Whether a function is globally 1-1. Let $f:U\subset \mathbb{R}^3\to \mathbb{R}^3$ is given by, $f(x,y,z)=(x+y+z,x^2+y^2+z^2,x^3+y^3+z^3)$ where $U=\{(x,y,z)\in \mathbb{R^3}: x>y>z\}$.


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*Is the function globally $1-1$?

*If $K\subset \mathbb{R}^3$ be compact then will $f^{-1}(K)$ comapct?


Here $$Df(x,y,z)=\begin{bmatrix} 1& 1 & 1\\ 2x& 2y & 2z\\ 3x^2& 3y^2&3z^2 \end{bmatrix}$$ determiant is non-zero, so it is locally $1-1$. But how do I determine whether it is globally $1-1$ or not?
 A: It is better to look at the problem as follows.

Claim: Let $a_1,a_2,\dots, a_n\in \mathbb{C}$.
  Consider the system of $n$ equations on $(x_1,\dots,x_n)$
$$p_k(x_1,\dots,x_n):=\sum_{i=1}^n{x_i}^k = a_k,$$
where $k$ ranges from $1$ to $n$.
  Then, up to permutation, the solution of the system is unique.

Proof: Let $S=(x_1,\dots,x_n)$ be a solution to the system and consider
$$q_S(X)=\prod_{i=1}^n(X-x_i).$$
There are constants $c_0,\dots,c_{n-1}\in \mathbb{C}$ such that $q_S(X)=X^n+c_{n-1}X^{n-1}+\dots+c_1X+c_0$.
By Vieta's formulas, we have that for $k=1,\dots,n-1$
$$e_k(x_1,\dots,x_n)={(-1)}^k\,c_{n-k},$$
where the $e_k$ are elementary symmetric polynomials.
By Newton's identities, the $e_k$ can be expressed in terms of the power sum symmetric polynomials $p_k$, whose values are prescribed by the hypotheses of the claim.
It follows that the constants $c_{n-k}$ depend only on the $a_1,\dots,a_n\in\mathbb{C}$.
This means that if $S_1$ and $S_2$ are solutions to our system, the polynomials $q_{S_1}(X)$ and $q_{S_2}(X)$ are equal.
In other words, there is a polynomial $q$ such that every solution $S=(x_1,\dots,x_n)$ to our system satisfies $q(x_i)=0$ for all $i=1,\dots,n$.
Up to permutation, this uniquely defines the $x_i$ as roots (with multiplicity) of $q$, which completes the proof. $\square$

We can now show that $f$ is in fact one-to-one.
Indeed, by the claim we have that $f(x_1,y_1,z_1) =$ $f(x_2,y_2,z_2)$ if and only if $\{x_1,y_1,z_1\}=\{x_2,y_2,z_2\}$.
Since the domain of $f$ requires that $x_i>y_i>z_i$, it follows that we must have $x_1=x_2$, $y_1=y_2$ and $z_1=z_2$.
