How to prove $(f \circ\ g) ^{-1} = g^{-1} \circ\ f^{-1}$? (inverse of composition) I'm doing exercise on discrete mathematics and I'm stuck with question:

If $f:Y\to Z$ is an invertible function, and $g:X\to Y$ is an invertible function, then the inverse of the composition $(f \circ\ g)$ is given by $(f \circ\ g) ^{-1} = g^{-1} \circ\ f^{-1}$.

I've no idea how to prove this, please help me by give me some reference or hint to its solution.
 A: You put your socks first and then your shoes but you take off your shoes before taking off your socks.
A: $f:Y \to Z$ and $g:X \to Y$ are invertible functions. We need to prove $(f\circ g)^{-1}=g^{-1}\circ f^{-1}$.
$$
(f\circ g)^{-1}\circ (f\circ g)=(g^{-1}\circ f^{-1})\circ(f\circ g)=(g^{-1}\circ f^{-1})\circ f)\circ g=(g^{-1}\circ (f^{-1}\circ f))\circ g=(g^{-1}\circ I_{Y})\circ g=g^{-1}\circ g=I_{X}
$$
Similarly,
$$
(f\circ g)\circ (f\circ g)^{-1}=(f\circ g)\circ (g^{-1}\circ f^{-1})=f\circ(g\circ (g^{-1}\circ f^{-1}))=f\circ((g\circ g^{-1})\circ f^{-1})\\=f\circ (I_{Y}\circ f^{-1})=(f\circ I_{Y})\circ f^{-1}=f\circ f^{-1}=I_{Z}
$$
$(g^{-1}\circ f^{-1})\circ(f\circ g)=I_{X}$ and $(f\circ g)\circ (g^{-1}\circ f^{-1})=I_{Y}$ proves $f\circ g$ is invertible with $(f\circ g)^{-1}=g^{-1}\circ f^{-1}$.
A: Heres a hint: The jacket is put on after the shirt, but is taken off before
it.
A: let $(x,y) \in (f \circ g) ^{-1} $ 
$\Leftrightarrow (y,x) \in (f \circ g)$ 
$\Leftrightarrow \exists t((y,t) \in g  \land (t,x) \in f )$ 
$\Leftrightarrow \exists t((t,y) \in g^{-1} \land (x,t) \in f^{-1} )$
$\Leftrightarrow (x,y) \in g^{-1} \circ f^{-1}$
A: $$\begin{align}
& \text{id} \\
=& f \circ f^\circ \\
=& f  \circ \text{id} \circ f^\circ \\
=& f \circ (g \circ g^\circ) \circ f^\circ \\
=& f \circ g \circ g^\circ \circ f^\circ \\
=& (f \circ g) \circ (g^\circ \circ f^\circ)
\end{align}$$
Therefore $(f \circ g)^\circ = g^\circ \circ f^\circ$.
A: Use the definition of an inverse and associativity of composition to show that the right hand side is the inverse of $(f \circ g)$.
A: This is straightforward. Take x in the domain of f. It goes to f(x) = y. And g takes y to z = g(y). Therefore $g^{-1}$ takes z to y and $f^{-1}$ takes y to x. Both sides of your equation takes $z$ to $x$. 
Please try to think more before asking. This was not hard, was it? :) 
