Prove (if f and g are permutations) that $(f \circ g)^{-1} = g^{-1} \circ f^{-1}$.
My teacher gave me the hint that it has something to do with identity mapping, but that doesn't help me at all.
Here's what I know, but am not sure how to use:
A permutation is a bijection, so it has an inverse that is also a bijection.
And I'm guessing you would use something like $(f \circ g)^{-1} \circ (f \circ g) = I_A$
On
start with
$(f \circ g)^{-1} \circ (f \circ g) = I_A$
As we know composition of function is Associative, ie. $(f \circ g)\circ h= f \circ (g\circ h)$
Applying this $=>((f \circ g)^{-1} \circ f) \circ g = I_A $
$=>((f \circ g)^{-1} \circ f) \circ g \circ g^{-1} = I_A \circ g^{-1} $
$=>((f \circ g)^{-1} \circ f) \circ I_A = g^{-1} $
$=>(f \circ g)^{-1} \circ f = g^{-1} $
$=>(f \circ g)^{-1} \circ f \circ f^{-1} = g^{-1} \circ f^{-1} $
$=>(f \circ g)^{-1} \circ I_A = g^{-1} \circ f^{-1} $
$=>(f \circ g)^{-1} = g^{-1} \circ f^{-1} $
To show $u$ and $v$ are inverse, we show that $u\circ v$ and $v\circ u$ are the identity function.
In this instance then we want to show $f\circ g$ and $g^{-1}\circ f^{-1}$ are inverses, or equivalently
$$(f\circ g)\circ (g^{-1}\circ f^{-1})={\rm Id}=(g^{-1}\circ f^{-1})\circ (f\circ g).$$
To show this, use (generalized) associativity of function composition with the rule that both compositions $u\circ u^{-1}$ and $u^{-1}\circ u$ are $\rm Id$ for any function $u$.
Note that both conditions $u\circ v=\rm Id$ and $v\circ u=\rm Id$ must be met for $u$ and $v$ to be inverse functions; it is possible for pairs of non-bijective functions to satisfy one but not the other.