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.
Best Answer
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
$$a=b$$