Answer:
since g:B→C is onto
suppose z∈C,there exists a pre-image in B
Let the pre-image be y
Hence y∈B such that g(y)=z
similarly f:A→B is onto
suppose y∈B,there exists a pre-image in A
Let the pre-image be x
Hence x∈A such that f(x)=y
Now gof(x)=g(f(x))=g(y)=z
So for every x in A,there is an image z in C
