Show that $\alpha$ is bijective and find $\beta: \mathbb{R} \rightarrow \mathbb{R}$ such that $(\beta \alpha)(a) = a$ for all $a \in \mathbb{R}$ 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 $\beta$ ?
 A: $\alpha(1) = 1.$
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.
A: 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$
