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I don't need help with the question whether or not the map is invertible, however when it comes to the support of $(Z_1,Z_2)$ being as underlined in green, I would understand if $Z_1, Z_2$ were independent, as I could then use the fact that the support of independent random variables is the cross product of their individual supports. Although this turns out to be the case I have no way of knowing this at the stage of the calculation outlined here.

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\begin{align} Z_1 & = X+Y \\[4pt] Z_2 & = \frac X {X+Y} \\[12pt] X & = Z_1 Z_2 \\[4pt] Y & = Z_1(1-Z_2) \end{align}

Suppose you've figured out that the support of $Z_1$ is $[0,\infty)$ and the support of $Z_2$ is $[0,1]$. Then certainly the support of $(Z_1,Z_2)$ is a subset of $[0,\infty)\times[0,1]$. The question is whether it includes all of $[0,\infty)\times[0,1]$. So suppose $(z_1,z_2)$ is some point in $[0,\infty)\times[0,1]$. The question of whether that point is in the support is the question of whether it is the case that every open neighborhood of that point has positive probability. And that comes down to whether every open neighborhood of $(x,y)=(z_1z_2,z_1(1-z_2))$ is assigned positive probability by the distribution of $(X,Y)$. And the answer to that isn't hard to find.

Of course, another way to do this is to actually find the density of $(Z_1,Z_2)$ and then observe that -- lo and behold -- $Z_1$ and $Z_2$ are independent.

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You can just simply check that for any $z \in {\mathbb R}_+ \times [0,1]$, you can obtain that value. One way to "almost" see this is to note that you can keep $Z_2$ fixed and scale $Z_1$ to any value you want by replacing $x$ and $y$ with $cx$ and $cy$ for arbitrary positive $c$.

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