find $A,B$ in way that $f:A\rightarrow B$ is bijective, when $f(x)=\sqrt{x^2-1}$.

Attempt to solve

map $f:A \rightarrow B$ is bijective when it's surjective and injective simultaneously. This map is injective when:

$$ \forall(x,y) \in A : x \neq y \implies f(x) \neq f(y) $$

This map is surjective when:

$$ \forall y \in B \exists x \in A : f(x)=y $$

if function $f(x)$ is "truly" monotonic it implies that function has to be injective.


$$ \forall x \in \mathbb{R} : \frac{d}{dx}f(x) \neq 0$$

$$ \frac{d}{dx}\sqrt{x^2-1}=\frac{x}{\sqrt{x^2+1}} = f'(x) $$

It is visible that only way $f'(x)$ could be zero is when $x=0$ but $\sqrt{-1}$ is not defined in $\mathbb{R}$

which implies that $f(x)$ is injective if we pick $A$ in a way:

$$ A\in \mathbb{R}\setminus[-1,1] $$

and $B$:

$$ B \in \mathbb{R} $$

we have map: $$ f:\mathbb{R}\setminus[-1,1] \rightarrow \mathbb{R} $$

Inverse function of $f(x)$ can be computed by solving y from following equation. Inverse function is map $f^{-1}:B \rightarrow A$

$$ \sqrt{x^2-1}=y $$

$f(x)$ is defined when : $x^2-1 \ge 0 \implies x^2\ge 1 \implies -1 \ge x \ge 1$

$$ x^2-1=y^2 $$

$$ x^2=y^2+1 \implies x=\pm \sqrt{x^2+1} $$

$$ h^{-1}(x)=\sqrt{x^2+1} $$

meaning $f(x)$ is surjective. Which implies $f(x)$ is bijective when:

$$ A\in \mathbb{R}\setminus[-1,1], B \in \mathbb{R} $$

  • 1
    $\begingroup$ Your argument with the derivative does not work. Note that $f(x)=f(-x)$ when $f$ is defined in $x$. But with your reasoning you may choose $A=[1,\infty)$ and then compute the image $f(A)$. $\endgroup$ – Severin Schraven Sep 20 '18 at 20:45
  • $\begingroup$ $A=B=\varnothing$. $\endgroup$ – Asaf Karagila Sep 21 '18 at 12:41

You have made a mistake here

$$\frac{d}{dx}\sqrt{x^2-1}=\frac{x}{\sqrt{x^2-1}} = f'(x)$$

therefore $f(x)$ is injective for $x>1$ or for $x<-1$.

then for the surjectivity it suffices to observe that $$\lim_{x\to \pm \infty} f(x) = \infty$$

then $f(x)$ is bijective for the following restrictions

  • $f_1:(-\infty,a]\to [0,\infty)$
  • $f_2:[b,\infty)\to [0,\infty)$

with $a\le -1$ and $b\ge 1$


We can take $$A=[1,+\infty)$$

$$(\forall x>1) \; f'(x)=\frac{x}{f(x)}$$

$f$ is continuous at $A$ and strictly increasing at $(1,\infty)$ thus it is strictly increasing at $A$.


$f$ is a bijection from $A$ in $B$.


When you only have to give an $A$ and $B$ such that $f: A\to B$, $f(x)=\sqrt{x^2-1}$ is bijective, why not take simply $A=\{1\}$ and $B=\{0\}$.

If you want to finde the maximal intervalls you might go like this:

We construct $f^{-1}$ and while we do so, we have to note when the operations we use are defined or equivialent transformations.

$x=\sqrt{(y-1)(y+1)}$ First of all $y\geq 1$ or $y\leq -1$, else the square root is not defined.

We square both sides. Thus $x\geq 0$:

$x^2=y^2-1\Leftrightarrow y^2=x^2+1$


This gives us two solutions.

If $y\leq -1$, that means $y\in (-\infty, -1]$, then $f^{-1}(x)=-\sqrt{x^2+1}$.

If $y\geq 1$, that means $y\in [1,\infty)$, then $f^{-1}(x)=\sqrt{x^2+1}$

This gives us of course the intervals which $f$ is a bijection on. We had $x\geq 0$ which means $x\in[0,\infty)$

And the two solutions are $f:[1,\infty)\to\mathbb{R}_{\geq 0}$ or $f:(-\infty,-1]\to\mathbb{R}_{\geq 0}$


You have two good choices for $A$ and for each case you have a choice for $B$

First choice: let $A= (-\infty,-1]$ while $B= [0,\infty)$ where your function is strictly decreasing.

Second choice: Let $A= [1,\infty)$ while $B= [0,\infty)$ where your function is strictly increasing.

  • 1
    $\begingroup$ There are many many choices. $\endgroup$ – hamam_Abdallah Sep 20 '18 at 20:54
  • $\begingroup$ You know what I mean by these choices. $\endgroup$ – Mohammad Riazi-Kermani Sep 21 '18 at 18:24

Since $f(x)=f(-x)$, for $|x|\ge1$, your set $A$ can only contain one among a number and its negative.

The most natural choice is $A=[1,\infty)$, but obviously not the only one; also the set $A'$ consisting of rationals $\ge1$ and the irrationals $<-1$ would do.

Both $A$ and $A'$ are “maximal”, in the sense that you cannot add any number in the domain of $\sqrt{x^2-1}$ to either $A$ or $A'$ and still have an injective function. There are infinitely many choices of such “maximal domains of injectivity”, most notably $(-\infty,-1]$; which one to choose is a matter of preferences and simplicity.

Let's stick with $A=[1,\infty)$. The equation $\sqrt{x^2-1}=y$ can only have a solution if $y\ge0$. In this case it becomes $x^2-1=y^2$, so $x=\sqrt{y^2+1}$, which exists (and is $\ge1$) for all $y\ge0$.

Hence your set $B$ is $[0,\infty)$.

  • $\begingroup$ @N.F.Taussig I tried to expand it $\endgroup$ – egreg Sep 21 '18 at 12:24

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