Let $f = {(1, a), (2, c), (4, d), (3, b)}$ and $g^{- 1} = {(2, a), (4, b), (1, c), (3, d)}$ , then show that $(g o f)^{- 1} = f^{- 1} o g^{- 1}$ .

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Solution

Given $f = {(1, a), (2, c), (4, d), (3, b)}$ and $g^{- 1} = {(2, a), (4, b), (1, c), (3, d)}$ .
Then we have $f^{- 1} = {(a, 1), (c, 2), (d, 4), (b, 3)}$ and $g = {(a, 2), (b, 4), (c, 1), (d, 3)}$ .
Now $g o f = {(1, 2), (2, 1), (4, 3), (3, 4)} \Rightarrow (g o f)^{- 1} = {(2, 1), (1, 2), (3, 4), (4, 3)}$ ......(1).
Again $f^{- 1} o g^{- 1} = {(2, 1), (4, 3), (1, 2), (3, 4)}$ $= {(2, 1), (1, 2), (3, 4), (4, 3)}$ ........(2).
From (1) and (2) we have $(g o f)^{- 1} = f^{- 1} o g^{- 1}$ ,

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