TITLE: Prove that a function is bijective and show that G is a group
QUESTION [5 upvotes]: Let $a\in\mathbb{R}^∗$ and $b\in \mathbb{R}$. Consider the function $f_{a,b} \in \operatorname{Fun}(\mathbb{R},\mathbb{R})$ given by $f_{a,b}(x)= ax + b$.
a) Show that $f_{a,b}$ is a bijection, and find its inverse function.
b) Let $G$ be the set of functions $\{f_{a,b}|a \in \mathbb{R}^∗
, b \in \mathbb{R}\}$. Show that $G$ is a group, where the group operation is composition of functions. (Thus $G$ is a subgroup of $\operatorname{Bij}(\mathbb{R}, \mathbb{R})$.)

My attempt to solve part a:
Let $x_1, x_2 \in \mathbb{R}$ and assume $x_1 \ne x_2$. Then $f_{a,b}(x_1) = ax_1+b$ and $f_{a,b}(x_2) = ax_2+b$. Thus $f_{a,b}(x_1) \ne f_{a,b}(x_2)$ and hence, $f_{a,b}$ is injective.
Since $a\in \mathbb{R}^∗$ and $b\in \mathbb{R}$, then $ax+b\in \mathbb{R}$ and thus $f_{a,b}(x)=ax + b\in \mathbb{R}$ and since it is a mapping from $\mathbb{R}$ to $\mathbb{R}$ then $f_{a,b}(\mathbb{R})=\mathbb{R}$ which means that $f_{a,b}$ is surjective.
Therefore, $f_{a,b}$ is at a time injective and surjective which means it is bijective.
It's inverse function:
$$f_{a,b}^{-1}(x)=\frac{x-b}{a}$$
Could you please check if I made any mistake in the proof and please give some insights on how to solve part b? thank you!

REPLY [5 votes]: Your proof of injectivity is ok, but for me it is not that obvious why $f_{a, b} (x_1) \neq f_{a,b} (x_2)$. Your surjectivity proof is wrong. Here is my version:
Injectivity: Let $ax_1 + b = a x_2 + b$. Then $ax_1 = a x_2$ and since $a \neq 0$, it follows $x_1 = x_2$.
Surjectivity: Let $z \in \mathbb{R}$. Choose $x_0 = \frac{z - b}{a}$. Then $f(x_0) = z$. This holds for every $z$, so $f$ is surjective.
So $f$ is bijective. The inverse map is 
$$f^{-1} : \mathbb{R} \longrightarrow \mathbb{R},\ x \longmapsto \frac{x - b}{a}.$$
Note that if you show $f \circ f^{-1} = f^{-1} \circ f = \mathrm{id}$ at the beginning, you are already done.
For proving that $G$ is a group, you have to show that 

$G$ is stable under composition.
Composition is associative, i. e., $f \circ (g \circ h) = (f \circ g) \circ h$.
In $G$ exists a neutral element $e$ such that $f \circ e = e \circ f = f$.
Every $f \in G$ has an inverse element $g \in G$, i. e., $f \circ g = g \circ f = e$. Note that you therefore should use what you have shown before.

I show you the first one and let the rest for you. So let $f_1, f_2 \in G$ with $f_1(x) = ax + b, \ f_2(x) = cx + d$ with $a,b, c, d \in \mathbb{R}$ and $a, c \neq 0$. Then 
$$f_1 \circ f_2 (x) = f_1 (f_2(x)) = f_1(cx + d) = a (cx + d) + b = acx + ad + b = \lambda x + \mu \in G, $$
with $\lambda = ac$ and $\mu = ad + b$.

REPLY [4 votes]: The core of your proof for injectivity is correct, but it's missing a detail. It feels like you pull, out of thin air, the result you desire. The result you desire is correct, but it is lacking the justification in other words. We see that $x_1 \ne x_2$ implies the following:
$$\begin{align}
x_1 \ne x_2 & \implies ax_1  \ne ax_2 \\
& \implies ax_1 + b \ne ax_2 + b\\
& \implies f(x_1) \ne f(x_2)
\end{align}$$
It's a small detail but it's worth keeping stuff like this mind: your justification is as important as your results oftentimes.

Your surjectivity proof leaves much to be desired. You seem to assume that, since the output of $f$ is in $\Bbb R$, it inherently means $f$ is surjective. This is not always the case, which we can see by tweaking the codomain, or the definition of the function. Some examples:

Instead of letting $f : \Bbb R \to \Bbb R$ let $f : \Bbb R \to \Bbb C$. The image of $f$ is basically the same, except there are visibly no non-real outputs, so obviously $f$ is not surjective. This is despite the fact that $\text{image}(f) = \Bbb R \subset \Bbb C$.
Let $f : \Bbb R \to \Bbb R$ be defined by $x \mapsto x^2$. There is no $x \in \Bbb R$ such that $f(x) = -1$, so not surjective. This is again despite the fact that $\text{image}(f) \subset \Bbb R$.
Consider a third, extreme example. Let $f : \Bbb Z \to \Bbb Z$ be defined by $f(x) = 1$. Thus, while each output of $f$ is an integer, this is clearly not surjective.

You need to show that $f$ (going back to your problem) has an "input" for every "output," i.e. that every element in the codomain has a pre-image. To phrase it a third way, you need to show that, for every $y \in \text{cod}(f)$, there exists some $x$ such that $f(x)=y$.
So let $f$ be defined by $f(x)=ax+b$ as stated. Let $y \in \Bbb R$. Then we can define the $x$ in question to be $x = (y-b)/a$ (let $f(x)=y,$ solve for $x$). You can justify this as being well defined (e.g. since $a \ne 0$ as $a \in \Bbb R^*$), thus giving us surjectivity.

An alternative way to show bijectivity (killing both of the above "birds" with a single proverbial stone) is to find the inverse function $f^{-1}(x)$. You have already found this to be $f^{-1}(x) = (x-b)/a$. 
On finding this inverse, find $f \circ f^{-1}$ and $f^{-1} \circ f$. If both of these compositions yield the respective identity functions, then you have that $f$ is a bijection. Indeed, we see
$$(f \circ f^{-1})(x) = a \cdot \frac{x-b}{a} + b = x$$
and
$$(f^{-1} \circ f)(x) = \frac{(ax+b)-b}{a} = x$$
giving bijectivity.