Does $f(f(x)) = x \Rightarrow f$ is bijective apply? As far as I remember a reverse function for some function $f$ exists iff an inverse function exists.
Can I therefore follow from $f(f(x)) = x$ ($f$ is its own inverse function) for some function $f$ that it is bijective without proving it is injective and surjective?
Example:
$f : \mathbb{R} \rightarrow \mathbb{R},\ f(x) = 5 - x$
 A: Since f(f(x)) exists it implies that f(x) must be in the domain of f (else we can't form $f(f(x))$. So f must map some set A to itself, i.e. $f:A \rightarrow A$. Examples using say R and C are therefore not valid: the domain of f is either R or C, in this case you would have $f: R \rightarrow C$ and $F: C \rightarrow R$ with $f = F|_R$ (i.e. f is $F$ restricted to R), and then $F(f(x)) = x$.
In general if $f:A \to B$ and $F:B \to A$ then $f = F \implies B = A$, and here $f = F$ is implied by $f(f(x)) = x$.
So with that understanding then f is a bijection.


*

*f must be onto (surjective) since for all $x \in A$ $f(x) $ is defined (under normal understanding)  and $=x$, so for all $x \in A$ there exists $x \in A$ such that $x = f(x)$

*f must be into (injective) as $f(y) = y$ so if $f(x) = f(y)$ then $x = y$.

A: Assuming $f: A \to A$...
If $f \circ g$ is bijective, then $f$ is a surjective and $g$ is injective. Here, $f \circ f$ is a bijective (in fact it is the identity mapping), so $f$ is surjective and $f$ is injective. Therefore, $f$ is bijective.
A: The self-inverse property implies injectivity but not surjectivity.
A: $f$ is injective, because if $f(x)=f(y)$, then $x=f(f(x))=f(f(y))=y$.
$f$ is surjective because every $x$ is the image of some $y$, namely $f(f(x))=x$ (provided that the image is the same set as the domain).
So... yes.
A: This depends on the domain and codomain chosen for the function. For example, suppose that $f : \Bbb{R} \to \Bbb{C}$ is defined by $f(x) = x$. Then $f(f(x)) = x$ for every $x \in \Bbb{R}$, but $f$ is not a bijection since it is certainly not surjective. You can say, however, that if $x \neq y$ are both in the domain of $f$, then $f(x) \neq f(y)$ since otherwise $f(f(x)) = f(f(y)) \implies x = y$.
A: You have plenty of answers and according to my view you accepted one
that is indeed okay. I just want to mention a subtle point here. Your
condition was: $$\forall x\; f\left(f\left(x\right)\right)=x$$ and
under that condition $f$ is injective but does not have to be surjective. 
If it would have been
slightly different: $$\forall x\; f\circ f\left(x\right)=x$$ then you are dealing with something else. This condition can only 'make sense' if
function $f\circ f$ is indeed defined. That means that its domain and
codomain must be the same. So then the answer
would have been: yes, $f$ is a bijection.
The first condition is not equivalent with: $f$ is its own inverse (as you argued in your question). The second condition is.
