Let $f: \mathbb{C}[x,y] \to \mathbb{C}[x,y]$,
$(x,y) \mapsto (p,q)$, with $p,q \in \mathbb{C}[x,y]$ 
satisfying $\operatorname{Jac}(p,q):=p_xq_y-p_yq_x \in \mathbb{C}-\{0\}$.
Such a polynomial map is called a Keller map, and the two-dimensional Jacobian Conjecture
says that such a map is injective and surjective.

I know that there are many papers trying to prove injectivity or surjectivity of Keller maps.

My question is quite basic, and I think it is dealt with in one paper or another or perhaps in algebraic geometry books, but I am not able to find the relevant references now.

Further assume that $p(0,0)=q(0,0)=0$.

> **Question:** If $f$ is not injective, then is it true that there exist $c \in \mathbb{C}-\{0\}$ such that $p(c,0)=q(c,0)=0$ or $p(0,c)=q(0,c)=0$?

I ask, in other words: Given a map $\tilde{f}: \mathbb{C}^2 \to \mathbb{C}^2$, defined by $(a,b) \mapsto (p(a,b),q(a,b))$ (where $p,q \in \mathbb{C}[x,y]$ have invertible Jacobian),
if we assume that $\tilde{f}$ is not injective and $(0,0) \mapsto (0,0)$, is it true that there exist $c \in \mathbb{C}-\{0\}$, with $(c,0) \mapsto (0,0)$ 
or $(0,c) \mapsto (0,0)$.

(What I remember vaguely is something like: 'If $f$ is not injective, then it is not injective at zero'; I am not sure about my specific question. I guess it was meant that instead of $f$ we can change variables= compose it with an automorphism, and then get what I asked?).

**What I have tried:**
Write:
$p=p_ny^n+\cdots+p_1y+p_0$,
where $p_i \in k[x]$
and 
$q=q_my^m+\cdots+q_1y+q_0$,
where $q_j \in k[x]$.
The assumption $p(0,0)=q(0,0)=0$ implies that $p_0(0)=0$
and $q_0(0)=0$. 
We have:
$p(x,0)=p_0$
and
$q(x,0)=q_0$.
I am asking if $p_0$ and $q_0$ have a common zero $c \in \mathbb{C}-\{0\}$
($0$ is a common zero of $p_0$ and $q_0$).




A relevant paper, for example, is [this][1].

Thank you very much! 


  [1]: https://arxiv.org/pdf/alg-geom/9305008.pdf
  [2]: https://math.stackexchange.com/questions/4215275/injectivity-of-keller-maps