$f\circ f=f^2$ Solution

Question

Let $X=\{1,2,3,4\}$

Define a function $f: X\to X$ such that $f$ is not equal to $Ix$ and is one to one. Find $f\circ f=f^2$ and $f$'s inverse.

Answer 1

There are $24=4!$ injective, surjective maps from $X$ to $X$. These are usually called permutations of $X$. You are being asked to find one of the $23$ maps different from the identity, and name it $f$. The second part of the question is to find what permutation $f\circ f$ is. (This permutation is most often denoted $f^2$.) The third part of the question is to find the permutation that is inverse to $f$.

Answer 2

If I understand the question right:

A identity mapping $f$ defined by: $f(1) = 1, f(2) = 2, f(3) =3, f(4) = 4$

This is a bijection, $f^2 = f$ and $f^{-1} = f$ is what we start from. Now, we just need to change $f$ a bit to not make it an identity mapping.

Simplest way to do that is to mix it up by shifting the outputs by one ie:

$f(1) = 2, f(2) = 3, f(3) =4, f(4) = 1$ Then $f^2(x) = f(x+1)$ and $f^{-1}(x) = f^3(x)$