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Let $a\in \mathbb{R}$, $a\neq 0$, and $E=\{(x,y,z)\in \mathbb{R}^3:a+x+y+z\neq 0\}$ and $f:E\rightarrow \mathbb{R}^3$ defined by $$ f(x,y,z)=\big(\frac{x}{a+x+y+z}, \frac{y}{a+x+y+z}, \frac{z}{a+x+y+z} \big) $$ Show that $f$ is injective.


I've been trying to prove it directly by showing that if $f(x_1,y_1,z_1)=f(x_2,y_2,z_2)\Rightarrow$ $x_1=x_2, y_1=y_2,z_1=z_2$. But I get a system of equations which I'm find difficult to deal with. Any help would be greatly appreciated.

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  • $\begingroup$ If you would insist, the polynomial equations obtained easily imply either $a+x+y+z=0$, or $a=0$ or $x_i=y_i$ for $i=1,2,3$. So we are done. $\endgroup$ – Dietrich Burde Nov 5 '18 at 12:37
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Adding the three components of the equation $f(x_1,y_1,z_1)=f(x_2,y_2,z_2)$ gives $$ \frac{x_1+y_1+z_1}{a+x_1+y_1+z_1} = \frac{x_2+y_2 + z_2}{a+x_2+y_2+z_2} $$ which implies $$ x _1+y_1+z_1 = x_2+y_2 + z_2 \, . $$ Then consider the equation $f(x_1,y_1,z_1)=f(x_2,y_2,z_2)$ again, and conclude that $(x_1,y_1,z_1)=(x_2,y_2,z_2)$.

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  • $\begingroup$ Why is that last part true? if $x_1 = x_2-1, y_1=y_2+1$, and $z_1=z_2$ then wouldn't that equation still be true and the conclusion false? $\endgroup$ – Joe Man Analysis Nov 5 '18 at 12:29
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    $\begingroup$ @JoeManAnalysis: Use the equation $f(x_1,y_1,z_1)=f(x_2,y_2,z_2)$ as well! You have that $\frac{x_1}{a+x_1+y_1+z_1} = \frac{x_2}{a+x_2+y_2+z_2}$. If the denominators are equal, $x_1 = x_2$ follows. $\endgroup$ – Martin R Nov 5 '18 at 12:31
  • $\begingroup$ ahhh, I checked it and you're correct. Thank you! I believe I understand it correctly now. $\endgroup$ – Joe Man Analysis Nov 5 '18 at 12:36
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$x_1f_2(x_1,y_1,z_1)=y_1f_1(x_1,y_1,z_1)$
$x_1f_2(x_2,y_2,z_2)=y_1f_1(x_2,y_2,z_2)$
$x_1y_2=x_2y_1$

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