A system of absolute value equalities

Background: I'm trying to show that the transformation $T:\Bbb R^n\to\Bbb R^n$ defined by $T(x_1,\dots,x_n) := (|x_2-x_1|,|x_3-x_2|,\dots,|x_1-x_n|)$ is (or is not, this is out of curiosity only) bijective (or if not, bijective up to sign? injective only? surjective only?). The problem reduces to two highly elementary systems of absolute value problems as follows.

Injectivity: Given that $$|x_2-x_1| = |y_2-y_1|\\ |x_3-x_2| = |y_3-y_2|\\ \dots\\ |x_1-x_n| = |y_1-y_n|$$ is it necessarily true that $x_1=y_1,\dots,x_n=y_n$?

Surjectivity: Given a sequence $c_1,\dots,c_n$ of numbers, can we necessarily find another sequence $x_1,\dots,x_n$ such that $$|x_2-x_1| = c_1\\ |x_3-x_2| = c_2\\ \dots\\ |x_1-x_n| = c_n$$

Taking an initial look at these somewhat similar problems, my immediate gut feeling is that surjectivity is somewhat more obvious than injectivity. I've never, however, done anything quite like these, can someone point me in the right direction?

• It cannot be bijective as $T(-x) = T(x)$. Also, $[T(x)]_k \ge 0$ for all $k$, so the point $(-1,0,...,0)$ is not in the range. – copper.hat Aug 3 '14 at 20:50
• I see. That's a shame, but thanks for the help. – theage Aug 3 '14 at 20:53
• Just a suggestion, for these sorts of things, try the simple cases first. In this case with $n=2$, $T$ maps onto the ray $\{ (x,x) \}_{x \ge 0}$ which shows that it cannot be injective or surjective. – copper.hat Aug 3 '14 at 20:55
• Another obvious reason why it can't be injective is to use $y_i = x_i + k$ for any real number $k$. – Calvin Lin Aug 3 '14 at 21:30
• @copper.hat Thoughts on describing the image of the function in the all positive "quadrant"? It's a straight line (half-ray) for $n=2$, and I believe 3 planes for $n =3$. – Calvin Lin Aug 3 '14 at 21:33