I want to prove that this piecewise function is bijective

$$f: \Bbb R \to (-1,1), \quad f(x) = \begin{cases}1-\frac{1}{1+x} & \text{ if } x\ge 0 \\ -1+\frac{1}{1-x} & \text{ if } x \lt 0 \end{cases}$$

My attempt:

a) Injectivity: $f$ is injective iff $f(x)=f(y) \implies x=y$

Let $x,y \in \mathbb{R}$. There are four cases to consider.

if $x,y \ge 0$ both, then

$$f(x)=f(y) \iff 1-\frac{1}{1+x} = 1-\frac{1}{1+y} \iff x=y$$

if $x,y \lt 0$ both, then

$$f(x)=f(y) \iff -1+\frac{1}{1-x} = -1+\frac{1}{1-y} \iff x=y$$

if $x \ge 0, y \lt 0$, then

$$f(x)=f(y) \iff 1-\frac{1}{1+x} = -1+\frac{1}{1-y} \iff y = \frac{3x}{1+2x}$$

which is impossible, since $y$ is negative but $x$ is positive. So this case never occurs. The case for $y \ge 0, x \lt 0$ is analogous and therefore also never occurs.

Therefore $f$ is injective.

b) Surjectivity: $f$ is surjective iff $\forall y \in (-1,1)\, \exists x \in \mathbb{R}: f(x)=y$

Let $y \in (-1,1)$. Since the piecewise definition of the function is such, that $f \ge 0$ for $x \ge 0$ and $f \lt 0$ for $x \lt 0$, we need to consider two cases

if $y \in [0,1)$, then

$y = 1-\frac{1}{1+x} \iff x = \frac{1}{1-y}-1$. So $x \ge 0$ for $y \in [0,1)$ and $f(x) = y$.

if $y \in (-1,0)$, then

$y = -1+\frac{1}{1-x} \iff x = 1 - \frac{1}{y+1}$. So $x \lt 0$ for $y \in (-1,0]$ and $f(x) = y$.

Therefore $f$ is surjective.

A few questions:

  • For a). Is my argument for the case $x \ge 0, y \lt 0$ correct? Is there an easier way to prove injectivity than to consider each case?
  • For b). What if the range of $f$ doesn't split nicely for each case, e.g. that for the first case the range is positive and for the second case its negative. How do I prove surjectivity then if the range is more complicated?

(1) The injectivity argument for $x\ge0,y>0$ is correct. However, a better way to show injectivity would be to consider the inverse functions of each component, showing that for each $y$ in the range there is only one formula that gives an appropriate (in-its-domain) $x$, and that this $x$ is unique.

(2) This is not much of a problem. Surjectivity is slightly easier to prove than injectivity: just find for each $y$ in the range at least one function-piece that gives an appropriate $x$.

  • $\begingroup$ I corrected this mistake and changed the question to a proof verification. I think this is better since the above was a careless mistake due to my blindness after too many hours of work and is probably not so helpful to other people. $\endgroup$ – philmcole Dec 3 '17 at 17:16
  • $\begingroup$ @philmcole OK, there should be the response above. $\endgroup$ – Parcly Taxel Dec 3 '17 at 17:29
  • $\begingroup$ Thanks! You meant the injectivity argument but with $y \lt 0$ I guess? Regarding the invesere function. Couldn't the inverse function then directly be used to show that $f$ is bijective with "$f$ is bijective iff it has an inverse"? $\endgroup$ – philmcole Dec 3 '17 at 18:27
  • $\begingroup$ @philmcole Indeed. $\endgroup$ – Parcly Taxel Dec 4 '17 at 1:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.