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 $p=Ay$ and $q=x+By$, $A,B \in A_1(k)$, $y$ does not divide $A$.

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:**
$1=[q,p]=[x+By,Ay]=[x,Ay]+[By,Ay]=-[Ay,x]+[By,Ay]$
$=-(A[y,x]+[A,x]y)+[By,Ay]=-(A+[A,x]y)+[By,Ay]=-A-[A,x]y+[By,Ay]$

Denote: $E:=[By,Ay]$. Then,
$E=[By,Ay]=B[y,Ay]+[B,Ay]y=-B[Ay,y]-[Ay,B]y=$
$-B(A[y,y]+[A,y]y)-[Ay,B]y=-B[A,y]y-[Ay,B]y=$
$-B[A,y]y-(A[y,B]y+[A,B]y^2)=$
$-B[A,y]y-A[y,B]y-[A,B]y^2$.

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

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

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

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Please see [this][1] question, in order to understand the motivation for my above question.
(I have also asked the above question in MSE, but have not received any  comments yet).

Any hints and comments are welcome!
 


  [1]: https://mathoverflow.net/questions/334897/a-non-commutative-analog-of-a-result-concerning-a-jacobian-pair
  [2]: https://math.stackexchange.com/questions/3275302/a-non-commutative-analog-of-a-known-result-concerning-a-jacobian-pair