solutiongiven:− f:A→ B and g:B→ C be functions.proof that:− if gof is onto, then g is onto.proof:− we will be show that g is onto in other word , g:B → C is onto (every element in C has preimage in B)let z ∈ C , there exists y ∈ A (becouse gof is onto)such that gof(y)=zg[f(y)]=zg(x)=z (because f:A → B , y∈ A , f(y)=x ∈ B )thus for every element z ∈ C , there exit x ∈ B, such that g(x)=zg is onto
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