The inverse of $f$ is a continuous function Let the function $y=f(x)=(1+x^2)sgn(x)$. Prove that $y^{-1}$ is a continuous function.
I know that the $y$ is not  continuous, but don't know how to treat the problem at first. Thank you for your help.
 A: Your notation is a bit confusing.  I think you mean that $f^{-1}$ (i.e. the inverse function, not the reciprocal) is a continuous function.  Thus you're writing $x = f^{-1}(y)$.  
There are three cases to consider: $x > 0$, $x < 0$ and $x = 0$.  When $x > 0$, $y = 1 + x^2 > 1$ and $x = \sqrt{y-1}$.  This is a continuous function on $(1,\infty)$.
Similarly, when $x < 0$, $y = -1 - x^2 < -1$ and $x = -\sqrt{-1 - y}$.  This is a continuous function on $(-\infty, -1)$.  Finally, when $x = 0$, $y = 0$.  Thus
$$ f^{-1}(y) = \cases{\sqrt{y-1} & for $x > 1$\cr
                      0 & for $x = 0$\cr
                      -\sqrt{-1-y} & for $x < -1$\cr
                     \text{undefined} & otherwise\cr}$$
Since the three sets $(-\infty, -1)$, $\{0\}$ and $(1,\infty)$ are separated,
$f^{-1}$ is continuous on its domain which is their union.
A: The function $f$ is discontinuous (only) at $0$, and since $1+x^2\geq1$ for all $x\in\mathbf R$, the range of $f$ is contained in $(-\infty,-1)\cup\{0\}\cup(1,+\infty)$. As $f$ is strictly increasing and tends to $\pm\infty$ as $x\to\pm\infty$, the range of $f$ is in fact equal to that set, and $f$ is a bijection from $\mathbf R$ to that range.
The inverse of $f$ is therefore a well defined function $ (-\infty,-1)\cup\{0\}\cup(1,+\infty)\to\mathbf R$; continuity is a local property, and therefore determined separately on each of these three connected components of the domain of $f^{-1}$. The component $\{0\}$ is discrete (it is an isolated point), so continuity there is vacuously satisfied. For the other two components, we are dealing with the inverse of a restriction of $f$ to $(-\infty,0)$ respectively to $(0,+\infty)$, and these restrictions are continuous and strictly increasing. You should know how to prove that the inverse of such a function is continuous, and this settles the continuity of $f^{-1}$ everywhere on its domain.
