How to prove $(f \circ\ g) ^{-1} = g^{-1} \circ\ f^{-1}$? (inverse of composition)

Question

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.

Answer 1

Use the definition of an inverse and associativity of composition to show that the right hand side is the inverse of $(f \circ g)$.

Answer 2

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? :)

Answer 3

You put your socks first and then your shoes but you take off your shoes before taking off your socks.

Answer 4

$$\begin{align} & \text{id} \\\\ =& f \circ f^{-1} \\\\ =& f \circ \text{id} \circ f^{-1} \\\\ =& f \circ (g \circ g^{-1}) \circ f^{-1} \\\\ =& f \circ g \circ g^{-1} \circ f^{-1} \\\\ =& (f \circ g) \circ (g^{-1} \circ f^{-1}) \end{align}$$

Therefore $(f \circ g)^{-1} = g^{-1} \circ f^{-1}$.

Answer 5

Heres a hint: The jacket is put on after the shirt, but is taken off before it.