If functions `f:A to B` and `g : B to A` satisfy `gof= I_(A),` then show that `f` is one-one and g is onto.

Answer

N/a

Explanation

Given that,
`f:A to B` and `g : B to A` satisfy `gof= I_(A)`
` :' gof =I_(A)`
`implies gof{f(x_(1))}=gof{f(x_(2))}`
`implies g(x_(1))=g(x_(2)) " " [:' gof = I_(A)]`
` :. x_(1)=x_(2)`
Hence, f is one-one and g is onto.