Showing $(f^{-1}∘g^{-1})=(g∘f)^{-1}$ 
If the functions $f$ and $g$ are both bijections then the in inverse of the composition function $(f∘g)$ will exist. Show that it will be $(f^{-1}∘g^{-1})=(g∘f)^{-1}$
For the proof assume $f:A→B$ and $g:B→C$

Here's the proof I have worked out so far:
From the problem I know that $f(a)=b$ and $g(b)=c$. By definition of inverse relation it's clear to see that $f^{-1}(b)=a$ and $g^{-1}(c)=b$.
The problem is asking me to prove: $(f^{-1}∘g^{-1})=(f∘g)^{-1}$, so:
$$(f^{-1}∘g^{-1}) = f^{-1}(g^{-1}(c)) = f^{-1}(b) = a$$
But I can't seem to prove the other side of the expression:
$$(g∘f)^{-1} = g(f(a))^{-1} = g^{-1}(b) ≠ a$$
I happen to also have the solution to this problem, which is as follows:

So we have that $a = f^{-1}(b) = f^{-1}(g^{-1}(c)) = (f^{-1}∘g^{-1})(c)$ by definition of composite.

But this solution makes no sense to me, simply because I can't figure out how to prove $f^{-1}(b) = a$.

Consider the expression:
$$(f^{-1} · g^{-1})·(g·f) = f^{-1} · (g^{-1}·(g·f)) = f^{-1} · ((g^{-1}·g)·f) = f^{-1} · (id·f) = f^{-1} ·f) = id$$
$id$ is the identity map.  Associativity property of composition.  Composing the identity function with any function $k$ yields $k$.
http://orion.math.iastate.edu/butler/2011/fall/201/201_hw9.pdf
