Question

Suppose that f is an invertible function from Y to Z and g is an invertible function from X to Y. Show that the inverse of the composition f $\circ$g is given by$(f\circ g{)}^{-1}=g^{-1}\circ f^{-1}$

Short answer

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

Step-by-step solution

Step: 1

Assuming $z\in Z$and x the corresponding image of the function ${\left(f\circ g\right)}^{-1}$.

${\left(f\circ g\right)}^{-1}\left(z\right)=x$

By inverse property,

$\begin{array}{r}(f\circ g)\left(x\right)=z\\ f\left(g\right(x\left)\right)=z\\ f^{-1}\left(z\right)=g\left(x\right)\\ g^{-1}\left(f^{-1}\left(z\right)\right)=x\\ \left(g^{-1}\circ f^{-1}\right)\left(z\right)=x\\ \Rightarrow (f\circ g{)}^{-1}=g^{-1}\circ f^{-1}\end{array}$