Let $a, b \in \mathbb{R}$. Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$ is given by $f(x) = ax+b$.
Under what conditions on $a$ and $b$ is $f$ a bijection?
Here I'm looking at two cases: if $a=0$, and if $a\neq 0$.
Case 1:
If $a=0, f(x) = b$.
Thus $f = \{(0,b)\}$.
Let us test whether this function is injective and/or surjective.
For any $x \in \mathbb{R}$, there is no unique $y \in \mathbb{R}$ since $f(x) = b$, $\forall x \in \mathbb{R}$.
However, $b$ is only one point in the infinite set, $\mathbb{R}$. So $\exists y \in \mathbb{R}$ such that $f(x) \neq y.$
So the function is neither injective nor surjective, and as such not bijective.
Case 2:
If $a \neq 0$, then $f(x) = ax + b$.
Thus $f = \{(x_1, ax_1+b), (x_2,ax_2+b),(x_3,ax_3+b),...\}$
Let us test whether this function is injective and/or surjective.
Assume $f(x_1) = f(x_2)$.
$\implies ax_1+b = ax_2+b$
$\implies ax_1 = ax_2$
$\implies x_1 = x_2$, so $f$ is injective.
Now we check for surjectivity.
If $x \in \mathbb{R} \land a \neq 0$, we can choose an $x = \frac{y-b}{a}$
$f(x) = a(\frac{y-b}{a})+b$
$\implies f(x) = (y-b)+b$
$\implies f(x) = y$, so $f$ is surjective.
Therefore $f$ is bijective $\iff a \neq 0$.
I am a little confused on the first case. For some reason my brain wants to tell me that $f$ is actually surjective. I'm visualizing a straight line going in both directions on a plot. Wouldn't that cover everything? Why is it a point and not an infinite line that maps all of $\mathbb{R}$?