This question is about the Jacobian conjecture for a special case. I will first explain the Jacobian conjecture (since it is something every mathematician should know about).


Let $k$ be an algebraically closed field.


Consider a map $$F: k^n \rightarrow k^n,$$
defined by $$F(x\_1,\ldots,x\_n)=(f\_1(x\_1,\ldots,x\_n),\ldots,f\_n(x\_1,\ldots,x\_n)),$$
where $f\_1,\ldots,f\_n$ are polynomials.


The Jacobian of $F$, which I denote $J$, is the determinant of the matrix $dF$ where the $(i,j)$-th entry of $dF$ is $\partial f\_i/\partial x\_j$. (The matrix $dF$ gives the induced map on the tangent bundle, or maybe it's the cotangent bundle; it doesn't matter for this question.)


Since $F$ is given by polynomials, the entries of $dF$ are polynomials and $J$ is a polynomial. Hence $F$ is a nonsingular map if and only if $J$ is a constant.


The inverse function theorem tells us that $F$ has a smooth inverse map if and only if $J$ is constant. The Jacobian conjecture says that this smooth map is in fact also given by polynomials (in the case where the original map $F$ is given by polynomials).


**Question**: I would like to know if the Jacobian conjecture is known (or trivial) for the special case where the matrix $dF$ is a triangular matrix with $1$'s on the diagonal. If it is not known, I would like to know if the full Jacobian conjecture is known to be equivalent to this special case.


Motivation: This is a possible strategy for proving that a particular family of maps I have constructed for a particular purpose is in fact invertible within the category of affine algebraic varieties.


EDIT: Clarifying in light of Tom's remark. The inverse function theorem just says that $F$ has a local inverse. The Jacobian conjecture is that $F$ has a global inverse which is given by polynomials.



Jacobian Conjecture for unit triangular matrices
More simply: 


(I'll write this down for the case $n=3$ because writing and reading subscripts makes me tired.)


Let $(u,v,w)=F(x,y,z)$. By hypothesis 
$u-x$ has derivative $0$ with respect to $x$, so 
$$u=x+P(y,z)$$
for some $P$. And $v-y$ has derivative $0$ with respect to $x$ and $y$, so 
$$v=y+Q(z)$$
for some $Q$. And $w-z$ has derivative $0$ with respect to all three variables, so 
$$w=z+R$$
for some constant $R$.


Now just write down the inverse: 
$$z=w-R$$
$$y=v-Q(z)=v-Q(w-R)$$
$$x=u-P(y,z)=u-P(v-Q(w-R),w-R).$$


