An algebra problem by Manuel Kahayon

Relevant wiki: Bijection, Injection and Surjection - Problem Solving - Hard

Here's a straightforward example. Let $f(x) = x$ when $x$ is not an integer and $f(x) = -x$ when $x$ is an integer. Clearly $f$ is a bijection which is neither non-increasing or non-decreasing and $f ( f(x)) = x$ is non-decreasing.

Relevant wiki: Bijection, Injection and Surjection - Problem Solving - Hard

f(x)=1/x when x != 0 and f(0)=0 seems to be a good example

What I've noticed here is the fact that, if $g(x)$ is a function which is a bijection between the nonnegative real numbers (with $g(0) = 0$ ), then its inverse is also a bijection between the nonnegative real numbers. This implies that $f(x) = -g(x)$ if $x \geq 0$ , $f(x) = g^{-1}(-x)$ otherwise gives us $f(f(x)) = x$ for all x.

Let $h(x) = x$ if $\lfloor x \rfloor \equiv 0 \pmod 2$ , $h(x) = x+2$ if $\lfloor x \rfloor \equiv 1 \pmod 4$ , $h(x) = x-2$ if $\lfloor x \rfloor \equiv 3 \pmod 4$ . We can see that by checking cases (or graphing) that $h(x)$ is a bijection between the nonnegative real numbers (with $g(0) = 0$ ), and $h(x)$ is nonmonotonic. Letting $g(x) = h(x)$ , we get our desired function $f(x)$ .

So the answer it $\boxed{ \text{Yes}}$ .

P.S. Here is the graph of $h(x)$ to see how I was inspired to construct such a function.

Well...

$f(x)=\cases{x~~(x\le0) \\ \dfrac{1}{x}~~(x>0)}$

Enough said.

There appears to be no shortage of such functions. We are told that $f$ is neither increasing nor decreasing, and yet it's bijective. Thus the function must have points of discontinuity. Since $f$ is bijective, $f \circ f$ is also bijective. We're told it's non-descreasing, but it's allowed to be increasing. So the obvious thing is to create a function $f$ such that $f(f(x)) = x$ .

This is easier to do if we consider a subset of the real line. On $(0,1)$ let $f(x) = 1-x$ . This function is a bijection of the interval $(0,1)$ onto itself. If we compose $f$ with itself, we see that $f(f(x)) = 1 - (1-x) = x$ .

We can extend $f$ to the entire real line by setting $f(x) = \left \lceil{x} \right \rceil -x + \left \lfloor{x} \right \rfloor$ . This function is locally decreasing but increasing in the sense that if $x - y > 1$ then $f(x) > f(y)$ .

The 2. is tricky, but can be read as "not ((not increasing) and (not decreasing))", which is simply "increasing or decreasing". f(x) = x is bijective, increasing and f(f(x)) = x is increasing.