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Let $k$ be an algebraically closed field of characteristic zero and let $f_1,\ldots,f_n \in k[x_1,\ldots,x_n]$ have an invertible Jacobian, namely, the determinant ot their Jacobian matrix belongs to $k^*$. Let $F: k^n \to k^n$ be the corresponding polynomial map $F: \bar{a} \mapsto (f_1(\bar{a}),\ldots,f_n(\bar{a}))$, $\bar{a}:=(a_1,\ldots,a_n)$, $a_1,\ldots,a_n \in k$.

When $n=2$, it is not difficult to obtain the following: If $F(x_1,0)=(x_1,0)$, then $F$ is bijective, see Lemma 2.3.

Is the $n \geq 3$ case also true, namely: If $F(x_1,0,\ldots,0)=(x_1,0,\ldots,0)$, then $F$ is bijective?

It seems reasonable to me; the proof should include generalizations of basic properties in dimension two, which can be found here.

Any comments and hints are welcome!

An important remark: If I am not wrong, I think that I can prove the generalized JC based on a positive answer to my above question (it is the only argument missing in my hopefully valid proof), so of course if someone knows how to prove the above, then they will be co-authors of my paper.

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    $\begingroup$ Let $f,g\in k[x,y]$ be such that their Jacobian is a non-zero constant. Let $k[x,y,z]\to k[x,y,z]$ be defined as $(f,g,z)$. Then $z(0,0,z)=z$, and their Jacobian is still a non-zero constant. So, if this map is bijective so is the original map given by $f,g$. Thus 2 variable Jacobian conjecture would be true. Similarly for any $n$. $\endgroup$
    – Mohan
    Commented Apr 22, 2018 at 18:52
  • $\begingroup$ Thank you. (I had in mind a much more complicated argument then your argument). $\endgroup$
    – user237522
    Commented Apr 22, 2018 at 19:44

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