Show that $\alpha$ is bijective and find $\beta: \mathbb{R} \rightarrow \mathbb{R}$ such that $(\beta \alpha)(a) = a$ for all $a \in \mathbb{R}$

Question

Define $\alpha: \mathbb{R} \rightarrow \mathbb{R}$ via

$\alpha (a)=\begin{cases} 4a-3& \text{ if } a\leq 1 \\ a^2 & \text{ if } a>1 \end{cases}$

Show that $\alpha$ is bijective and find $\beta: \mathbb{R} \rightarrow \mathbb{R}$ such that $(\beta \alpha)(a) = a$ for all $a \in \mathbb{R}$.

My attempt:

If $a,b\leq1$, we have $4a-3=4b-3$, so $a=b$

If $a,b>1$, then $a^2=b^2$, so $a=b$

now if I consider the cases $a\leq1, b>1$ and $a>1,b\leq1$, I have that $b=\sqrt{4a-3}$ and $a=\sqrt{4b-3}$, which only happens if $a,b=1$

How can I interpret this?

How can I conclude that $\alpha$ is indeed injective, surjective?

Who would $\beta$ ?

Answer 1

$\alpha(1) = 1.$

Take any $c < 1.$

Set $a = \frac{c + 3}{4} \implies a < 1 \implies$
$\alpha(a) = \left(4 \times \frac{c + 3}{4}\right) - 3 = c.$

Take any $c > 1.$

Set $a = \sqrt{c} \implies a > 1 \implies \alpha(a) = a^2 = c.$

Thus, for all values of $c$, there exists a value for $a$ such that $\alpha(a) = c.$

Therefore $\alpha$ is a surjective function.

If $a < 1,~$ then $~\alpha(a) < 1. \tag1$

If $a > 1,~$ then $~\alpha(a) > 1. \tag2$

Therefore, the only value for $a$ such that $\alpha(a) = 1$ is $a = 1. \tag3$

To show that $\alpha$ is an injective function, I need to show that $[\alpha(a) = \alpha(b)] \implies [a = b].$

$\underline{\text{case 1}}$

$\alpha(a) = \alpha(b) = 1.$
By (3), $a = 1 = b.$

$\underline{\text{case 2}}$

$\alpha(a) = \alpha(b) < 1.$
By (1) and (2), $a < 1$ and $b < 1.$ Therefore, $4a - 3 = 4b - 3 \implies a = b.$

$\underline{\text{case 3}}$

$\alpha(a) = \alpha(b) > 1.$
By (1) and (2), $a > 1$ and $b > 1.$ Therefore, $\sqrt{a} = \sqrt{b} \implies a = b.$

Thus in all three cases, $[\alpha(a) = \alpha(b)] \implies [a = b].$

Therefore, $\alpha$ is an injective function.

Answer 2

The function is injective: In any case $(a,b\leq1), (a,b>1), (a\leq 1 \; \text{and}\; b>1)$ and $(a>1 \; \text{and}\;b\leq1)$ we have that $\alpha(a)=\alpha(b)$ implies that $a=b$, so $\alpha$ is injective.

The function is surjective: If $c\in \mathbb{R}$,Then

For $a \leq1, a=\frac{c+3}{4}$

$\alpha(\frac{c+3}{4})=c$

And for $a>1, a=\sqrt{c}$

$\alpha(\sqrt{c})=c$

That is, for any $c\in \mathbb{R}$ there exists $a\in \mathbb{R}$ such that $\alpha(a)=c$. So $\alpha$ is surjective and thus, $\alpha$ is bijective.

In fact, we will use $\beta: \mathbb{R}\rightarrow \mathbb{R}$ given by

$\beta(c)=\begin{cases} \frac{c+3}{4}& \text{ if } c\leq 1 \\ \sqrt{c} & \text{ if } c>1 \end{cases} $

We have that if $a\leq 1$

$ \beta(\alpha(a))=\frac{(4a-3)+3}{4}=a$

And if $a>1$

$ \beta(\alpha(a))=\sqrt{a^2}=a$