Question

# Consider $f:\left\{ 1,2,3 \right\}\to \left\{ a,b,c \right\}$ and $g:\left\{ a,b,c \right\}\to \left\{ apple,ball,cat \right\}$ defined as $f(1)=a,f(2)=b,f(3)=c,g(a)=apple,g(b)=ball,g(c)=cat$ . Show that $f,g,g\circ f$ are invertible. Find ${{f}^{-1}},{{g}^{-1}}$ and ${{\left( g\circ f \right)}^{-1}}$ . Show that ${{\left( g\circ f \right)}^{-1}}={{f}^{-1}}\circ {{g}^{-1}}$ .

Hint: At first we have to check if the functions $f,g$ are one to one and onto or not. A function is invertible only if the function is both one to one and onto.

The function f is defined as:
$f:\left\{ 1,2,3 \right\}\to \left\{ a,b,c \right\}$.
$f(1)=a,f(2)=b,f(3)=c$ .
Here $\left\{ 1,2,3 \right\}$ is the domain of the function f. $\left\{ a,b,c \right\}$ is the codomain of the function f.
We know that a function is said to be one to one if every different element of the domain has different images.
Here image of 1 is a. Image of 2 is b. Image of 3 is c. Therefore, every different element of the domain has a different image. Hence, f is one to one.
We know that a function is said to be onto if for every element of the codomain, we can find out at least one preimage from the domain.
The preimage of a is 1. Preimage of b is 2. Preimage of c is 3.
Therefore, every element of the codomain has a preimage. Hence, f is a onto function.
Therefore, f is both one to one and onto. So, f is invertible.
Similarly, the function g is defined as:
$g:\left\{ a,b,c \right\}\to \left\{ apple,ball,cat \right\}$
$g(a)=apple, g(b)=ball, g(c)=cat$
Here $\left\{ a,b,c \right\}$ is the domain of the function. $\left\{ apple,ball,cat \right\}$ is the codomain of the function.
The function g is one to one as image of a is apple, image of b is ball, image of c is cat. Therefore every element of the domain has a different image.
The function g is onto as preimage of apple is a, preimage of ball is b, preimage of cat is c. Therefore every element of the codomain has one preimage in the domain.
Hence, g is both one to one and onto. So, g is invertible.
Now, $(g\circ f):\left\{ 1,2,3 \right\}\to \left\{ apple,ball,cat \right\}$ is defined as:
As $\left( g\circ f \right)\left( 1 \right)=g\left( f\left( 1 \right) \right)=g\left( a \right)=apple$
\begin{align} & (g\circ f)(2)=g\left( f\left( 2 \right) \right)=g(b)=ball \\ & (g\circ f)(3)=g\left( f\left( 3 \right) \right)=g(c)=cat \\ \end{align}
$g\circ f$ is one to one as every element of the domain $\left\{ 1,2,3 \right\}$ has different image.
$g\circ f$ is onto as every element of the codomain $\left\{ apple,ball,cat \right\}$ has a preimage in the domain.

Therefore $g\circ f$ is invertible.
We know that if a function $f$ maps one element $x$ to $y$, then the inverse function maps the image $y$ to $x$. That is:
$f\left( x \right)=y\Rightarrow {{f}^{-1}}\left( y \right)=x$

Therefore,
\begin{align} & f(1)=a\Rightarrow {{f}^{-1}}\left( a \right)=1 \\ & f(2)=b\Rightarrow {{f}^{-1}}\left( b \right)=2 \\ & f(3)=c\Rightarrow {{f}^{-1}}\left( c \right)=3 \\ \end{align}

Hence,
${{f}^{-1}}:\left\{ a,b,c \right\}\to \left\{ 1,2,3 \right\}$ , such that:
${{f}^{-1}}\left( a \right)=1,{{f}^{-1}}\left( b \right)=2,{{f}^{-1}}\left( c \right)=3$
Similarly,
\begin{align} &g\left( a \right)=apple\Rightarrow {{g}^{-1}}\left( apple \right)=a \\ &g\left( b \right)=ball\Rightarrow {{g}^{-1}}\left( ball \right)=b \\ &g\left( c \right)=cat\Rightarrow {{g}^{-1}}\left( cat \right)=c \\ \end{align}

Therefore,
${{g}^{-1}}:\left\{ apple,ball,cat \right\}\to \left\{ a,b,c \right\}$ , such that:
\begin{align} & {{g}^{-1}}(apple)=a \\ & {{g}^{-1}}\left( ball \right)=b \\ & {{g}^{-1}}\left( cat \right)=c \\ \end{align}
Similarly,
\begin{align} & \left( g\circ f \right)\left( 1 \right)=apple\Rightarrow {{\left( g\circ f \right)}^{-1}}\left( apple \right)=1 \\ & \left( g\circ f \right)\left( 2 \right)=ball\Rightarrow {{\left( g\circ f \right)}^{-1}}\left( ball \right)=2 \\ & \left( g\circ f \right)\left( 3 \right)=cat\Rightarrow {{\left( g\circ f \right)}^{-1}}\left( cat \right)=3 \\ \end{align}
Therefore,
${{\left( g\circ f \right)}^{-1}}:\left\{ apple,ball,cat \right\}\to \left\{ 1,2,3 \right\}$ , such that:
\begin{align} & {{\left( g\circ f \right)}^{-1}}\left( apple \right)=1 \\ & {{\left( g\circ f \right)}^{-1}}\left( ball \right)=b \\ & {{\left( g\circ f \right)}^{-1}}\left( cat \right)=c \\ \end{align}

Now,
\begin{align} & \left( {{f}^{-1}}\circ {{g}^{-1}} \right)\left( apple \right)={{f}^{-1}}\left( {{g}^{-1}}\left( apple \right) \right)={{f}^{-1}}\left( a \right)=1 \\ & \left( {{f}^{-1}}\circ {{g}^{-1}} \right)\left( ball \right)={{f}^{-1}}\left( {{g}^{-1}}\left( ball \right) \right)={{f}^{-1}}\left( b \right)=2 \\ & \left( {{f}^{-1}}\circ {{g}^{-1}} \right)\left( cat \right)={{f}^{-1}}\left( {{g}^{-1}}\left( cat \right) \right)={{f}^{-1}}\left( c \right)=3 \\ & \\ \end{align}

Therefore,
\begin{align} & {{\left( g\circ f \right)}^{-1}}\left( apple \right)=\left( {{g}^{-1}}\circ {{f}^{-1}} \right)\left( apple \right) \\ & {{\left( g\circ f \right)}^{-1}}\left( ball \right)=\left( {{g}^{-1}}\circ {{f}^{-1}} \right)\left( ball \right) \\ & {{\left( g\circ f \right)}^{-1}}\left( cat \right)=\left( {{g}^{-1}}\circ {{f}^{-1}} \right)\left( cat \right) \\ \end{align}
Hence, ${{\left( g\circ f \right)}^{-1}}=\left( {{f}^{-1}}\circ {{g}^{-1}} \right)$

Note: We generally make mistakes to find out the inverse function. Always remember:
$f\left( x \right)=y\Rightarrow {{f}^{-1}}\left( y \right)=x$