Let $k$ be a field of characteristic zero, and let $A_1(k)$ be the first Weyl algebra, namely, the associative non-commutative $k$-algebra generated by $x$ and $y$ subject to the relation $yx-xy=1$.

Suppose that the following map $f$ is a $k$-algebra endomorphism of $A_1$:
$(x,y) \mapsto (f(x):=p,f(y):=q)$,
where $p=uy$ and $q=x+vy$, $u,v \in A_1(k)$, $y$ does not divide $u$,
namely, $u \notin A_1y$.

As a $k$-algebra endomorphism of $A_1(k)$, we have $[q,p]=1$; indeed, just apply $f$ to $yx-xy=1$.

> Is it true that $f$ is actually an automorphism of $A_1(k)$?
In particular, is it true that necessarily $p=- y$?

**My partial answer:**
$1=[q,p]=[x+vy,uy]=[x,uy]+[vy,uy]=-[uy,x]+[vy, uy]$
$=-(u[y,x]+[u,x]y)+[vy,uy]=-(u+[u,x]y)+[vy,uy]=-u-[u,x]y+[vy,uy]$

Denote: $E:=[vy,uy]$. Then,
$E=[vy,uy]=v[y,uy]+[v,uy]y=-v[uy,y]-[uy,v]y=$
$-v(u[y,y]+[u,y]y)-[uy,v]y=-v[u,y]y-[uy,v]y=$
$-v[u,y]y-(u[y,v]y+[u,v]y^2)=$
$-v[u,y]y-u[y,v]y-[u,v]y^2$.

So we have $E=-v[u,y]y-u[y,v]y-[u,v]y^2$.
Then, $1=-u-[u,x]y-v[u,y]y-u[y,v]y-[u,v]y^2$.

Write $u=a_ny^n+\cdots+a_1y+a_0$, $a_j \in k[x]$, $a_0 \neq 0$ (since we have assumed that $y$ does not divide $u$).
We see that $a_0=-1$.

Now, the highest $(0,1)$-term of $-u-[u,x]y-v[u,y]y-u[y,v]y-[u,v]y^2$ should be zero, and by considerations of $(0,1)$-degrees, it equals the $(0,1)$-highest term of  $-[u,v]y^2$.

**********************************************************************

**Motivation:** Please see [this][1] question, in order to understand the motivation for my above question.


Any hints and comments are welcome!
(I have also asked the above question in [MSE][2], but have not received any comments yet). 


  [1]: https://mathoverflow.net/questions/334897/a-non-commutative-analog-of-a-result-concerning-a-jacobian-pair
  [2]: https://math.stackexchange.com/questions/3281015/is-a-specific-endomorphism-of-a-1-an-automorphism