Question

Let$\mathit{f}\mathbf{:}\mathit{B}\mathbf{\to }\mathit{C}$ and$\mathit{g}\mathbf{:}\mathit{C}\mathbf{\to }\mathit{D}$ be bijective functions. Then the composite function$\mathit{g}\mathbf{}\mathit{o}\mathbf{}\mathit{f}\mathbf{}\mathbf{:}\mathit{B}\mathbf{\to }\mathit{D}$ is bijective by Exercise 27. Prove that ${\left(gof\right)}^{\mathbf{-}\mathbf{1}}\mathbf{=}{\mathit{f}}^{\mathbf{-}\mathbf{1}}\mathit{o}\mathbf{}{\mathit{g}}^{\mathbf{-}\mathbf{1}}$.

Short answer

Thus, the given statement is proved ${(\mathrm{g}\mathrm{o}\mathrm{f})}^{-1}=f^{-1}o{g}^{-1}$.

Step-by-step solution

The composite function  is bijective

Consider the provided statement to prove that ${(\mathrm{g}\mathrm{o}\mathrm{f})}^{-1}=f^{-1}o{g}^{-1}$.

As it is given that $f:B\to C$and $f:C\to D$be bijective function then the composite function $g\mathit{}o\mathit{}f\mathit{}\mathit{:}\mathit{}B\mathit{\to }D$is also bijective.

Now, it is sufficient to prove that,

$\begin{array}{lll}(f^{-1}o& g^{-1})o& \left(gof\right)=I,\\ \left(gof\right)& o(f^1& 0g^{-1})=I_{,}\end{array}$

Now, the solve as follows:

localid="1660545496144" $$\begin{gathered} (f^{-1}\mathrm{o}{g}^{-1})\mathrm{o}\left(gof\right)=\left\{(f^{-1}\mathrm{o}{g}^{-1})og\right\}of \\ =\left\{f^{-1}\mathrm{o}\left({\mathrm{g}}^{-1}\mathrm{o}\mathrm{g}\right)\right\}of \\ =\left\{f^{-1}o{l}_y\right\}of \\ =\left\{f^{-1}\right\}of \\ =l \\ \mathrm{Hence},\mathit{(}{f}^{\mathit{-}\mathit{1}}\mathit{}o\mathit{}{g}^{\mathit{-}\mathit{1}}\mathit{)}o\mathit{}\mathit{(}\mathit{g}\mathit{}\mathit{o}\mathit{}\mathit{f}\mathit{)}\mathit{=}{l}_x\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{\left(}1\mathit{\right)} \end{gathered}$$

Proving that (g o f)-1=f-1o g-1

Now, the similarly solve as follows:

$$\begin{gathered} \left(gof\right)\mathrm{o}(f^{-1}\mathrm{o}{g}^{-1})=\left\{(gof)o{f}^{-1}\right\}o{g}^{-1} \\ =\left\{\mathrm{g}\mathrm{o}\left(fo{f}^{-1}\right)\right\}o{g}^{-1} \\ =\left\{go\left(I_x\right)\right\}o{g}^{-1} \\ =\left\{go{g}^{-1}\right\} \\ \mathit{(}g\mathit{}o\mathit{}f\mathit{)}o\mathit{(}{f}^{\mathit{-}\mathit{1}}\mathrm{o}\mathit{}{g}^{\mathit{-}\mathit{1}}\mathit{)}\mathit{=}{l}_z\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{\left(}2\mathit{\right)} \end{gathered}$$

Now, from equation (1) and equation (2),

${\left(gof\right)}^{-1}\mathit{=}{f}^{\mathit{-}\mathit{1}}\mathrm{o}\mathit{}{g}^{\mathit{-}\mathit{1}}$

Hence, the statement is proved ${\mathit{(}g\mathit{}o\mathit{}f\mathit{)}}^{-1}\mathit{=}{f}^{\mathit{-}\mathit{1}}\mathrm{o}\mathit{}{g}^{\mathit{-}\mathit{1}}$.