Prove that if an inverse function exists, then it is unique. I have been trying to solve this proof for some time in preparation for a test, though I'm not sure if I am going about this the correct way.
Prove that if an inverse function exists, then it is unique.

I am attempting proof by contradiction.
Let $f$ be a function, with an inverse.
Let $a$ be an inverse of $f$.
Let $b$ also be an inverse of $f$.
$$f\circ a = x$$
$$f\circ b = x$$
$$(f\circ b)(f\circ a)  = x(f\circ a)$$
$$(f\circ b)x  = x(f\circ a)$$
$$(f\circ b)  = (f\circ a)$$
because $(f\circ b) = x$ and $(f\circ a) = x$, $a = b$, thus proving that if an inverse of $f$ exists, it is then unique.
I'm not sure why, but something feels a bit off with my reasoning, or at least the way I have explained it.
If anyone could shed some light on a better way for me to explain this, I would greatly appreciate it.
 A: Can't you go with contradiction in the following sense?
Let $f:A\rightarrow B$, and $g:B\rightarrow A$ and $h:B\rightarrow A$ both be the inverses of $f$. Assume $g\neq h$. Then, there is $b\in B$ such that $g(b)\neq h(b)$.
As $f$ is bijective, then for all $b\in B$ there is $a\in A$ such that $f(a)=b$. As both, $g$ and $h$ are inverse functions of $f$, then it must hold that $g(f(a))=a$ and $h(f(a))=a$, but $f(a)=b$, so we have $g(b)=a=h(b)$, which contradicts our hypothesis.
A: If $a$ and $b$ are both inverse functions of $f$, then:
$$a \circ f= f \circ a = Id$$
$$b \circ f= f \circ b = Id$$
Therefore,
$$f \circ a= f \circ b $$
Composing by left side,
$$a \circ (f \circ a)=a \circ (f \circ b) $$
By associativity
$$(a \circ f) \circ a=(a \circ f) \circ b $$
Since $a \circ f= Id$, then
$$Id \circ a = Id \circ b$$
which means 
$$a=b$$
A: Let $f:V\to W$, and both $f^{-1}_{A}:W\to V$ and $f^{-1}_{B}:W\to V$ are its inverse.
So 
$$f^{-1}_A(f(x))=I_V(x)=x=f^{-1}_B(f(x)),$$
and
$$f(f^{-1}_A(y))=I_W(y)=y=f(f^{-1}_B(y)).$$
So
$$f^{-1}_B\circ I_W=f^{-1}_B\circ (f\circ f_A^{-1})=(f^{-1}_B\circ f)\circ f^{-1}_A=I_V\circ f^{-1}_A.$$
This finish the proof.
