# Are monotonic and bijective functions the same?

The question is simple : consider the family of monotonic functions ; $m(x) : \mathbb{R} \rightarrow \mathbb{R}$, and the familly of bijective functions ; $b(x) : \mathbb{R} \rightarrow \mathbb{R}$. Are they actually the same? If not, I would like to see some simple counter-examples.

• $f(x)=1$ is monotonous but not bijective
– Surb
Jan 3, 2017 at 15:05
• Nitpick: You should be saying "monotone" or "monotonic", not "monotonous".
– MPW
Jan 3, 2017 at 15:25
• @MPW: Agreed. Hence my edit to the question.
– J W
Jan 3, 2017 at 15:31

$$f(x)=1$$

is monotonic, but clearly not bijective.

You are probably asking about strictly monotonic functions (that way you can get injectivity), but the answer is still no.

$$f(x)=e^x$$ is monotonic, but not bijective.

$$f(x)=\begin{cases}x & x>0\lor x<-1\\ -x-1 & -1\leq x\leq 0\end{cases}$$

is bijective, but not monotonic.

You might mean strictly monotonic continuous functions, in which case the answer is still no ($$f(x)=e^x$$ is strictly monotonic and continuous, but not bijective), however, it is true that the other type of counterexample cannot be found, i.e.

Every continuous bijective function from $$\mathbb R$$ to $$\mathbb R$$ is strictly monotonic.

Edit for the question posed in comments:

You are making a mistake a lot of math students make, and it's usually the fault of the teachers not emphasizing it enough. The thing is:

A FUNCTION IS DEFINED BY THREE THINGS:

1. The domain.
2. The codomain.
3. The "action".

So, if I want to truly mathematically correctly define some function, I can say:

$$f$$ is the function from $$A$$ to $$B$$ defined by $$f(x)=...$$

Note, it is important to note both from where the function is mapping, to where it is mapping, and how it is mapping.

Example:

• The function $$f:\mathbb R\to\mathbb R$$ defined by $$f(x)=e^x$$ is a function.
• The function $$g:\mathbb R\to(0,\infty)$$ defined by $$g(x)=e^x$$ is a function.

IMPORTANT:

$$f$$ and $$g$$ are not the same function. I cannot stress this enough. $$f$$ and $$g$$ map all numbers to the exact same number, but because their codomains are different, they are, by definition, different functions. It is true that if we restrict the codomain of $$f$$ to $$(0,\infty)$$, we get $$g$$, but it is not true that $$f$$ is the same function as $$g$$.

Why?

You may think this is unnecessary, but it is very necessary if you want any meaningful definition of the word surjective. Why? Well, remember:

A function $$f:A\to B$$ is surjective if, for every $$b\in B$$, there exists some $$a\in A$$ such that $$f(a)=b$$.

Now, take any function $$h:A\to B$$. And define $$B'=f(A)=\{f(a)|a\in A\}$$. Then, this statement is true:

• For every $$b\in B'$$, there exists some $$a\in A$$ such that $$h(a)=b$$.

So, is $$h$$ all of a sudden surjective? Just because we restricted its codomain? NO. If we restrict $$h$$ to $$B'$$, we get a different function, and the restricted function is surjective, but $$h$$ may not be.

Similarly, our function $$g$$ mapping from $$\mathbb R$$ to $$(0,\infty)$$ is surjective, but the function $$f$$ is not.

• why is it not bijective. explanation please.
– user394255
Jan 3, 2017 at 15:04
• @A.Molendijk Because there exists no $x\in\mathbb R$ such that $f(x)=-1$.
– 5xum
Jan 3, 2017 at 15:06
• @5xum, hmm it's not totally clear to me yet. You have $y = e^x$, as a mapping from A to B, and it can be inverted : $x = \ln{y}$, which is a mapping of B to A. For each value of $x \in A$, there is a unique value of $y \in B$. But also, for each value of $y \in B$, there is a unique value of $x \in A$, so the mapping is bijective, isn't ?
– Cham
Jan 3, 2017 at 15:25
• @Cham The function is a bijective function from $\mathbb R$ to $(0,\infty)$, but that's a different function (because it has a different codomain).
– 5xum
Jan 3, 2017 at 15:27
• @5xum "...but because their domains are different...". I presume "codomain" is meant here. Feb 6, 2019 at 5:50

Let $f: \mathbb{R} \to \mathbb{R}$ be the zero function. It is monotonous yet not injective, since if $x_1 \neq x_2$ does not imply that $f(x_1) \neq f(x_2)$.

I guess you really mean strictly monotone functions, since constant functions are monotone and obviously not bijective. But even so, it is not enough. (Recall: a function $f:X\to Y$ between topological spaces is said to be monotone if $f^{-1}(y)$ is connnected for each $y\in Y$. Monotone functions can have "flat" spots where they are locally constant.)

It's a simple matter to construct a large number of nonbijective monotone functions. If $f:\mathbb R\to\mathbb R$ is monotone, then $E_f:\mathbb R\to\mathbb R$, where $E_f(x)\equiv e^{f(x)}$, is a monotone function which is not surjective (so not bijective).

• Merry christmas :)
– Surb
Jan 3, 2017 at 15:32