Wre given a function $f: X \to X$ on a finite set $X$ such that $f(f(x)) = f(x)$ for all $x \in X$, and $f$ is not the identity function.
A fundamental property of functions mapping a finite set to itself is that injectivity implies surjectivity, and vice versa. If $f: X \to X$ (where $X$ is finite) is injective, it must also be surjective (and thus bijective). If it is surjective, it must also be injective (and thus bijective). Because of this, it is impossible for $f$ to be injective but not surjective, or surjective but not injective. This immediately eliminates options A and B.
Now we'll test for bijectivity.
Let's see what happens if $f$ is both injective and surjective (bijective). If $f$ is bijective, it has an inverse function, $f^{-1}$. We are given the equation:$f(f(x)) = f(x)$
If we apply the inverse function $f^{-1}$ to both sides of the equation, we get:$f^{-1}(f(f(x))) = f^{-1}(f(x))$$f(x) = x$
This implies that if $f$ is bijective, it must map every element to itself, which is the exact definition of the identity function. However, the problem explicitly states that $f$ is not the identity function. Therefore, $f$ cannot be bijective. This eliminates option C.
Since $f$ cannot be injective (because it would then be bijective and therefore the identity function) and it cannot be surjective (for the same reason), $f$ must be neither injective nor surjective.
We can easily construct a valid example to prove this is possible. Let $X = \{1, 2\}$. Define $f$ such that $f(1) = 1$ and $f(2) = 1$.
Check the condition: $f(f(1)) = f(1) = 1$, and $f(f(2)) = f(1) = 1$. The condition holds.
It is not the identity function because $f(2) \neq 2$.
It is not injective because $f(1) = f(2) = 1$.
It is not surjective because $2$ is never mapped to.
Conclusion : The only possible scenario is that $f$ is neither injective nor surjective.
The correct option is D.
