Is $f: \Bbb{Z} \to \Bbb{Z}, f(x) = 3x + 6$ a bijection? Why?

Question

$f: \Bbb{Z} \to \Bbb{Z}, f(x) = 3x + 6$

Is $f$ a bijection? If no, explain why it isn’t. If yes, find an expression computing $f^{−1}(y)$ for $y \in \Bbb{Z}$.

How to approach that question?

Answer 1

It's easy to see that the function's value is a multiple of 3. Is every integer a multiple of 3? No. So which property does that violate?

Answer 2

Since $f(x)=3(x+2),$ the image of $f$ is not all of $\mathbb{Z}$ but only the multiples of $3$. So there can be no inverse map from $\mathbb{Z}$ to $\mathbb{Z}.$ But $f$ does happen to be an injection.

Answer 3

We can also use a different approach since it is easy to invert the function.

Suppose for a contradiction that $f$ is a bijection, which implies $f$ is invertible. Then inverse function $f^{-1}$ of $f$ is $$f^{-1}(x) = \frac{x-6}{3},\ x\in\mathbb{Z}, f^{-1}:\mathbb{Z}\to\mathbb{Z}$$ But this is not even a function since $x$ values that are not multiple of $3$ are not even mapped to an element in codomain. So we have a contradiction as required.