In a group, is every element in that group the inverse of some element in the same group? In a group, $<G,*>$, is every element in that group the inverse of some element in the same group ?
I mean, if I define $f:G \to G$ such that $f(x)=x^{-1}$, is f a bijective function ?
If so, how can we prove it ?

Since $x^{-1}$ is unique, $f$ is one-to-one, but I don't know how to show the surjectivity of the function.
 A: Yes.
It is part of the axioms of a group that for each $x\in G$ there is an inverse $x^{-1}$. You can check from the definition of the inverse that $x=(x^{-1})^{-1}$ - this gives surjectivity.
A: 
Proposition. Let $f\colon X\to Y$, $g\colon Y\to Z$ be functions.
$(i)$  If $g\circ f$ is surjective, $g$ is surjective as well.
$(ii)$ If $g\circ f$ is injective, $f$ is injective as well.

Proof. Exercise.

Corollary. Let $f\colon X\to X$ be involution, i.e. $f\circ f = \operatorname{id}_X$. Then $f$ is bijective.

Now, since $x\cdot x^{-1} = x^{-1}\cdot x = e_G$, $\forall x\in G$, by uniqueness of inverse we have $(x^{-1})^{-1} = x$. Thus, function $x\mapsto x^{-1}$ is involution and hence bijective by the above corollary.
A: By definition, a group is closed under inversion. This means that for every element $g\in G$, there is an inverse $g^{-1}\in G$. Applying this to $g^{-1}$ implies there is an inverse, call it $h$, for $g^{-1}$. So $gh^{-1}$ is the identity, and therefore $g=h$. Thus we have shown that $g$ is the inverse of some other element of the group, namely $g^{-1}$.
A: Injective:
If $y^{-1} = x$ and $z^{-1} = x$ then $y = z$.  
Pf: $y^{-1} = x \& z^{-1} = x \implies yx = 1 \& zx = 1 \implies y = yxx^{-1}=1x^{-1} = x^{-1} \& z=zxx^{-1}=1x^{-1}=x^{-1}\implies y = z = x^{-1}$
Surjective:
For every $x$,  $(x^{-1})^{-1} = x$.  Pf:  $(x^{-1})x = x(x^{-1}) = 1$.  So $x = (x^{-1})^{-1}$.
