Let f:A=B such that f(x)=y and g:B=C such that g(y)=z. name the function defined from A to C.​

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

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step-by-step explanation:

it is true

step-by-step explanation:

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