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Let f: X->Ybe an invertible function. Sh...

Let `f: X->Y`be an invertible function. Show that f has unique inverse. (Hint: suppose `g_1( and g)_2`are two inverses of f. Then for all `y in Y ,fog_1(y)=I_Y(y)=fog_2(y)`Use one oneness of f ).

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Let if possible, g and h be two inverses of `f`.
`therefore `For all `y in Y`
`" " (fog) (y) = 1_Y(y) = (foh) (y)`
`rArr " " f[g(y) ] = f[h(y)]`
`rArr " " g(y) = h(y) " " ( because f` is one-one and onto)
`rArr " " g = h`
Therefore, the inverse of `f` is unique. Hence Proved.
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