Show that if f and g are one-to-one functions such that f (compose)g is defined, then (f (compose)g)^-1= g^-1(compose)f^-1
This is because $\displaystyle (f\circ g)(g^{-1} \circ f^{-1}) = f \circ g \circ g^{-1} \circ f^{-1} = \text{identity}$.
Therefore, $\displaystyle (f\circ g)^{-1} = g^{-1}\circ f^{-1}$.