For this function, is f is injective?

Question

For $f(f(x))=x$ where $f: X →X,~∀x ∈ X$. Is $f$ is injective? From what I can see, since the condition $f: X \to X$ is satisfied by all elements of $X$, $f(x)$ must $= x$. Also, I know if $g$ is an inverse of $f$, then $g(f(x))=x$, but should I use this property to prove injection?

Answer 1

Firstly it's not correct to "define $f(f(x))=x$". All you can say is "If $f : X \to X$ for some collection $X$ such that $f(f(x))=x$ for any $x \in X$, then ... [what you can deduce about such an $f$ goes here] ...". Also, it is false that such an $f$ must satisfy $f(x) = x$ even for one $x \in X$. I'm sure you can figure out a counter-example where $X$ has just $2$ elements.

So what do you want to prove? If you want to prove that such an $f$ is injective, you want exactly the following logical structure.

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For any collection $X$ and function $f : X \to X$ such that $f(f(x)) = x$ for every $x \in X$:

  For any $a,b \in X$:

    If $f(a) = f(b)$ then:

      $a = b$.

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You should then realize that this is easy to obtain as follows:

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For any collection $X$ and function $f : X \to X$ such that $f(f(x)) = x$ for every $x \in X$:

  For any $a,b \in X$:

    If $f(a) = f(b)$ then:

      $f(f(a)) = f(f(b))$ [by substitution].

      $a = b$.

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