if a function is one-to-one and onto then it is invertible.
 

Hi,

For each ordered pair (a,b) of real numbers with $a \ne 0$, let $\alpha_{a,b} : \mathbb{R} \rightarrow \mathbb{R}$ be defined by $\alpha_{a,b}(x) = ax+b$. Let $A$ denote the set of all such mappings. Then composition is an operation on A, $\alpha_{a,b} \circ \alpha_{c,d} = \alpha_{ac, ad+b}$.

Consider the operation $\circ$ on the set A of the above exercise
Prove that each $\alpha_{a,b} \in A$ is invertible by verifying that it is one-to-one and onto. (with $a \ne 0$).

So, this is my attempt:

Assuming $\alpha_{a,b}$ both one-to-one and onto, we'll go to show that $\alpha_{a,b}$ is invertible describing an inverse.
Assuming $y = ax+b \in \mathbb{R}$ in the codomain, since it is onto, there is at least an element $x \in \mathbb{R}$ in the domain such that $\alpha_{a,b}(x) = y$
But, since $\alpha_{a,b}$ one-to-one, then that element must be unique. We consider another function $\beta(y) = x$, this can be done for each element $y \in \mathbb{R}$ in the codomain, therefore we obtain a function $\beta : \mathbb{R} \rightarrow \mathbb{R}$ inverse of $\alpha_{a,b}$.

Can you help me please? What do you think about it? Many thanks!

Apply the definition of bijection.

Than try to figure out if it's possible to do the same with $Y^2=R^2-X^2$

Can you see that you've just assumed the thing you are supposed to prove?

by definition of injective function, if each elements in the set of images in the codomain (not all the codomain), are images of only one element in the codomain

To verify that the function is injective I have also tried this:
$\alpha_{a,b} = ax+b \mbox{ is injective if } \\
x_1 \ne x_2 \Rightarrow \alpha_{a,b}(x_1) \ne \alpha_{a,b}(x_2), (x_1, x_2 \in \mathbb{R}) \\
ax_1 + b \ne ax_2 + b \\
ax_1 - ax_2 \ne 0 \\
x_1 - x_2 \ne 0 \, (\forall a \ne 0) \\
x_1 \ne x_2$