If $f:\mathbb{R}\to \mathbb{R}$ is an invertible function, is it necessary that the function has to be strictly monotonic?

Question

If $f:\mathbb{R}\to \mathbb{R}$ is an invertible function, is it necessary that the function has to be strictly monotonic without any additional condition?

For invertibility to hold, we have to ensure that it's a bijective function on $\mathbb{R}$.

Now, let's say, a function is continuous and has a convex up form starting from $-\infty$ and is finally asymptotic at $y=5$. At point $x=3$, it has a jump discontinuity such that $(x,y)=(3,8)$. Can such a function satisfy the conditions of $f$ in question? I don't think so as the codomain is $\mathbb{R}$ and therefore, it has to be surjective on the codomain I guess. Like, we won't be able to find $f^{-1}(12)$. Right? Or, there's no relationship with the range and codomain here?

Can this be a suitable example to the function in question? (Please see the picture below) In the picture, the green coloured circles represent open intervals and the filled-blue circles represent closed intervals.

Answer 1

Blockquote If $f:\Bbb R\to\Bbb R$ is an invertible function, is it necessary that the function has to be strictly monotonic without any additional condition?

The answer to this question is obvious: NO. See the counterexample: $$f(x)=\begin{cases}\dfrac{1}{x}&\text{for }x\ne 0,\\[1ex] 0&\text{for }x=0.\end{cases}$$

If we assume continuity, then an invertible function is strictly monotonic.

Answer 2

Your function is clearly injective and surjective, hence is bijective, hence invertible, so everything is OK.

The strictly monotone bit is needed when your function is continuous by IVT.

Answer 3

The answer is NO

$A=\mathbb{R}-${$1$}

$B=\mathbb{R}-${$5$}

$C=B-${$1$}

$f(x)=x.1_{C }+5.1_{\{1\} }+1_{\{5\} }$

$f(x)$ is invertible as it is one-to-one but still not monotonic

Answer 4

Yet another simple example: $f(0)=1,f(1)=0,$ and $f(x) = x$ elsewhere.

Here's a more ambitious example: There exists a bijection $f:\mathbb R \to \mathbb R$ that is nowhere monotonic, meaning there is no interval of positive length on which $f$ is monotonic.

To do this, let $I_1,I_2,\dots$ be the open intervals with rational endpoints. We can inductively choose $x_n<y_n$ in $I_n$ such that all the points $x_1,y_1,x_2,y_2,\dots$ are distinct. Let $E= \{x_1,y_1,x_2,y_2,\dots \}.$ Define $f:E\to E$ by setting $f(x_n)=y_n, f(y_n)=x_n$ for each $n.$ On $\mathbb R \setminus E,$ define $f(x)=x.$ Then $f$ is a bijection of $\mathbb R$ to $\mathbb R.$ If $I$ is any interval of positive length, then $f$ is strictly increasing on $I\setminus E.$ But there must be some $I_n \subset I.$ Hence $x_n<y_n$ belong to $I,$ and $f(x_n)>f(y_n).$ Thus $f$ has the advertised property.