Let $f: \mathbb R^n \to \mathbb R$ be a smooth function. Then the first derivative $f^{(1)}$ makes sense as a function $\mathbb R^n \to \mathbb R^n$, and the second derivative makes sense as a function $f^{(2)}: \mathbb R^n \to \{\text{symmetric }n\times n\text{ matrices}\}$. I would like either a proof or a counterexample to the following claim:



> 
> **Claim:** Suppose that for every $x\in \mathbb R^n$, $f^{(2)}(x)$ is invertible as an $n\times n$ matrix. Then $f^{(1)}: \mathbb R^n \to \mathbb R^n$ is one-to-one.
> 
> 
> 


Some comments:


1. If you replace $\mathbb R$ by $\mathbb C$ and "smooth" by "algebraic", then the only such functions are (inhomogeneous) quadratic in $x$, since over $\mathbb C$ all non-constant functions have zeros. Then $f^{(1)}$ is (inhomogeneous) linear, and so one-to-one.
2. If you replace "invertible" by "positive (or negative) definite", then $f$ is convex, and so the claim follows. In particular, the claim is true when $n=1$.
3. This is a special case of the following more general question. If $g: \mathbb R^n \to \mathbb R^n$ is smooth, then its derivative makes sense as a function $g^{(1)}: \mathbb R^n \to \{n\times n\text{ matrices}\}$. If $g^{(1)}(x)$ is invertible for all $x\in \mathbb R^n$, is $g$ necessarily one-to-one? Of course, then $g$ is *locally* a diffeomorphism, but I don't know if it is globally. I don't think it is.


Oh, and I have no idea what the best tags are.



If the second derivative of a function on $\mathbb R^n$ is everywhere nondegenerate, does it follow that the first derivative is an injection?
The counter example given in the comments by Brian Conrad (and Dylan Thurston) is very nice. However, it oscillates wildly at $\infty$, and I believe that if you assume some nice properties at $\infty$ you will get a possitive answer.


Translating $f$ by a linear function does not change the assumptions on $f$ and so the claim is equivalent to the fact that all such $f$ has no or a unique critical points.


If we add assumptions such that e.g. the Conley index (or homotopy index) is well-defined and a sphere then this modified claim would follow. Indeed, if two or more critical points existed they would by assumptions be non-degenerate with same Morse index and a small pertubation would yield a Conley index which is a vedge of two or more spheres - a contradiction.


