If f maps Z on to Z is an isomorphism, prove that f is the identity map. Solution f:Z->Z is an isomorphism let ,x Z and y Z (codomain) ; = belongs to hence, y=f(x) ;y is as function of x isomorphism means it is onto and one-one onto: \"every\" yZ (codomain) ,there exist xZ ,such that f(x)=y ...1 one-one: for \"any\" yZ(codomain) ,there exist at most one xZ such that y=f(x) ...2 from 1 and 2, we get \"every\" yZ(codomain) has at most one xZ such that y=f(x) and f:Z->Z. which means every element of set Z maps to the same element in Z(codomain), since domain and range are same i.e Z=Z. hence identity map..