TITLE: Injectivity of a function $\frac{x}{1+|x|}$
QUESTION [5 upvotes]: Let $f: \mathbb R \to  (-1,1)$
$x \in $ $\mathbb R$,  $f(x) = \frac{x}{1+|x|} $
Show that f is (1-1)

What I did : $x,y \in $ $\mathbb R$
$f(x)=f(y)$ iff $$\frac{x}{1+|x|} =\frac{y}{1+|y|}\Rightarrow\left| \frac{x}{1+|x|} \right| =\left| \frac{y}{1+|y|} \right| \\\Rightarrow \frac{|x|}{1+|x|} =\frac{|y|}{1+|y|}\Rightarrow |x|=|y|$$ $\Rightarrow$ $x=y$ or $x=-y$
If $x=-y,\frac{y}{1+|y|} =\frac{-y}{1+|y|}\Rightarrow y=-y$
It is only valid for unique real value of $y$ that is $0$ namely $f(x) \neq f(y)$ for other values of $y$. Thus $x=y$
I think it is not sufficient for injectivity, can someone correct me please?

REPLY [4 votes]: You obtained that $|x|=|y|$ then replacing this in   the first equality you get 
$$\frac{x}{1+|x|} =\frac{y}{1+|y|} \Longleftrightarrow\frac{x}{1+|x|} =\frac{y}{1+|x|}\Longleftrightarrow x=y$$