $l_0 = 0$.
If $r = Y$ then $\psi(l) = 0 = \psi(l_0)Yt$ so $\psi(l_0)Y = 0$ which is impossible.
If $r = X$ then $\psi(l_0)$ is a polynomial in $X$ (after removing $t$ if necessary) which again implies it must be zero.
Questions:
Is the proof the $\psi$ is injective correct? It feels a little funky. How can it be made better?
To show the existence of the homomorphism $\psi$ does it indeed suffice to show $\phi$ factors through $U(\mathfrak{g})$?