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Answer:

$\left ( gof \right )^{-1}=f^{-1}og^{-1}$

Given:

$f=\left \{ \left ( 1,a \right ),\left ( 2,b \right ),\left ( c,3 \right ) \right \},g=\left \{ \left ( a,apple \right ),\left ( b,ball \right ),\left ( c,cat \right ) \right \}$, clearly $f$ and $g$ are bijections.

Hint:

Here, $f$$g$  and $fog$ are invertible.

Solution:

Now, $f^{-1}=\left \{ \left ( a,1 \right ),\left ( b,2 \right ),\left ( 3,c \right ) \right \},g^{-1}=\left \{ \left ( apple,a \right ),\left ( ball,b \right ),\left ( cat,c \right ) \right \}$

So, $f^{-1}og^{-1}=\left \{ \left ( apple,1 \right ),\left ( ball,2 \right ),\left ( cat,3 \right ) \right \}$

$f:\left \{ 1,2,3 \right \}\rightarrow \left \{ a,b,c \right \}\: and\: g:\left \{ a,b,c \right \}\rightarrow \left \{ apple,ball,cat \right \}$

So, $gof:\left \{ 1,2,3 \right \}\rightarrow \left \{ apple,ball,cat \right \}$

$\Rightarrow \left ( gof \right )\left ( 1 \right )=g\left [ f\left ( 1 \right ) \right ]=g\left ( a \right )=apple$

$\left ( gof \right )\left ( 2 \right )=g\left [ f\left ( 2 \right ) \right ]=g\left ( b \right )=ball$

$\left ( gof \right )\left ( 3 \right )=g\left [ f\left ( 3 \right ) \right ]=g\left ( c \right )=cat$

$\therefore$ $\left ( gof \right )=\left \{ \left ( 1,apple \right ),\left ( 2,ball \right ),\left ( 3,cat \right ) \right \}$

Clearly, $gof$ is a bijection.

So, $gof$ is invertible.

$\left ( gof \right )^{-1}=\left \{ \left ( apple,1 \right ),\left ( ball,2 \right ),\left ( cat,3 \right ) \right \}$

$From\: \left ( i \right ),\: \left ( iii \right ),$  we get,

$\left ( gof \right )^{-1}=f^{-1}og^{-1}$

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