Question:
$ \ f : \mathbb{R} \to \mathbb{R}$
$ f(x) = \begin{cases} \frac{1}{x-1} , x<1 \\ \sqrt{x-1} , x \ge 1 \end{cases}$
Prove that $ \ f$ is a bijection and find its inverse.
My attempt:
Injective:
Notice that $\ f(x) < 0, \forall x < 1$ and $\ f(x) \ge 0, \forall x \ge 1$. (###)
Suppose $ f(a) = f(b) \implies a,b < 1$ or $\ a,b \ge 1$ by (###)
If $\ a,b < 1 \implies \frac{1}{a-1} = \frac{1}{b-1} \implies b-1 = a-1 \implies a = b$
If $ \ a,b \ge 1 \implies \sqrt{a-1} = \sqrt{b-1} \implies a - 1 = b -1 \implies a = b$
Surjective:
Pick any $ \ y \in \mathbb{R}$.
If $ \ y\ge 0 \implies y^2 \ge 0 \implies y^2 +1 \ge 1$. So, $ \ f(y^2 +1) = \sqrt{y^2} = |y| = y$, since $\ y \ge 0$.
If $ \ y< 0 \implies \frac{1}{y} < 0 \implies \frac{1}{y} + 1 < 1.$ So, $ \ f(\frac{1}{y} + 1) = y$
Inverse:
I tried to find the inverse using the proof for surjective .
$ f(x) = \begin{cases} x^2 + 1 , x \ge 0 \\ \frac{1}{x} + 1 , x < 0 \end{cases}$
Is this the correct approach to prove a piecewise function is a bijection?