Showing $x \to \frac{2x-1}{1-\vert 2x - 1 \vert}$ is onto from $(0,1)$ to $\Bbb{R}$. So I want to show $x \to \frac{2x-1}{1-\vert 2x - 1 \vert}$ is onto from $(0,1)$ to $\Bbb{R}$. Here is what I have thus far:
Let $r \in \Bbb{R}$, then setting
$$r = \frac{2x-1}{1-\vert 2x - 1 \vert}$$
we have that
$$x_1=\frac{2r+1}{2r+2},x_2=-\frac{1}{2r-2}$$
For $r \in [\frac{1}{2},\infty)$ $x_1 \in (0,1)$ but $x_2 \not\in (0,1)$.
For $r \in (-\infty, -\frac{1}{2}]$ I have that $x_2 \in (0,1)$ but $x_1 \not\in (0,1)$.
And when $r=0$ they're both equal to $\frac{1}{2}$. But at values between $0$ and $1/2$ and between $-1/2$ and $0$ they seem to both be in the interval. What am I missing? I know the mapping is also $1-1$ so I shouldn't have both $x_1,x_2 \in (0,1)$ simultaneously.
 A: To summarize the discussion in the comments:
In this case it is possible to construct an explicit inverse, though it is not necessary.
To do it:  note that our function is negative for $0<x<.5$ and positive for $.5<x<1$, and note that $2x-1$ has constant sign over those two open intervals (making it possible to simplify the function).  We'll therefore split the domain into those two intervals, and the range into the negatives and the positives, to simplify the computation.
To write down the inverse, split $\mathbb R$ into $(-\infty, 0)$ and $(0, \infty)$ (of course the inverse takes $0$ to $\frac 12$).  Then for the positive reals we seek to solve $$\frac {2x-1}{2-2x}=r\implies x=1-\frac 1{2(r+1)}$$ and for the negative reals we seek to solve $$\frac {2x-1}{2x}=r\implies x= \frac 1{2(1-r)}$$ Note that in each case, it is clear that the given inverse does in fact map to the correct interval in $(0,1)$.
In general, however, it is much easier to avoid writing out an explicit inverse (in many cases, it is not even possible to write an explicit inverse).  Here, for instance, we can split the domain into $(0,.5)$ and $(.5,1)$ (of course $.5$ is mapped to $0$) and then note that, on $(.5,1)$ our function is $$\frac {2x-1}{2-2x}=\frac {2x-2+1}{2-2x}=-1+\frac 1{2-2x}$$
It is clear that all such values are positive and they go to $+\infty$ as $x\to 1^{-}$ so all positive reals are hit.
Similarly, on $(0,.5)$ our function is $$\frac {2x-1}{2x}=1-\frac 1{2x}$$ and it is easy to see that all such values are negative and, as the function goes to $-\infty$ as $x\to 0^{+}$ we see that all negative values are hit, so we are done.  Note, the explicit form of the function on both intervals instantly shows that it is injective as well.
