In a lecture we are given the following example and I don't understand how if the codomain is $\mathbb{Z}$ this can be a bijection, considering that integers are infinite and in this example the $Ran(f)$ is restricted:
EXAMPLE: $f:[m..n] \mapsto \mathbb{Z}$ where $f(x) = x+p$. $f$ is injective and $Ran(f) = [(m+p)..(n+p)]$ By the variant of the bijection rule $|[m..n]| = |[(m+p)..(n+p)]|$.
Thank you!
I understand the $Ranf$, but - is $f$ a bijection?
Any injective function $f:X\to Y$ induces on a natural way the bijective function $g:X\to\mathsf{ran}f$ prescribed by $x\mapsto f(x)$.
This allows the conclusion that $|X|=|\mathsf{ran}f|$ whenever $f$ is injective.
The function $f$ prescribed in your question is not bijective, but the "bijection rule" is applied on the induced bijective function.