Inverse image of an open set with a function Let $f: (\mathbb{R},|.|)\to (\mathbb{R},|.|)$ defined by $f(x)=\dfrac{1}{1+x^2}$
I want to find $f^{-1}(]a,b[)$ 
What I do: 
$$f^{-1}(]a,b[)=\{y\in \mathbb{R}, \frac{1}{1+y^2}\in ]a,b[\}$$
that is : $$a<\frac{1}{1+y^2}<b$$ 


*

*If $a>0$  then $$\frac1b-1<y^2<\frac1a-1$$
after that we must suppose two sases $\frac1b\leq 1 $ or $\frac1b>1$
Is there an other method ?
i suppose that $a<b$
$$
f^{-1}(]a,b[)=
\begin{cases}
\emptyset,~ b<0\\
]-\infty,-\sqrt{\frac1b-1}[\cup]\sqrt{\frac1b-1},+\infty[,~ a\leq0 ~\text{and}~b<1\\
\mathbb{R},~ b>1,a\leq0\\
\mathbb{R}\setminus{0},~ b=1,a\leq0\\
\emptyset,~b\geq1,a>1\\
]-\sqrt{\frac1a-1}, \sqrt{\frac1a-1}[,~b\geq1,a<1\\
]\sqrt{\frac1b-1},\sqrt{\frac1a-1}[,~ b<1, a>0
\end{cases}
$$
Thank you

I think a good way to approach the problem is to think first about the function $f$:
First of all, since $x^{2} \geq 0$ then $1+x^{2} \geq 1$ which leads to $ f(x) = \frac{1}{1+x^{2}} \leq 1$. This means that if $a > 1$ then $f^{-1}(]a,b[) = \emptyset$.
We also have that $1+x^{2} \geq 1 > 0$ so $f(x) = \frac{1}{1+x^{2}} >0$. This means that if $b \leq 0$ then we also have $f^{-1}(]a,b[) = \emptyset$.
Furthermore, $\forall a,a' \in ]\infty, 0]$ and $b>0$ we have $f^{-1}(]a,b[) = f^{-1}(]a',b[)$ and $\forall a < 1$ and $b \in ]1, \infty]$ we have $f^{-1}(]a,b[) = f^{-1}(]a,b'[)$
With this results, we can then separate the problem in appropriate cases:
$\underline{\text{Case} \,\, 1:}$ $0<a<b \leq 1$
$$a < f(x) < b \,\,\Leftrightarrow \,\, a < \frac{1}{1+x^{2}} < b \,\,\overset{a,b >0}{\Longleftrightarrow}\,\, \frac{1}{b} < 1+x^{2} < \frac{1}{a} \,\,\Leftrightarrow \,\,\frac{1}{b}-1 < x^{2} < \frac{1}{a}-1$$ 
Now, since $a, b\leq 1$, we have that $\frac{1}{b}-1 \geq 0$ and $\frac{1}{a}-1 \geq 0$ and we can take square roots to have 
$$ \frac{1}{b}-1 < x^{2} < \frac{1}{a}-1 \,\, \Leftrightarrow \,\, \sqrt{\frac{1}{b}-1} < |x| < \sqrt{\frac{1}{a}-1}$$
(notice that the fact of the square root being an increasing injective function allows us to take square roots and not changing the grater-than and less-than symbols and to have the equivalence symbol) 
We finally have that 
$$ x \in f^{-1}(]a,b[) \,\,\Leftrightarrow \,\, x\in \,\, ]\sqrt{\frac{1}{b}-1},\sqrt{\frac{1}{a}-1}[ \quad \text{or} \quad x\in \,\, ]-\sqrt{\frac{1}{a}-1},-\sqrt{\frac{1}{b}-1}[$$
We put now our attention on the limit cases:
$\underline{\text{Case} \,\, 2:}$
$\underline{\text{Subcase} \,\, 1:}$ $a=0$ and $b \leq 1$
$a=0 < \frac{1}{1+x^{2}}$ so the condition $a<f(x)<b$ is equivalent to $f(x)<b$.
$\underline{\text{Subcase} \,\, 2:}$ $a>0$ and $b > 1$
We have $\frac{1}{1+x^{2}} \leq 1<b$ so the condition $a<f(x)<b$ is equivalent to $a<f(x)$.
We can summarize the problem in the next cases:
The empty cases:
$a > 1\, \Rightarrow \,f^{-1}(]a,b[) = \emptyset$
$b \leq 0 \, \Rightarrow \,f^{-1}(]a,b[) = \emptyset$
The whole case:
$a \leq 0$ and $b >1$ $ \, \Rightarrow \,f^{-1}(]a,b[) = \mathbb{R}$
The partial cases:
$a \leq 0$ and $0 < b \leq 1$ $ \, \Rightarrow \,f^{-1}(]a,b[) = ]-\infty,-\sqrt{\frac{1}{b}-1}[ \,\, \bigcup \,\, ]\sqrt{\frac{1}{b}-1},\infty[$
$1 > a > 0$ and $ b > 1$ $ \, \Rightarrow \,f^{-1}(]a,b[) = ]-\sqrt{\frac{1}{a}-1},\sqrt{\frac{1}{a}-1}[ $
The restrictive case:
$0<a<b \leq 1$ and $0 < b \leq 1$ $ \, \Rightarrow \,f^{-1}(]a,b[) = ]-\sqrt{\frac{1}{a}-1},-\sqrt{\frac{1}{b}-1}[ \,\, \bigcup \,\, ]\sqrt{\frac{1}{b}-1},\sqrt{\frac{1}{a}-1}[$
Notice that the good choice of the limit points ($0$ for $a$ and $1$ for $b$) is the key.
The proof may appear very long but if you draw the function (maybe with WolframAlpha) you will understand it much better.
I hope it helps you!
