Proving a vector field without explicit inverse surjective

Question

Consider the vector field given by \begin{align} F: \mathbb{R}^3 \mapsto \mathbb{R}^3, (x,y,z)\mapsto (y+e^z,z+e^x,x+e^y). \end{align} I want to show that $F$ is surjective/onto. After some trying, I found that there probably exists no explicit right inverse that would prove the claim. The strongest result that I have been able to prove so far is that, for every $(u,v,w)\in (\mathbb{R}^-)^3$, there exists $(x,y,z)$ such that $f(x,y,z)=(u,v,w)$. The argumentation is as follows: Consider $X=(-\infty,w]\times(-\infty,u]\times(-\infty, v]$ and $G: X \mapsto X, (x,y,z)\mapsto (w-e^y,u-e^z, v-e^x)$. Since $G$ is $C^1$, we have, for example in the euclidean norm, the lipschitz constant $max(e^u,e^v,e^w)<1$, which means the banach fixed point theorem is applicable. The fixed point of $G$ is then our desired preimage. Is it possible to generalize this argument for arbitrary $(u,v,w)$ or another way to show the surjectivity?

Answer 1

There is another way. First check that the Jacobian of $F$ is everywhere invertible (i.e. $\text{det}(J(F)(x,y,z))\neq 0$ for all $(x,y,z)\in \mathbb{R}^3$), then use the inverse function theorem to show that $\text{im}(F)$ is an open subset of $\mathbb{R}^3$. Then try to show that $\text{im}(F)$ is also a closed subset of $\mathbb{R}^3$ (Try to show that $F$ is a proper map which implies that it is also a closed map, hence $\text{im}(F)$ is closed). Since $\mathbb{R}^3$ is connected the only non-empty open and closed subset of $\mathbb{R}^3$ is $\mathbb{R}^3$, so we get that $\text{im}(F)=\mathbb{R}^3$.