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Consider f : {1, 2, 3}->{a , b , c}given...

Consider `f : {1, 2, 3}->{a , b , c}`given by `f(1) = a`, `f(2) = b` and `f(3) = c`. Find `f^(-1)`and show that `(f^(-1))^(-1)= f`.

Text Solution

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`f={(1,a),(2,b),(3,c)}`
`f^(-1)={(a,1),(b,2),(c,3)}`
Hence `f^(-1)(a)=1,f^(-1)(b)=2,f^(-1)(c)=3`
Now
`f^(-1)={(a,1),(b,2),(c,3)}`
`(f^(-1))^(-1)={(1,a),(2,b),(3,c)}`
`=f`
Hence `(f^(-1))^(-1)=f`
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